doc 903b b.p.s. ix s.a. i maths chapterwise 5 printable worksheets with solution 2014 15
DESCRIPTION
IX S.a. I Maths ChapterwiseTRANSCRIPT
BRILLIANT PUBLIC SCHOOL , SITAMARHI
(Affiliated up to +2 level to C.B.S.E., New Delhi)Affiliation No. - 330419
CBSE Board Level IX S.A.- I Maths Chapterwise
Printable Worksheets with Solution
Session : 2014-15
Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301Website: www.brilliantpublicschool.com; E-mail: [email protected]
Ph.06226-252314, Mobile: 9431636758, 9931610902
MATHEMATICS (Class-IX)
Index: S.A.-I
CBSE Chapter-wise Solved Test Papers
1. 001
2. 022
3. 040
4. 045
5. 079
6. 106
7.
Number System
Polynomials
Introduction to Euclid’s Geometry
Lines and Angles
Triangles
Coordinate Geometry
Heron’s Formula 119
CBSE TEST PAPER-01
CLASS - IX MATHEMATICS (Number System)
1. Which of the following rational numbers have terminating decimal
representation
3 2( ) ( )
5 1 34 0 2 3
( ) ( )2 7 7
i i i
i i i i v
[1]
2. How many rational numbers can be found between two distinct rational
numbers?
(i)Two (ii) Ten (iii) Zero (iv) Infinite
[1]
3. The value of ( )( )2 3 2 3+ − in
(i)1 (ii)-1
(iii) 2 (iv) none of these
[1]
4. (27)-2/3 is equal to
(i) 9 (ii) 1/9
(iii) 3 (iv) none of these
[1]
5. Simplify: 3 42 3× [2]
6. Find the two rational numbers between ½ and 1/3. [2]
7. Find two irrational numbers between 2 and 3. [2]
8. Multiply ( ) ( )3 5 6 2by− + [2]
9. Express 0.8888… in the form p/q. [3]
10. Simply by rationalizing denominator
7 3 5
7 3 5
+−
[3]
11.
Simplify
211 4
2625−−
[3]
12. Visualize 3.76 on the number line using successive magnification. [3]
13. Prove that
1 1 11
1 1 1b a c a a b c b a c b cx x x x x x− − − − − −+ + =+ + + + + +
1
CBSE TEST PAPER-01
CLASS - IX MATHEMATICS (Number System)
[ANSWERS]
Ans 01. (i)
Ans 02. (iv)
Ans 03. (i)
Ans 04. (ii)
Ans 05 3 42 3×1 13 4
1 4 1 143 12 12 12
1 3 1 134 12 12 12
1 1 1 1 13 4 12 12 12
1
12
2 ,3
The LCM of 3 and 4 is 12
2 2 =(2 ) 16
3 3 =(3 ) 27
2 3 =16 27 (16 27)
(432)
∴ = =
= =
× × = ×
=
Ans 06. First rational number
1 1 1 1 3 2 5
2 2 3 2 6 12
1 5 1Ans , and
2 12 3Second rational number
1 1 5 1 6 5 11=
2 2 12 2 12 24
5 11Ans
12 24and
+ = + ⇒ ⇒
+ + ⇒ ⇒
Ans 07. Irrational number between 2 and 3 is 2 3 6× =
Irrational number between 2 and 6 is
2
( )
1 1 1 12
2 4 4 4
1 1 1 112 44 4 4 44
2 6 2 6 2 6
2 6 4 6 (24) 24
×× = × = ×
= × = × = =
Ans 08. ( )( )3 5 6 2− +
( ) ( )3 6 2 5 6 2
18 3 2 6 5 5 2
18 3 2 6 5 10
= − − +
= + − − ×
= + − −
Ans 09. Let x = 0.8888
x=0.8 (1)
10 10 0.8888
10 8.8888
10 8.8 (2)
10 8.8 0.8
9 8
8
9
x
x
x
x x
x
x
− − −= ×=
= − − − −
− = −=
=
Ans 10. 7 3 5 7 3 5
7 3 5 7 3 5
+ +×− +
(Rationalizing by denominator)
( )( )
( )2 22
2 22
7 3 5 7 3 5 2 7 3 5
49 3 37 3 5
+ + + × +=
− ×−
49 9 5 42 5 49 45 42 5
49 27 22
+ × + + += =−
94 42 5 94 425
22 22 22
+= = +
47 215
11 11= +
Ans11.
211 42625
−−
3
( )
21214
4
1 122 2
1 1 1
25625
maam
−−
−
= = = 1
241
25
− × =
1
2
1 12
2
1 1 15
25 55
−
−×−
= = = =
Ans12.
Ans13. 1 1 1
1 1 1b a c a a b c b a c b cx x x x x x− − − − − −+ ++ + + + + +
1
1 . .b a c ax x x x− −+ ++
1
1 . .− −+ +a b c bx x x x+
1
1 . .a c b cx x x x− −+ +1 1 1
. . . . . . . . .a a b a c a b b a b c b c c a c b cx x x x x x x x x x x x x x x x x x− − − − − − − − −+ ++ + + + + +
( ) ( ) ( )1 1 1
a a b c b a b c c a b cx x x x x x x x x x x x− − −+ +
+ + + + + +
( ) ( ) ( )a b c
a b c a b c a b c
x x x
x x x x x x x x x+ +
+ + + + + +
1a b c
a b c
x x x
x x x
+ += =+ +
4
CBSE TEST PAPER-02
CLASS - IX MATHEMATICS (Number System)
1. Every natural number is
(i) not an integer (ii) always a whole number
(iii) an irrational number (iv) not a fraction
[1]
2. Select the correct statement from the following
( ) ( )
( ) ( )
7 4 2 3i > ii <
9 5 6 9-2 -4 -5 -3
iii > iv <3 5 7 4
[1]
3. 7.2 is equal to
( ) ( ) ( ) ( )68 64 65 63i ii iii iv
9 9 9 9
[1]
4. 0.83458456……………is
(i) an irrational number (ii) rational number
(iii) a natural number (iv) a whole number.
[1]
5. Evaluate ( ) ( )3 4125 1250i ii [2]
6. Find rationalizing factor of 300 [2]
7. Rationalize the denominator
1
5 2+and subtract it from 5 2− [2]
8. Shove that 7 3− is irrational [2]
9. Represent 3 on number line [3]
10. Simplify ( ) ( )2 2
3 2 2 3 3 2 2 3+ − [3]
11. Express 2.4178 in the from
p
q
[3]
12. Simplify ( )
1 322 2327 9 .3
−− ÷[3]
13. It 5 2.236= and 3 1.732= Find the value of
2 7
5 3 5 3+
+ −[5]
5
CBSE TEST PAPER-02
CLASS - IX MATHEMATICS (Number System)
[ANSWERS]
1. (ii)
2. (iii)
3. (iii)
4. (i)
5. (i) ( ) ( )11
33 33125 5 5 5 5 5= × × = =
(ii) ( ) ( )11
44 441250 2 5 5 5 5 2 5= × × × × = ×11
4 4442 5 5 2= × × = ×
6. 300 2 2 3 5 5= × × × ×2 22 3 5= × ×
2 5 3 10 3= × = Rationalizing factor is 3
7. 1 5 2
5 2 5 2
−×+ −
( ) ( )2 2
5 2 5 2 5 2
5 2 35 2
− − −= = =−−
Difference
5 25 2
3
−= − −
=5 2
5 23 3
− − +
=5 2
5 23 3
− − −
= ( )2 5 2 2 25 2
3 3 3− = −
6
8. suppose 7 3 is rational −let 7 3 (x is a rational number)
7 3
x is a rational no. 3 is also rational number
x+3 is rational number
but 7 is irrational number which is controduction
7 3 is irrational number
x
x
− =
= +
∴
∴ −
9 OA=AB=1 unitTake
.
0
2 2 2
2
1 1
01
2 2 21 1 1 1 1 1
2 2 21 1
and = 90
In OAB=OB =1 +1
OB =2
OB= 2
OA1= 2 1.41
take A B = 1unit
and =90
In O B A [OB =OA +A B
OB =( 2) 1 3
A
OB
Now
A
OB
∠∆
∴ = =
∠
∆
+ =
10. ( ) ( )2 2
3 2 2 3 3 2 2 3+ −
( )( )( ) ( )3 2 2 3 3 2 2 3 3 2 2 3 3 2 2 3+ + − −
= ( )( )( ) ( )3 2 2 3 3 2 2 3 3 2 2 3 3 2 2 3+ − + −
( ) ( ) ( ) ( )2 2 2 2
3 2 2 3 3 2 2 3 = − −
=[ ][ ]9 2 4 3 9 2 4 3× − × × − ×
=[ ][ ]18 12 18 12− −= 6 6 36× =
11. Let 2.4178p
q=
2.4178178178p
q=
Multiplying by 10
10 24.178178p
q=
7
Multiplying by 1000
10,000 1000 24.178178p
q= ×
10,000 24178.178178p
q=
10000 24178.178178 24.178178p p
q q− = −
9999 24154p
q=
24154
9999
p
q=
12. ( )1 322. 2327 9 3
−− ÷
( )( )
3223
1
2
3 3 3 3
3 3
−× × ×
×
1mm
aa
− =
=( )
( )
2 33 3 2
12 2
3 3
3
−×
132
32
11
3
3 3 1
3 33
−−
+= = =
4 33
1 1
813
= =
13. 2 7
5 3 5 3+
+ −
( ) ( )( )( )
2 5 3 7 5 3 2 5 2 3 7 5 7 3
5 35 3 5 3
− + + − + +=−+ −
9 5 5 3 2 2.236 5 1.732
2 2
+ × + ×= =
4.472 8.66 13.1326.566
2 2
+= = =
8
CBSE TEST PAPER-03
CLASS - IX MATHEMATICS (Number System)
1. A terminating decimal is
(i) a natural number (ii) a rational number
(iii) a whole number (iv) an integer.
[1]
2. The
p
qform of the number 0.8 is
8 8 1( ) ( ) ( ) ( ) 1
10 100 8i ii iii iv
[1]
3. The value of 3 1000 is
(i) 1 (ii) 10 (iii) 3 (iv) 0
[1]
4. The sum of rational and an irrational number
(i) may be natural (ii) may be irrational
(iii) is always irrational (iv) is always rational
[1]
5. Find two rational numbers between 7 and 5. [2]
6. Show that 5 2+ is not a rational number. [2]
7. Simplify 2( 5 2)+ [2]
8.
Evaluate
5
2
3
2
11
11
[2]
9. Find three rational numbers between 2.2 and 2.3 [3]
10. Give an example of two irrational numbers whose
(i) sum is a rational number
(ii) product is a rational number
(iii) Quotient is a rational number.
[3]
11. If
52 1.414 and 3 1.732, find the value of
2 3= =
+[3]
12. Visualize 2.4646 on the number line using successive magnification . [3]
13. Find the value of
3 7, 5 =2.236 and 2 1.414
5 2 5 2If+ =
+ −
[5]
9
CBSE TEST PAPER-03
CLASS - IX MATHEMATICS (Number System)
[ANSWERS]
1. (ii) 2. (i) 3. (ii) 4. (iii)
5. First rational number = [ ]1 127 5 6
2 2+ = = =7, 6 ,5
Second rational number = [ ]1 1 137 6 13
2 2 2+ = × = = 6 and
13
2
6. let 5 2+ is rational number.
Say 5 2 x+ = � 2 5x= −
x is a rational number 5 is rational number
5x∴ − is also rational number
But 2 is irrational number
Which is a contradiction
5 2∴ + Is irrational number
7. ( ) ( ) ( )2 2 2
5 2 5 2 2 5 2+ = + + × 5 2 2 10 7 2 2= + + = +
8.
55 322 2
3
2
1111
11
−=
mm
n
aa n
a
= −
∵
25 32311 11
−
= =11=
9. Three rational numbers between 2.2 and 2.3 are 2.212341365.... and
2.2321453269....
10. (i) 2 2.+ and 2 2−
Sum 2 2 2 2 4+ − = rational number
(ii) 3 2 and 6 2
Product 3 2 6 2 18 2 36× = × = rational
(iii) 2 125 and 3 5
Quotient 251252 125 2 225
3 5 33 5= = =
2 105
3 3× =
10
11. 5 2 3
2 3 2 3
−×+ −
(Rationalising by denominator)
=( )
( ) ( )2 2
5 2 3 5 2 3
2 32 3
− −=−−
= [ ]5 1.414 1.732− −
= 5 0.318 1.59− × − =
12.
13. 3 7
5 2 5 2+
+ −
=( ) ( )( )( )
3 5 2 7 5 2
5 2 5 2
− + +
+ −
=3 5 3 2 7 5 7 2
5 2
− + +−
=10 5 4 2
3
+
=10 2.236 4 1.414
3
× + ×
= 22.36 5.656+= 28.016
11
CBSE TEST PAPER-04
CLASS - IX MATHEMATICS (Number System)
1. The rational number not lying between
3
5 and
2
3 is
( ) ( )
( ) ( )
49 50
75 7547 46
75 75
A B
C D
[1]
2. 0.123 is equal to
(A) 122
990 (B)
122
100
(C) 122
99 (D) None of these
[1]
3. The number ( )2
1 3+ is
(A) natural number (B) irrational number
(C) rational number (D) integer
[1]
4. The simplest form of 600 is
( ) ( )( ) ( )
10 60 100 6
20 3 10 6
A B
C D
[1]
5. Find four rational numbers between
3 4
7 7and
[2]
6. Writ the following in decimal form ( ) ( )36 2
100 11
i ii[2]
7. Express 2.4178 in the form
a
b
[2]
8. Multiply 3 by 3 5 [2]
9. Rationalize the denominator of
1
4 2 3+[3]
10. Visualize the representation of 5.37 on the no. line 3 decimal places [3]
11. Shone that 5 2 is not rational number [3]
12. Simplify 3 3 33 250 7 16 4 54+ − [3]
13. Simplify
2 5 2 5
2 5 2 5
+ −+− +
[5]
12
CBSE TEST PAPER-04
CLASS - IX MATHEMATICS (Number System)
[ANSWERS]
1. (B) 2. (A)
3. (B) 4. (D)
5.
3 10 30
7 10 70× = and
4 10 40
7 10 70× =
Take any four rational numbers between 30
70 and
40
70 i.e. rational numbers between
3
7 and
4
7 are
31 32 33 34 35, , , ,
70 70 70 70 70
6. (i)36
0.36100
= (ii) 2
0.1811
=
7. 2.4178x =
10 24.178x = ______(1)
10 24.178178178...x =
1000 10 1000 24.178178178....x× = ×
10,000 24178.178178....x =
10000 24178.178x = _______(2)
Eq (2) – eq(1)
10,000 24178.178 24.178x x− = −
9990 24154x =
24154
9990x =
24154 120772.4178
9990 4995= =
3
7
4
7
31 32 33 34 35, , , ,
70 70 70 70 70
13
8. 3 and 3 5
Or 1
23 and 1
35
LCM of 2 and 3 id 6
( ) ( )1 31 1 1
32 32 6 63 3 3 27×
= = =
( ) ( )1 1 2 1 1
23 3 2 6 65 5 5 25×
= = =
( ) ( ) ( )1 1 1
3 6 6 63 5 27 25 27 25× = × = ×
=1
66675 675=
9. 1 4 2 3
4 2 3 4 2 3
−×+ −
=
( )2
4 2 3 4 2 3 4 2 3
16 4 3 16 1216 2 3
− − −= =− × −−
=( )
2
2 2 34 2 3
4 4
−− =
=2 3
2
−
10.
14
11. let 5 2 is rational no
5 2x = ( x is rational)
25
x =
x is rational no.
5 is rational no.
5
x∴ is rational no.
But 25
x = and 2 is irrational no.
Which is a contradiction
5 2∴ is irrational number
12. 3 3 33 250 7 16 4 54+ −
= 33 33 5 5 5 2 7 2 2 2 2 4 3 3 3 2× × × + × × × − × × ×
= 3 3 33 5 2 7 2 2 4 3 2× + × − ×
=15 3 3 32 14 2 12 2+ −
=[ ] 3 315 14 12 2 17 2+ − =
15
CBSE TEST PAPER-05
CLASS - IX MATHEMATICS (Number System)
1. The value of 0.23 + 0.22 is
( ) ( )( ) ( )
0.45 0.44
0.45 0.44
A B
C D
[1]
2. The value of
1 4
3 32 2−
× is
( ) ( )( ) ( )
12
23 None of these
A B
C D
[1]
3. 16 13 9 52÷ is equal to
( ) ( )
( ) ( )
3 9
9 88
None of these9
A B
C D
[1]
4. 8 is an
(A) natural number (B) rational number
(C) integer (D) irrational number
[1]
5. Find the value of
2 5
5
+ it 5 2.236= and 10 3.162=
[2]
6. Convert 0.25 into rational [2]
7. Simplify ( )( )3 3 2 2 2 3 3 2+ + [2]
8.
Simplify
3 4
2 29 .91
92
− [2]
9. Simplify
5 13 48 4 3
2 3− +
[3]
10. If
10.142857.
7= find the value of
2 3 4, ,
7 7 7
[3]
11. Find 6 rational no. between
6
5and
7
5
[3]
12. Show how 4 can be represented on the number line [3]
13. Find a and b if
3 66
3 2 6a b
− = −+
[5]
16
CBSE TEST PAPER-05
CLASS - IX MATHEMATICS (Number System)
[ANSWERS]
1. 0.23 0.232323...=
0.22 0.222222...=
0.23 0.22 0.454545...+ =
0.45=
2.
31 4 1 4 1 433 3 3 3 32 2 2 2 2
−− −−× = = =
1 12
2− =
3. 16 13 9 52÷
8
4
1616 13 16 13 1
9 52 9 29 52= × = ×
=8
9
4. 8 is an irrational number
4 2 2 2× =∵
5. 2 5 5 10 5
55 5
+ +× =
=3.162 5 8.162
1.63245 5
+ = =
6. let ( )0.25........x i=
0.252525...x =
17
Multiplying both sides by 100
100 x =25.252525…
100 x = ( )25.25........... ii
Eq(ii) - eq(i)
100 25.25 0.25x x− = −
99 25x =
25
99x =
7. ( )( )3 3 2 2 2 3 3 2+ +
( ) ( )3 3 2 3 3 2 2 2 2 3 3 2+ + +
= 6 3 9 3. 2 4 2. 3 6.2× + + +
=18 9 6 4 6 12+ + +
= ( )30 9 4 6 30 3 6+ + = +
8.
3 4 3 4
2 2 2 2
1 1
2 2
9 ,9 9
9 9
− −
= .m n m na a a +=
1
2
1 1 1 1
2 2 2 2
9 9 1
9 9 9
−
+= = =
1m ma a− =
2
2
1 1
99
= =
9. 5 1
3 48 4 32 3
− +
=5 3
3 16 3 4 32 3 3
× − +×
5 13 4 3 . 3 4 3
2 3= × − +
=5
12 3 3 4 36
− +
18
=5 5
12 4 3 16 36 6
− + = −
=91
36
10. 1
0.1428577
=
22 0.142857
7= ×
0.285714=
33 0.142857
7= ×
0.428571=
44 0.142857
7= ×
0.571428=
11. 6 7
,5 5
is divided into
10 equal parts than the value of each part
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 7, , , , , , , , ,
5 5 5 5 5 5 5 5 5 5
∴6 rational no. between 6
5 and
7
5
61 62 63 64 65 66, , , , , ,
50 50 50 50 50 50
12. Take 1OA AB= = unit
And 90A∠ = �
In OAB∆2 2 2 21 1 2OB OB= + = =
2OB =
1 2OB OA= =o A A
1A2
A3
�2 �4�3
B B1
B2
19
Take 1 1 1A B =
And 1 90A∠ = �
In ( )22 2
1 1 1 2 1OB A OB∆ = = +
1 3OB =
1 2 3OB OA= =
Take 2 2 1a b = unit
2 90A∠ = �
( )22 2 2
2 23 1 3 1OB OB= + = = +
2 4 2OB = =
13. 3 6
63 2 6
a b− = −
+
Rationalizing L.H.S. on denominator
3 6 3 2 66
3 2 6 3 2 6a b
− −× = −+ −
( )22
9 6 6 3 6 2 66
3 2 6a b
− − + × = −−
9 9 6 126
9 24a b
− + = −−
9 6 216
15a b
− + = −−
73
5 5
9 6 216
15 15a b
−/ + = −− −/
3 76 6
5 5a b− = −
Comparing coefficient on both sides
3 7,
5 5a b= + = +
20
CBSE TEST PAPER-01
CLASS - IX Mathematics (Polynomials)
1. Which of the following expression is a polynomials
(a) 3 1x − (b) 2x +
(c) 22
1x
x− (d) 5 1t t+ −
[1]
2. A polynomial of degree 3 in x has at most
(a) 5 terms (b) 3 terms
(c) 4 terms (d) 1 term
[1]
3. The coefficient of x2 in the polynomial 2x3 + 4x2 + 3x + 1 is
(a) 2 (b) 3
(c) 1 (d) 4
[1]
4. The monomial of degree 50 is
(a) x50 + 1 (b) 2x50
(c) x+50 (d) 50
[1]
5. Divide f(x) by g(x) & verify that the remainder f(x) = x3 + 4x2 – 3x – 10, g(x) = x + 4 [1]
6. Find the value of K it x – 2 is factor of 4x3 + 3x2 – 4x + K [2]
7. Factorise the polynomial x3 + 8y3 + 64 z3 – 24xyz [2]
8. Without actually Calculating the cubes, find the value of (-12)3 + (7)3 + (5)3 [2]
9. If x – 3 and
1
3x − are both factors of px2 + 5x + r , then show that P = r
[2]
10. Using suitable identify expand
35 3
4 4x
+
[3]
11. Using factor theorem factorise f(x) = x2 – 5x + 6 [3]
12. Without actual division, prove that the polynomial 2x3 + 4x2 + x – 34 is exactly
divisible by (x – 2)
[3]
13. If x2 – bx + c = (x + p) (x – q) then factorise x2 – bxy + cy2 [5]
21
3 24 4x x x+ +3 2
3 10
4
x
x x
− −
+
3x
− −
− 10
3x
−
−
2 3
12
2
x −
−+ +
CBSE TEST PAPER-01
CLASS - IX Mathematics (Polynomials)
1. (a)
2. (b)
3. (d)
4. (b)
5. Dividend = x3 + 4x2 – 3x – 10, divisor = x + 4
Quotient = x3 – 3 Remainder = 2
Dividend = Divisor × quotient + Remained
= (x + 4) (x2 - 3) + 2
= x3 - 3x + 4x2 – 12 + 2
= x3 + 4x2 – 3x – 10
Verified
6. x – 2 is factor of 4x3 + 3x2 – 4x + k
x – 2 = 0 ⇒ x = 2 3 24(2) 3(2) 4 2 0
32 12 8 0
44 8 4
36 _ 0
36
k
k
k
k
k
∴ + − × + =+ − + =− + =
== −
7. 3 3 38 64 24x y z xyz+ + −3 3 3
2 2
2 2 2
(2 ) (4 ) 3. .(2 )(4 )
( 2 4 )[ (2 2 4 ) 2 2 4 4 ]
( 2 4 )( 4 16 2 8 4 )
x y z x y z
x y z x y z x y y z x z
x y z x y z xy yz xz
+ + −= + + + + + − × − × − ×= + + + + − − −
8. 3 3 3 3a b c abc+ + =
3 3 3
0
( 12) (7) (5) 3 12 7 5
1260
12 7 5 12 12 0
a b c+ + =− + + = × − × ×
= −− + + = − + =∵
22
9. 1
3 and are factors of3
x x− −∵
2
2
15 3,
3
zero of 5
Px x r x x
Px x r
+ + ∴ = =
+ +2(3) 5 3 0
9 15 4 0
P r
P
∴ + × + =+ + =
2
9 14 (1)
1 15 0
3 3
P r
P r
+ = − − − − − − − − − − −
+ × + =
50
9 315 9
09
Pr
P r
+ + =
+ + =
9 15 (2)
9 9
P r
P r P r
+ = − − − − − − − − − −+ = +
(1) & (2)
9 9
By
P r
p r
==
Hence prove
10.
35 3
2 4x
+
3 3 3
3 3 3
( ) 3 ( )
5 3 5 3 5 3 5 33
2 4 2 4 2 4 2 4
a b a b ab a b
x x x x
+ = + + +
+ = + × × +
3
32
32
125 27 45 5 3
8 64 8 2 4
125 27 225 135
8 64 16 4
125 225 135 27
8 16 4 64
xx x
xx x
xx x
= + + +
= + + +
= + + +
11. f(x) = x2 – 5x + 6
Put x = 1 2
2
(1) 1 5 1 6 2 0
Put 2 (2) 2 5 2 6 4 10 6 0
f
x f
= − × + = ≠= = − × + = − + =2x∴ − is factor of f(x)
23
2
2
32 5 6
2
3 6
3 6
0
xx x x
x x
x
x
−− − +
−− +
− +− ++ −
2 5 6 ( 2)( 3)x x x x∴ − + = − −
12. Let ( ) 3 22 4 34f x x x x= + + −
2x − Is factor of ( )f x
2x = Zero of ( )f x
( ) 3 22 2 4 2 2 34
16 16 2 34
34 34 0
f z = × + × + −= + + −= − =
13. We have ( )( )2x bx c x p x q− + = + −
( )2 2x bx c x p q x pq− + = + − −
Equating coefficient of x and constant
and b p q c pq− = − = −Substituting there values of b and c in
2 2 ,x bxy cy− + We get
( )
( ) ( )( ) ( )
2 2
2 2
x p q xy pqy
x pxy qxy pqy
x x py py x py
x py x qy
+ − −
+ − −+ − +
+ −
24
CBSE TEST PAPER-02
CLASS - IX Mathematics (Polynomials)
1. Which of the following expression is a monomial
(a) 3 + x (b) 4x3
(c) x6 + 2x2 + 2 (d) None of these
[1]
2. A linear polynomial
(a) May have one zero (b) has one and only one zero
(c) May have two zero (d) May have more than one zero
[1]
3. If P(x) = x3 – 1, then the value of P(1) + P(-1) is
(a) 0 (b) 1
(c) 2 (d) - 2
[1]
4. when polynomial x3 + 3x2 + 3x + 1 is divided by x + 1, the remainder is
(a) 1 (b) 0
(c) 8 (d) – 6
[1]
5. Factorise x2 + y – xy – x [1]
6. Show that 5 is a zero of polynomial 2x3 – 7x2 – 16x + 5 [2]
7. Using remainder theorem find the remainder when f(x) is divided by g(x)
f(x) = x24 – x19 – 2 g(x) = x + 1
[2]
8. Find K if x + 1 is a factor of P(x) = Kx2 – 2 x + 2 [2]
9. Find the values of m and n if the polynomial 2x3 +mx2 + nx – 14 has x – 1 and x + 2
as its factors.
[2]
10. Factorise 1 – a2 – b2 – 2ab [3]
11. Expand
21 1
12 3
a b − +
[3]
12. Verify each of the following identifies
(i) x3 + y3 = (x + y) (x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
[3]
13. Factorise (2x – 3y)3 + (3y – 4z)3 + (4z – 2x)3 [5]
25
CBSE TEST PAPER-02
CLASS - IX Mathematics (Polynomials)
1. (a) 2. (b) 3. (d) 4. (b)
5. x2 + y – xy – xx2 – x + y – xy = x2 – x – xy + y = x (x – 1) – y (x – 1) = (x – 1) (x – y)
6. Put x = 5 in 2x3 – 7x2 – 16 x + 153 22 5 7 5 16 5 5
250 175 80 5
255 255 0
× − × − × += − − += − =
5x∴ = is zero of polynomial 2x3 – 7x2 – 16x + 5
7. When f(x) is divided by g(x)Then remainder f(-1) F(-1) = (-1)24 – (-1)19 – 2 = 1 – (-1) – 2 = 1+ 1 – 2 = 0
8. Here 2( ) 2 2P x Kx x− +1x +∵ is factor of P(x)
2
( 1) 0
( 1) 2( 1) 2 0
2 2 0
(2 2)
P
K
K
K
∴ − =
− − − + =
+ + =
= − +
9. x – 1 and x + 2 is factor of 2x3 + mx2 + nx – 14x = 1, x = -2
3 22(1) (1) (1) 14 0
2 14 0
12 0
m n
m n
m n
∴ + + − =+ + − =+ − =
m n+3 2
12 (1)
2(2) (2) (2) 14 0
16 4 2 14 0
m n
m n
= − − − − −+ + − =
+ + − =
4 2 2 0
4 2 2
2
m n
m n
m n
+ + =+ = −
+ 1 (2)= − − − − − − −soving eq.(1) and eq.(2)
13 13m m− = ⇒ = −
26
Put 13 (1) 13 12
12 13 25
m in n
n
= − − + == + =
10. 2 21 2a b ab− − −2 2 2 21 ( 2 ) 1 ( )
(1 )(1 )
a b ab a b
a b a b
− + + = − += + + − −
11. 2
1 11
2 3a b
− +
2 221 1 1 1 1 1
1 2 2 1 2 12 3 2 3 3 2
a b a b b a− − = + − + + × × + × × + × ×
2 2 21
4 9 3 3
a b ab ba= + + − − +
12. (i) 3 3 2 2( )( )x y x y x xy y+ = + − +Taking R.H.S
2 2 2 2 2 2
3 2
( )( ) ( ) ( )x y x xy y x x xy y y x xy y
x x y
+ − + = − + + − +
= − 2xy+ 2yx+ 2xy− 3
3 3 . . .
. . . . .
y
x y L H S
L H S R H S
+
= + ==
Verified (ii) 3 3 2 2( )( )x y x y x xy y− = − + +
2 2 2 2
3 2
. . ( ) ( )R H S x x xy y y x xy y
x x y
= + + − + +
= + 2xy+ 2yx− 2xy− 3
3 3
. . .
y
x y
L H S
−
= −=
. . . . . .L H S R H S= Verified
13. Let 2 3 , 3 4 , 4 2a x y b y z c z x= − = − = − 2then a b c x+ + = 3y− 3y+ 4z− 4z+ 2x− 0
3 3 3 3a b c abc
=
∴ + + =
( ) ( ) ( ) ( )( ) ( )( )( ) ( )( ) ( ) ( )
3 3 32 3 3 4 4 2 3 2 3 3 4 2 2 2
3 2 3 3 4 2 2 2
6 2 3 3 4 2
x y y z z x x y y x z x
x y y x z x
x y y x z x
− + − + − = − − −
= − − −
= − − −
27
CBSE TEST PAPER-03
CLASS - IX Mathematics (Polynomials)
1. The value of K for which x – 1 is a factor of the polynomial 4x3 + 3x2 – 4x + K is
(a) 0 (b) 3
(c) – 3 (d) 1
[1]
2. The factors 12x2 – x – 6
(a) (3x – 2) (4x + 3) (b) (12x + 1) (x – 6)
(c) (12x – 1) (x + 6) (d) (3x + 2) (4x – 3)
[1]
3. x3 + y3 + z3 – 3xyz is
(a) (x + y – z)3 (b) (x – y + z)3
(c) (x + y + z)3 – 3xyz (d) (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
[1]
4. The expended form of (x + y – z)2 is
(a) x2 + y2 + z2 + 2xy + 2yz + 2zx (b) x2 + y2 – z2 + 2xy – 2yz – 2xz
(c) x2 + y2 + z2 + 2xy – 2yz – 2zx (d) x2 + y2 + z2 + 2xy + 2yx + 2xz
[1]
5. Find the integral zeroes of the polynomial x3 + 3x2 – x – 3 [1]
6. Check whether 7+ 3x is a factor of 3x2 + 7x [2]
7. Factorise 22 4
3x x
x− − [2]
8. Evaluate (101)2 by using suitable identify [2]
9. Find m and n it x – 1 and x – 2 exactly divide the polynomial x3 + mx2 – nx + 10 [2]
10. Using identify (a + b)3 = a3 + b3 + 3ab (a + b) drive the formula a3 + b3 = (a + b)
(a2 – ab + b2)
[3]
11. Factorise (i) 64y3 + 125z3 (ii) 27m3 – 343n3 [3]
12. Without actually calculating the cubes. Find the value of (26)3 + (-15)3 + (11)3 [3]
13. Factorise: 12(y2 + 7y)2 – 8 (y2 + 7y) (2y – 1) – 15(2y – 1)2 [5]
28
CBSE TEST PAPER-03
CLASS - IX Mathematics (Polynomials)
1. (c) 2. (d) 3. (d) 4. (c)
5. Given polynomial 3 2( ) 3 3P x x x x= + − −2
2
( ) ( 3) 1( 3)
( 3)( 1)
p x x x x
x x
= + − += + −For zeros ( ) 0p x =
2( 3)( 1) 0
( 3( 1)( 1) 0
3, 1, 1
x x
x x x
x x x
+ − =+ + − =
= − = − =Zeros of polynomial -1, 1, -3.
6. Let 2( ) 3 7p x x x= +7 + 3x is factor of p(x)
Remainder = 0
Remainder = 7
3P −
27 7
3 73 3
= − + −
3= 49
9× 49
9
0
−
=Hence 7 + 3x is factor of p(x)
7. 23 4
2 3x x− −
3 42
2 3
−× = −
We factorise by splitting middle term
2
2 1 1
3 42
2 33 4 4
12 3 3
3 41
2 3
x x x
x x x
x x
− + = −
− + −
= − + −
= + −
29
8. 2 2(101) (100 1)= +2 2 2
2 2 2 2
( ) 2
100, 1
(101) (100 1) 100 2 100 1 1
10000 200 1
10201
a b a ab b
here a b
+ = + += == + = + × × +
= + +=
9. Let ( ) 3 2 1p x x mx nx= + − +x – 1 and x – 2 exactly divide p(x)
3 2
(1) 0 (2) 0
(1) 1 1 1 10 0
1 10 0
11 0
p and p
p m n
m n
m n
∴ = == + × − × + =
+ − + =− + =
3 2
1 (1)
(2) 2 2 2 10 0
8 4 2 10 0
4 2 18
m n
p m n
m n
m n
− = − − − − − − − − − −= + × − × + =
+ − + =− = −
{ }2 9 2
(2)
m n dividing by− = − − − − −− + + − − − − − − − −subtracting eq. (2) form (1). we get
2
2
m
m
− = −=
put m = 2 in eq. (1). we get
2 11
11 2
n
n
− = −− = − −
13n+ = +
10. Solution given 3 3 3( ) 3 ( )a b a b ab a b+ = + − +3 3 3
2
2 2
( ) 3 ( )
( )[( ) 3 ]
( )[ 2 3 ]
a b a b ab a b
a b a b ab
a b a b ab ab
∴ + = + − += + + −= + + + −
2 2
2 2
( )( )
( )( )
a b a b ab
a b a ab b
= + + −= + − +
13
2
n
m
==
30
11. Solution
(i) 64y3 + 125z3
3 3
2 2
3 3 2 2
2 2
(4 ) (5 )
(4 5 )[(4 ) 4 5 (5 ) ]
( )( )
(4 5 )(16 20 5 )
y z
y z y y z z
a b a b a ab b
y z y yz z
++ − × +
∴ + = + − +
= + − +
(ii) 27m3 – 343n3 3 3
2 2
3 3 2 2
2 2
(3 ) (7 )
(3 7 )[(3 ) 3 7 (7 ) ]
[ ( )( )]
(3 7 )(9 21 49 )
m n
m n m m n n
a b a b a ab b
m n m mn n
= −= − − × +
− = − + +− − +
∵
12. Solution Let a = 26, b = -15, c = -11
a + b + c = 26 – 15 – 11 = 0
Then a3 + b3 + c3 = 3abc 3 3 3(26) ( 15) ( 11) 3 26 15 11
12870
+ − + − = × × − × −=
13. 2Let 7 , 2 1a y y b y= + = −2 2 2
2 2
2 2
12( 7 )2 8( 7 )(2 1) 15(2 1)
12 8 15
12 18 10 15
Then y y y y y y
a ab b
a ab ab b
+ − + − − −= − −= − + −
2
6 (2 3 ) 5 (2 3 )
(2 )(6 5 )
7 2 1
a a b b a b
a b a b
put a y y and b y
= − + −= − +
= + = −2 2
2 2
2 2
[2( 7 ) 3(2 1)][6( 7 ) 5(2 1)
[2 14 6 3][6 42 10 5]
(2 8 3)(6 52 5)
y y y y y y
y y y y y y
y y y y
= + − − + + −= + − + + + −= + + + −
31
CBSE TEST PAPER-04
CLASS - IX Mathematics (Polynomials)
1. The value of 1023 is
(a) 1061208 (b) 1001208
(c) 1820058 (d) none of these
[1]
2. (a-b)3 + (b-c)3 + (c-a)3 is equal to
(a) 3abc (b) 3(a-b) (b-c) (c-a)
(c) 3a3b3bc3 (d) [a-(b+c)]3
[1]
3. The zeroes of the polynomial P(x) = x (x-2) (x+3) are
(a) 0 (b) 0, 2, 3
(c) 0, 2, -3 (d) none of these
[1]
4. If (x+1) and (x-1) are factors of Px3+x2-2x+9 then value of p and q are
(a) p = -1, q = 2 (b) p = 2, q = -1
(c) p = 2, q = 1 (d) p = -2, q = -2
[1]
5. Factorise 8a3-b3-12a2b+6ab2 [2]
6. Evaluate 993 [2]
7. Find the value of k, if x-1 is factor of P(x) and P(x) = 3x2+kx+ 2 [2]
8. Expand
32
13
x +
[2]
9. Find the values of m and n so that the polynomial x3-mx2-13x+n has x-1 and x+3
as factors.
[3]
10. Prove that x2+6x+15 has no zero. [3]
11. Factorise 3 (x+y)2 - 5(x+y) + 2 [3]
12. The volume of a cuboid is given by the expression 3x3-12x.find the possible
expressions for its dimensions.
[3]
13. Factorise x6 + 3y6 - z6+6x2y2z2 [5]
32
CBSE TEST PAPER-04
CLASS - IX Mathematics (Polynomials)
[ANSWERS]
Ans01. (a) Ans02. (a) Ans03. (c) Ans04. (b)
Ans05. 3 3 2 2 8 12 6a b a b ab− − +3 3
3 3
3
(2a) 6 (2 )
(2a) 3(2 ) (2 )
(2a-b)
(2a-b) (2a-b) (2b-b)
b ab a b
b a b a b
= − − −= − − −==
Ans06. 3 399 (100 1)= −
We know that 3 3 2 2 3( ) 3 3a b a a b ab b− = − + −
3 3 2 2 3
100, b = 1
(100-1) 100 3 100 1 3 100 1 1
= 1000000-30000+300-1
= 1000300-30001
= 970299
Take a == − × × + × × −
Ans07. x-1 is factor of P(x)
P(1) = 0
3 1+k 1+ 2 0
3+k+ 2 0
k = -(3+ 2)
∴
× × =
=
Ans08. 3 3 3( 3) 3 ( )a a b ab a b+ = + + +
3 33 3 2
2a = , b =1
3
2 2 2 2 8 4 21 1 3 1 1 = 1
3 3 3 3 27 3 3
x
xx x x x x x
+ = + + × × + + + +
Ans09. Let polynomial be 3 2( ) 12
1 factor of P(x)
P(1) = 0
P x x mx x n
x is
= − − +−
∴
33
Ans10. 2x 6 15x +2 2
2
2
2
2
= x 2 3 3 6
= (x+3) 6
( 3) is positive & 6 is positive
(x+3) 6 has no. zero.
x 6 15 has no. zero.
x
x
x
+ × + ++
+∴ +
+ +
Ans11. 23( ) 5( ) 2x y x y+ − + +
2 2
2
= z
3z 5 2 3 3 2 2
= 3z (z-1) -2 (z-1)
= (3z-2) (z-1)
Put z= x+y
3(x+y) 5( ) 2 [3( ) 2] [x+y-1]
= [3x+3y-2]
Let x y
z z z z
x y x y
+− + = − − +
∴ − + + = + − [x+y-1]
Ans12. The volume of cuboid is given by 3 2
3 2
3x 12 3 ( 4) = 3x (x+2) (x-2)
Dimensions of the cuboid are given by 3x, (x+2) and (x-2)
P(1) = 1 1 13 1 0
= 1-m-13+n = 0
= -m+n =12
x x x
m n
− = −
− × − × + =
3 2
(1)
x+3 is factor of P(x)
P(-3) = 0
P(-3) = (3) ( 3) 13 ( 3) 0
= -27-9m+39+n = 0
= -9m+n 12 =0 (2)
= -9m+n =
m n
∴− − − × − + =
-12Subtracting eq. (2) from (1)
8m = 24, m = 3
Put m = 3 in eq (1)
m = 3 and n = 15
Ans13. 6 6 6 2 2 28 6x y z x y z+ − +2 3 2 3 2 3 2 2 2= (x ) (2 ) ( ) 3( ) (2y ) (-z )y z x+ + − −2 2 2 2 3 2 2 2 2 2 2 2 2 2 2
2 2 2 4 4 4 2 2 2 2 2 2
[x ] [(x ) (2 ) ( ) 2 2 ( ) ( )
(x 2 ) (x 4 2 2 )
y z y z x y y z x z
y z y z x y y z x z
= + − + + − − × − − − × −= + − + + − + +
34
CBSE TEST PAPER-05
CLASS - IX Mathematics (Polynomials)
1. If x+y+x = 0, then 3 3 3 isx y z+ +(a) xyz (b) 2xyz (c) 3xyz (d) 0
[1]
2. The value of (x-a)3 + (x-b)3 + (x-c)3 – 3 (x-a) (x-b) (x-c) when a + b + c= 3x, is
(a) 3 (b) 2 (c) 1 (d) 0
[1]
3. Factors of x2 + 3 2 4 arex +
(a) ( 2 2) (x- 2)x + (b) ( 2 2) (x+ 2)x +
(c) ( 2 2) (x+ 2)x − (d) ( 2 2) (x- 2)x −
[1]
4. The degree of constant function is
(a) 1 (b) 2 (c) 3 (d) 0
[1]
5. Factorise 3 3 327 9x y z xyz+ + − [2]
6. Evaluate 105 95× [2]
7. Using factor theorem check whether g(x) is factor of p(x)
P(x) = 3 24 6, g(x) = x-3− + +x x x
[2]
8. Expand
32
3x y −
[2]
9. Using remainder theorem factorise3 23 3x x x− − +
[3]
10. If 3 2 6y ay by+ + + is divisible by y – 2 and leaves remainder 3 when divided by
y – 3, find the values of a and b.
[3]
11. Factorise x6 – 64 [3]
12. The volume of a cuboid is given by the algebraic expression ky2-6ky+8k. Find
the possible expressions for the dimensions of the cuboid.
[3]
13. Factorise: ( )
3 331 5 3 3 2
2 527 3 4 4 3
x y y z z x− + + + − +
[5]
35
CBSE TEST PAPER-05
CLASS - IX Mathematics (Polynomials)
[ANSWERS]
Ans01. (b) Ans02. (c) Ans03. (b) Ans04. (d)
Ans05. 27X3 +Y3 +Z3-9XYZ
= (3X) 3 +Y3 +Z3 -3(3X) (Y) (Z)
= [3X+Y+Z] [(3X) 2 +Y2 +Z2 -3XY –YZ-3X2]
= (3X+Y+Z) (9X2+Y2+Z2 – 3XY –YZ -3X2)
Ans06. 105×95
= (100+5) (100-5)
=1002-52 [ ∴ (a+b) (a-b) =a2-b2]
=10000 -25 = 9975
Ans07. Given g(x) =X-3, X-3=0
Put x=3 in p (x)
P (3) =33 -4×32 +3+6
= 27+9 -4×9 =36-36 =0
Remainder =0
∴ By factor theorem g(x) is factor of P (X)
Ans08. 32
( )3
x y−
3 3 3
3 3 3
3 3
3 3 2 2
( ) = a -b -3ab(a-b)
2 Hence a=X , b=
32 2 2 2
(x- ) - ( ) 3 ( )3 3 3 3
8 2 =X 2 ( )
27 38 4
=X - 2x +27 3
a b
y
y x y X Y X Y
y XY x Y
y Y Xy
∴ −
∴ = − × −
− − −
−
Ans09. X3 -3X2 – X+3
Coefficient of X3 is 1
Constant =3
3×1 = 3
∴We can Put X= ± 3 and ( X ) and check
Put= X=1
36
13 – 3 × 12 - 1+3
1 – 3 – 1 + 3 = 0
Remainder =0 3 21 3 3X is factor of x x x− − − +∵
2
3 2
3 2
2
2
x -2x-31 3
x
-2x - x
-2x +2x-3x + 3
-3x + 30
x X x
x
− −−
3 2 2
2
x 3 3 ( 1) (x 2 3)
= (X-1) (x 3 3)
= (X-1) [X(X-3) +1 (X-3)]
= (X-1) (X-3) (X+1)
x x x x
x x
∴ − + = − − −− + −
Ans10. Let
P(y) = y3 + ay2 + by + 6
P(y) is divisible by y – 2
Then P (2) = 0 3 22 2 2 6 0
8 4 2 6 0
4a+2b = -14
2a+b = 7 (i)
If P (y) is divided by y-3 remainder is 3
P (3)
a b
a b
+ × + × + =+ + + =
∴3 2 3 3 3 6 3
9a + 3b = -30
3a + b = -10 (ii)
a b+ × + × + =
eq (i) - eq (ii)
-a = 3 and a = -3
Put a = -3 in eq (i)
2 -3 + b = -7
-6 + b = -7
b = -7
×
+ 6
b = -1
37
Ans11. 6 64x −2 3 2 3
2 2 4 2
4 2
= (x ) - (2 )
= (x -2 ) [x +4x +16]
= (x+2) (x-2) (x +4x +16)
Ans12. Given volume of cuboid 2
2
6 8
= k [y 6 8]
ky ky k
y
− +− +
2 [ 4 2 8]
= k [y (y-4) -2 (y-4)]
= k (y-2) (y-4)
Thus dimension of cuboid
k, (y-2) and (y-4)
k y y y− − +
Ans13. Given expression can be written as 3 3 3
3 3 3
3
1 5 3 3 2(2 5 )
3 3 4 4 3
1 5 5 (2 5 ) ,
3 3 43 2
4 3
2 5 5 5 3 20
3 3 3 4 4 3
3
,
1 5 3(2 5 )
27 3 4
x y y z z x
Let x y a y z b
and z x C
a b c x y y z z x
a b c abc
Thus
x y y z
+ + − + + − −
−+ = + =
− − =
+ + = + − + − − =
∴ + + =
−+ + +3 3
3 2
4 3
1 5 5 3 2 3 (2 5 )
3 3 4 4 3
5 5 3 2 (2 5 )
3 4 4 3
20 15 9 8 (2 5 )
12 12
1 (2 5 ) (20 9 ) (9 8 )144
z x
x y y z z x
x y y z z x
y z z xx y
x y y z z x
− +
− − = + + −
− = − + + +
− + + = − +
= + − +
38
CBSE TEST PAPER-01
CLASS - IX Mathematics (Introduction to Euclid’s geometry)
1. The edges of a plane surface are.
(a) Line (b) points.
(c) Angles (d) planes.
[1]
2. Given four points such that No three of them are collinear, then there exists.
(a) 2 lines (b) 4 lines
(c) 6 lines (d) 5 lines
[1]
3. One and only one line passes through. – Distinct points.
(a) one (b) two
(c) Three (d) four.
[1]
4. If equals are added to equals. The wholes are –
(a) equal (b) not equal
(c) Doubled (d) none of these.
[1]
5. If A, B and C are three points on a line and B is between A and C, then prove that.
AC- BC = AB
[2]
6. In fig AC = BD, then prove that AB=CD [2]
7. In fig AB =CD prove that AC=BD [2]
8. How would you write Euclid’s fifth postulate [2]
9. If a point C lies between two points A and B such that AC=BC, then prove that AC
=1
2 AB Explain by drawing the figure.
[3]
10. Prove that An equilateral triangle can be constructed on any given line segment. [3]
39
11. If AB= PQ and PQ =XY then prove that AB = XY. Explain by drawing the fig. [3]
12. Give a definition for each of the following are there other terms which need to
be defined first? What are they?
[3]
13. Which of the following statements are true and which are false Explain.
(I) Only one line pass. Through a single point
(ii) There are an infinite number of line which passes through two distinct
points.
(iii) A terminated line can be produced indefinitely on both sides.
(iv) If two circles are equal, then their radii are equal.
(v) In fig AB=PQ and PQ=XY, then AB=XY
[5]
40
CBSE TEST PAPER-01
CLASS - IX Mathematics (Introduction to Euclid’s geometry)
[ANSWERS]
Ans01. (A)
Ans02. (C)
Ans03. (b)
Ans04. (a)
Ans05. Solution, in fig
AB coin cider with AC –BC
So AC-BC=AB [things which coincide with one another are equal to one another]
Ans06. Solution: AC=BD [given]
Also BC=BC [Things which coincide with one another are equal to one another]
∴ AC-BC = BD-BC
AB=CD [If equals are subtracted from equal then remaining are equal]
Ans07. Solution: AB=CD [given]
Also BC=BC [things which coincide with one another are equal to one another]
∴ AB+BC=CD+BC [it equals are added from equal then remaining are equal]
Ans08. For every line L and for every point p not lying on L, there exists a unique line m
passing through p and parallel to L.
41
Ans09.
Solution:
In fig point C is between A and B
AC+CB = AB [things which coincide with one another are equal to one another]
But AC=BC
∴ AC+AC=AB
2AC=AB
AC =1
2AB
Ans10. Solution: using Euclid’s third postulate.
Draw circles with centre d at the points A
and B and radius equal to the length of the
segment AB cutting each other at C.
Now, AB=AC [radii of the same circle]
BC=AB [radii of the same circle]
∴ AC=BC
Ans11. Solution,
Given AB=PQ and PQ =XY
AB =XY [things which are equal to the same things are equal to one another]
Ans12. (i) Two lines in a plane are parallel if they are equidistant, distance should be
defined before the above definition.
(ii) Two lines are perpendicular to each other if they make a right angle. Angle
and right angle are two terms to be defined before the above definition.
42
Ans13. (i) False: An infinite number of lines can pass through a single point.
(ii) False: There is a unique line joining two distinct points.
(iii) True: There days a terminated line is called a line segment but Euclid
called it a terminated line. So a line segment can be produced
indefinitely on both sides and line segment AB when produced on both
sides becomes line AB.
(iv) True: In equal circles, when the region bounded by one circle in
superimposed on the other then the circles coincide i.e. their centre and
boundaries coincide. As such their radii are equal.
(v) True: by Euclid’s axiom, things which are equal to the same things are
equal to one another.
43
CBSE TEST PAPER-01
CLASS - X Mathematics (Lines and angles)
1. Measurement of reflex angle is
(i) 90� (ii) between 0 and 90� �
(iii) between 90 and 180� � (iv) between 180 and 360� �
[1]
2. The sum of angle of a triangle is
(i) 0� (ii) 90� (iii) 180� (iv) none of these
[1]
3. In fig if x= 30� then y=
(i) 90� (ii) 180�
(iii) 150� (iv) 210�
[1]
4. It two lines intersect each other then
(i) vertically opposite are equal (ii) corresponding angle are equal
(iii) alternate interior angle are equal (iv) none of these
[1]
5. In fig lines x y and m n intersect at 0 If
poy =90 and a:b =2:3 find c∠ �
[2]
6. In fig find the volume of x and y then Show that ABIICD [2]
7. What value of x would make AOB a line if AOC=4x and BOC=6x+30∠ ∠ ° [2]
44
8. In fig POQ is a line. Ray OR is perpendicular
to line PQ. OS is another ray lying between
rays OP and OR. Prove that
1 ROS= ( QOS- POS)
2∠ ∠ ∠
[3]
9. In fig lines P and R intersected at 0, if x = 45� find x, y and u [2]
10. Prove that sum of three angles of a triangle is 180� [3]
11. It is given that XYZ =64∠ � and X Y is produced to point P, draw a fig from the
given information. If ray Y Q bisects ∠ Z Y P, find ∠ XYQ and reflex ∠ QYP.
12. In fig if PQIIST, ∠ PQR = 110� and ∠ RST=130� find ∠ QRS.
13. In fig the side AB and AC of △A B C Are
produced to point E And D respectively. If
bisector BO And CO of ∠ CBE And ∠ BCD
respectively meet at point O, then prove
that ∠ BOC =1
902
− ∠� BAC
45
CBSE TEST PAPER-01
CLASS - X Mathematics (Lines and angles)
[ANSWERS]
Ans01 (iv)
Ans02. (iii)
Ans03. (iii)
Ans04. (i)
Ans05. Given in fig. ∠ POY=90�
a: b: 2: 3
Let a=2x and b =3x
a + b + ∠ POY=180� ( ) xoy is a line∵
2x+3x+ 90�=180�
5x=180 -90� �
5x= 90�
x= 90
185
=�
�
a= 36 , b=54∴ � �
MoN is a line.
b+C=180�
54 +C�
=180�
180 54 126
126
C
Ans C
= − ==
� � �
�
Ans06. ( )50 180 pairx by liner+ =� �
( )
180 50
130
130 vertically opposite angles are equal
x
x
y
= −=
=
� �
�
�
∵
46
Ans07. AOC=4x And BOC =6x+30given ∠ ∠ �
( ) AOC+ BOC =180 linear pair
4 6 30 180
10 180 30
10 150 15
By
x x
x
x x
∠ ∠
+ + == −= = =
�
� �
� �
Ans08. ( )1. . =
2R H S QOS POS∠ − ∠
( )
( ) ( )
( )
1 = ROS+ QOR - POS
21
= ROS +90 POS 12
POS+ ROS=90
By 1
1 =
21
= 2 ROS= R2
ROS POS ROS POS
∠ ∠ ∠
∠ − ∠ →
∠ ∠∴
+ ∠ + ∠ − ∠
× ∠ ∠
�
�
∵
OS
= L.H.S
Hence proved
∠
Ans09. 45X = �
( )
( )
Z=45 vertically opposite angles are equal
X+y=180
45 =180
y=180 45
y=135
y=u
u=135
By linear pair
y
vertically opposite angles
∴
+−
�
�
� �
�
�
�
∵
Ans10. ; ABCgiven △
To prove : 180
: through A draw XYIIBC
Proof : XYIIBC
A B C
construction
∠ + ∠ + ∠ = �
∵
47
( )( )
2= 4 1 Alternate interior angle
And 3= 5 2 Angles are equal
Adding eq (1) And eq (2)
2+ 3= 4+ 5
Adding both sides 1
∴ ∠ ∠ →
∠ ∠ →
∠ ∠ ∠ ∠∠
∵
( ) 1+ 2+ 3= 1+ 4+ 5
1+ 2+ 3=180 1, 4, and 5 forms line
A+ B+ C=180
∠ ∠ ∠ ∠ ∠ ∠∠ ∠ ∠ ∠ ∠ ∠
∠ ∠ ∠
�
�
∵
Ans11. ( )180 pairXYZ PYZ linear∠ + ∠ = �
( ) 64 180 64
PYZ=180 64
pyz =116
1 116 ZYQ= 58
2 2 XYQ= XYZ+ ZYQ
=64 58 122
PYZ given XYZ
ZYP
⇒ + ∠ = ∠ =
∠ −
∠
∠ ∠ = =
∠ ∠ ∠+ =
� � �
� �
�
�
� � �
Also reflex QYP= XYP
=122 180
=302
XYQ struight∠ ∠ + ∠+� �
�
Ans12. Through point R Draw line KLIIST
( )
PQIIST
STIIKL PQIIKL
PQIIKL
PQR+ 1=180
int 180
110 1 180
1=70
Sum of erior angle on the same side of transverral is
∴
∴ ∠ ∠
+ ∠ =∠
�
�
� �
�
∵
∵
2+ RST=180
2+130 180
2=50
Similarly ∠ ∠∠ =
∠
�
� �
�
48
1+ 2+ 3=180
70 50 3 180
3=180 120
3=60
∠ ∠ ∠+ + ∠ =
∠ −
∠
�
� � �
� �
�
QRS=60∠ �
Ans13. : Ray BO bisects CBESolution ∠
( ) ( )
( )
( )
1 C B O = C B E
21
= 1 8 0 C B E + y = 1 8 02
= 9 0 12
C s e c
1 B C O =
21
= 1 8 02
y
y
S im i la r ly r a y o b i ts B C D
B C D
Z
∴ ∠ ∠
− ∠
− →
∠
∠ ∠
−
� �
�
�
∵
= 9 02
Z−�
( )
BOC
BOC+ BCO+ CBO=180
BOC+90 90 1802 21
BOC=2
But X+Y+Z=180
Y+Z
In
Z Y
Y Z
∠ ∠ ∠
∠ − + − =
∠ +
�
� � �
�
△
( )=180
1 BOC= 180 90
2 21
BOC= 90 BAC2
X
XX
−
∠ − = −
∠ − ∠
�
� �
�
49
CBSE TEST PAPER-02
CLASS - X Mathematics (Lines and angles)
1. The measure of Complementry angle of 63o is
(a) 30o (b) 36o
(c) 27o (d) none of there
(1)
2. If two angles of a triangle is 30o and 45o what is measure of third angle
(a) 95o (b) 90o
(c) 60o (d) 105o
(1)
3. The measurement of Complete angle is
(a) 0o (b) 90o
(c) 180o (d) 360o
(1)
4. The measurement of sum of linear pair is
(a) 180o (b) 90o
(c) 270o (d) 360o
(1)
5. The exterior angle of a triangle is 110o and one of the interior opposite angle is
35o. Find the other two angles of the triangle.
(1)
6. Of the three angles of a triangle, one is twice the smallest and another is three
times the smallest. Find the angles.
(2)
7. Prove that if one angle of a triangle is equal to the sum of other two angles, the
triangle is right angled.
(2)
8. In fig. sides QP and RQ of PQR∆ are produced to points S and T respectively.
If 135 and PQT=110 , find PRQo oSPR∠ = ∠ ∠ .
(2)
50
9. In fig the bisector of and BCA ABC∠ ∠ intersect each other at point O prove
that 1
902
oBCO A∠ = + ∠
(2)
10. The side BC of ABC∆ is produced from ray BD. CE is drawn parallel to AB show
that .ACD A B∠ = ∠ + ∠ Also prove that 180oA B C∠ + ∠ + ∠ = .
(3)
11. Prove that if a transversal intersect two parallel lines, then each pair of
alternate interior angles is equal.
(3)
12. In given fig. AB // CD. Determine a∠ . (5)
13. In fig. > R Q∠ ∠ and M is a point on QR such that PM is the bisector of QPR∠ if
the perpendicular from P on QR meets QR at N, then prove that
( )1
2MPN Q R∠ = ∠ − ∠
(5)
51
CBSE TEST PAPER-02
CLASS - X Mathematics (Lines and angles)
[ANSWERS]
Ans01. (c) Ans02. (d)
Ans03. (d) Ans04. (a)
Ans05. The exterior angle of a triangle is equal to the sum of interior opposite angles.
ACD= A+ B
110= A+35
A=110 35
A=75
C=180-( A+ B)
C=180-(75 35 )
C=70
o
o o
o
o o
o
∴ ∠ ∠ ∠
∠∠ −∠∠ ∠ ∠∠ +
∠
Ans06. Let the smallest angle be xo
Then other two angles are 2xo and 3xo
xo+2xo+3xo=180o [sum of three angle of a triangle is 180o]
6xo +=180o
x =180
6 =30o
angles are 30o, 60o and 90o
Ans07. Given in B= A+ CABC∆ ∠ ∠ ∠To prove: ABC∆ is right angled.
Proof: 180oA B C∠ + ∠ + ∠ = (1) [Sum of three angle of a ∆ ABC is 180o]
A C B∠ + ∠ = ∠ (2)
From (1) and (2)
180
2 B=180
B=90
o
o
o
B B∠ + ∠ =∠∠
52
Ans08. 180oPQT PQR∠ + ∠ = 110 180
180 110
PQR=70
SPR= PQR+ PRQ
135 70
PRQ=135 70
PRQ=65
o o
o o
o
o o
o o
o
PQR
PQR
Also
PRQ
+ ∠ =∠ = −∠∠ ∠ ∠
= + ∠
∠ −∠
Ans09. Given A ABC∆ such that the bisectors of and BCAABC∠ ∠ meet at a point O
To Prove 1
902
oBOC A∠ = + ∠
Proof: In BOC∆1 2 180oBOC∠ + ∠ + ∠ = (1)
In ABC∆ 180
2 1 2 2 180o
A B C
A
∠ + ∠ + ∠ =∠ + ∠ + ∠ =
�
[BO and CO bisects CB and∠ ∠ ]
1 2 =902
1+ 2=902
o
o
A
A
∠⇒ + ∠ + ∠
∠∠ ∠ −
[Divide by 2]
Ans10. ∵ AB // CE and Ac intersect them 1 4∠ = ∠ (1) [Alternate interior angles]
Also AB//CE and BD intersect them 2 5∠ = ∠ (2) [Corresponding angles]
Adding eq (1) and eq (2)
1 2 4 5
A B ACD
∠ + ∠ = ∠ + ∠∠ + ∠ = ∠
Adding C∠ on both sides, we get
180o
A B C C ACD
A B C
∠ + ∠ + ∠ = ∠ + ∠∠ + ∠ + ∠ =
53
Ans11. Given: line AB//CD intersected by transversal PQ To Prove: (i) 2 5∠ = ∠ (ii) 3 4∠ = ∠
Proof: 1 2∠ = ∠ (i) [Vertically Opposite angle] 1 5∠ = ∠ (ii) [Corresponding angles]
By (i) and (ii)
2 5∠ = ∠ Similarly 3 4∠ = ∠
Hence Proved
Ans12. Through O draw a line l parallel to both AB and CD Clearly
1 2
1 38
2 55
55 38
93
o
o
o o
o
a
a
a
∠ = ∠ + ∠
∠ =∠ =∠ = +∠ =
[Alternate interior angles]
Ans13. sec PM bi ts QPR∠∵ QPM MPR∠ =∵
180 [By angle sum property of ]
PQN+ QPN+90 180
90 (i)
o
o o
o
In PQN
PQN QPN PNQ
PQN QPN
∆
∠ + ∠ + ∠ = ∆
⇒ ∠ ∠ =∠ = − ∠
In ∆ PNR
180 [By angle sum property]
PRN+ NPR+90 180
PRN=90 (2)
Subtracting eq (2) from (1)
o
o o
o
PRN NPR PNR
NPR
∠ + ∠ + ∠ =∠ ∠ =
∠ − ∠
( )PQN - PRN=(90 ) 90o QPN NPQ∠ ∠ − ∠ − ° − ∠
54
NPR - QPN
=( NPM+ MPR)-( QPM- NPM)
=2 NPM+( MPR- QPM)
=2 NPM+( MPR- MPR) [ PM bisects QPM
] QPM MPR
= ∠ ∠∠ ∠ ∠ ∠∠ ∠ ∠∠ ∠ ∠ ∠
∴∠ = ∠∵
Hence
2
2
1 NPM= [ ]
21
MPN= [ ]2
Hence Proved
PQN PRN NPM
Q R NPM
or Q R
or Q R
∠ − ∠ = ∠∠ − ∠ = ∠
∠ ∠ − ∠
∠ ∠ − ∠
55
CBSE TEST PAPER-03
CLASS - IX Mathematics (Lines and angle)
1. The difference of two complementary angles is 040 . The angles are
(a) 0 065 ,35
(b) 0 070 ,30
(c) 0 025 ,65
(d) 0 070 ,110
[1]
2. Given two distinct point P and Q in the interior of ABC∠ , then AB����
will be
(a) in the interior of ABC∠
(b) in the interior of ABC∠
(c) on the ABC∠
(d) on the both sides of BA����
[1]
3. The complement of ( )090 a− is
(a) - 0a (b) ( )090 2a+
(c) ( )090 a− (d) 0a
[1]
4. The number of angles formed by a transversal with a pair of lines is
(a) 6 (b) 3
(c) 8 (d) 4
[1]
5. In the given figure POR and QOR∠ ∠ from a linear pair if a – b = 080 . Find the
value of ‘a’ and ‘b’.
[2]
56
6. If ray OC stands on a line AB such that AOC BOC∠ = ∠ , then stands that
090AOC∠ =
[2]
7. In the given figure show that AB//EF [2]
8. In figure if AB//CD, 0 050 127 APQ and PRD find x and y∠ = ∠ = . [2]
9. In the given figure ∆ ABC is right angled at A. AD is drawn perpendicular to BC.
Prove that BAD ACB∠ = ∠[3]
10. In ∆ ABC 0 045 , 55 secB C and bi tor A∠ = ∠ = ∠ meets BC at a point D. find [3]
57
ADB and ADC∠ ∠
11. In figure two straight lines AB and CD intersect at a point O. It
0 0(45 )BOD x and AOD x∠ = ∠ = − . Find the value of x hence find
(a) BOD∠ (b) AOD∠
(c) AOC∠ (d) BOC∠
[3]
12. The side BC of a ∆ ABC is produced to D. the bisector of ∠ A meets B in L as
shown if fig. prove that ∠ ABC+ ∠ ACD=2 ALC∠
13. In fig M and N are two plane mirrors perpendicular to each other; prove that the
incident ray CA is parallel to reflected ray BD.
>
>
58
CBSE TEST PAPER-03
CLASS - IX Mathematics (Lines and angle)
[ANSWERS]
Ans01. (A)
Ans02. (C)
Ans03. (D)
Ans04. (C)
Ans05. 0
0
0
0
0
0 0
0 0 0
180 (1) [By lineas pair]
a-b=180 (2)
2a= 260 [ Adding e.q (1) And (2)]
a=130
a=130 in eq (1)
130 180
b=180 130 50
a b
Put
b
+ = →→
+ =− =
Ans06.
0
0
0
0
[Given]
AOC+ BOC=180 [BY lines pair]
AOC+ AOC=180
2 AOC=180
AOC=90
AOC BOC∠ = ∠
∠ ∠∠ ∠
∠∠
Ans07.
0 0 0
B C D = B C E + E C D
= 3 6 3 0 6 6
A B C D [A lte rn a te in te rio r an g le s a re e q u a l
A B C
∠ ∠ ∠+ = = ∠
∴ ∏
59
0 0
0 0 0
0
ECD=30 FEC=150
ECD+ FEC=30 150 180
Hence EF CD [Sum of consecutive interior Angle is 180 ]
CD EF
then AB EF
Again and
AB CDand
∠ ∠
∴ ∠ ∠ + =∏
∏ ∏∏
Ans08.
0
0 0
PQ is a transversal
APQ = PQD [ Pair of alternate angles]
50 X
Also AB CD and PR is a transversal
APR= PRD
50 127
AB CD and
Y
∏∠ ∠
=∏∠ ∠
+ =0 0 0 Y= 127 50 77− =
Ans09.
0
0
0 0
0
0
0
AD
90
ABD
ABD+ BAD+ ADB=180
ABD+ BAD+90 180
ABD+ BAD=90
BAD=90 (1)
But A+ B+ C=180 in ABC
BC
ADB ADC
from
ABD
⊥∴∠ = ∠ =
∆∠ ∠ ∠∠ ∠ =
∠ ∠∠ − ∠ →
∠ ∠ ∠ ∆
∵
0 0
0
B+ C=90 A=90
C=90 (2)
from (1) and (2)
BAD= C
BAD= ACB H
B
∠ ∠ ∴ ∠∠ − ∠ →
∠ ∠∠ ∠ ence proved
Ans10.
0 0
ABC
A+ B+ C=180 [Sum of three angle of a is 180 ]
In ∆
∠ ∠ ∠ ∆
60
00 0
0 0 0
0 0
0
0 0 0
A+45 55 180
A=180 100 80
AD bisects A
1 1 1= 2= 80 40
2 2 Now in ADB, We have
1+ B+ ADB=180
40 45 180
ADB=180
A
ADB
⇒ ∠ + =∠ − =
∴ ∠
∠ ∠ ∠ = × =
∆∠ ∠ ∠
⇒ + + ∠ =⇒ ∠ 0 0 0
0
0 0
0 0 0
0 0
85 95
Also ADB+ ADC=180
95 180
ADC=180 95 85
Ans ADB=95 ADC=85
ADC
and
− =∠ ∠
+ ∠ =
∠ − =∠ ∠
Ans11.
0
0
0
0
0
BYlinear pair
180 4 5
180 5 5
5x=185
185 x= 37
5 AOD=4x-5=4 37-5=148-5
=143
BOC=143 AOD and BO
ADB AOD DOB
x x
x
∠ = ∠ + ∠= − +
+ =
=
∴ ∠ ×
∠ ∴ ∠ ∠0
C
BOD=x=37 verti cally copposite Angles∠
37BOD AOC∠ = ∠ = °
61
Ans12.
ABC we have
ACD= B+ A (1) [ Exterior angle property]
ACD= B+2L1 [ A is the bisector of A =2L1]
In AB
ALC= B+ BA [Exterior angle property]
A
In ∆∠ ∠ ∠ →
⇒ ∠ ∠ ∴ ∠ ∠∆ ∠
∠ ∠ ∠ ∠∠ ∠C= B+ 1
2 A LC = 2 B+2L1 (2)
Subtracting (1) from (2)
2 ALC- ACD= B
2 ALC= B+ ACD
ACD+ ABC = 2 ALC
∠ ∠⇒ ∠ ∠ →
∠ ∠ ∠∠ ∠ ∠
∠ ∠ ∠
Ans13. Draw AP ⊥ M and BQ ⊥ N
0
0
0 0
0
a M N
BOA = 90
BQ AP
In BOA 2+ 3+ BOA=180 [ ]
2+ 3+90 180
2+ 3=90
1= 2 and 4= 3
BQ N and AP M nd
By angle Sum property
Also
∴ ⊥ ⊥ ⊥∴ ∠⇒ ⊥∆ ∠ ∠ ∠
⇒ ∠ ∠ =∴ ∠ ∠∠ ∠ ∠ ∠
0
0 0 0
0
0
1+ 4= 2+ 3=90
( 1+ 4)+ ( 2+ 3) =90 90 180
( 1+ 2)+( 3+ 4)=180
or CAB+ DBA=180
⇒ ∠ ∠ ∠ ∠∴ ∠ ∠ ∠ ∠ + =
⇒ ∠ ∠ ∠ ∠∠ ∠
CA BD [ By sum of interior angle of same side of transversal] ∴ ∏
>
>
>
62
CBSE TEST PAPER-04
CLASS - IX Mathematics (Lines and angle)
1. In fig L1 ∏ L2 And 01 52 ∠ = the measure of ∠ 2 is.
(A) 038
(B) 0128
(C) 052
(D) 048
[1]
2. In fig x= 030 the value of Y is
(A) 010
(B) 040
(C) 036
(D) 045
[1]
3. Which of the following pairs of angles are complementary angle?
(A) 0 025 ,65
(B) 0 070 ,110
(C) 0 030 ,70
(D) 0 032.1 , 47.9
[1]
4. In fig the measure of ∠ 1 is.
(A) 0158
(B) 0138
(C) 042
(D) 048
[1]
5. Prove that if two lines intersect each other then vertically opposite angler are
equal.
[2]
6. The measure of an angle is twice the measure of supplementary angle. Find its
measure.
[2]
63
7. In fig ∠ PQR = ∠ PRQ. Then prove that ∠ PQS= ∠ PRT. [2]
8. In the given fig ∠ AOC = ∠ ACO and ∠ BOD = ∠ BDO prove that AC�DB [2]
9. In fig lines XY and MN intersect at O If ∠ POY= 90�and a:b=2:3 find ∠ C [3]
10. In fig PT is the bisector of ∠ QPR in ∆ PQR and PS ⊥ QR, find the value of x [3]
64
11. The sides BA and DC of a quadrilateral ABCD are produced as shown in fig show
that ∠ X+ ∠ Y = ∠ a+ ∠ b
[3]
12. In the BO and CO are Bisectors of ∠ B and ∠ C of ∆ ABC, show that
∠ BOC= 90�+1
2∠ A.
[3]
13. In fig two straight lines PQ and RS intersect each other at o, if ∠ POT= 075 Find
the values of a,b and c
[3]
65
CBSE TEST PAPER-04
CLASS - IX Mathematics (Lines and angle)
[ANSWERS]
Ans01. (b) Ans02. (b) Ans03. ( a ) Ans04. (c)
Ans05.
Given: AB and CD are two lines intersect each other at O.
To prove: (i) 1 2∠ = ∠ and (ii) 3 4∠ = ∠
Proof: 0
0
1 4 180 ( ) [ ]
4 2 180 ( ) [ ]
1 4 4 2 [ ( ) ( )]
1 2
,
3
i Bylinearpair
ii Bylinearpair
By eq i and ii
Similarly
∠ + ∠ = →∠ + ∠ = →∠ + ∠ = ∠ + ∠∠ = ∠
∠ = ∠4
Ans06. Let the measure be 0x
Then its supplement is 0 0180 x−According to question
( )0 0 0
0 0 0
0
0
2 180
360 2
3 360
120
x x
x x
x
x
= −
= −
==
The measure of the angle is 0120 .
66
Ans07. [ ]PQS PQR PRQ PRT Bylinearpair∠ + ∠ = ∠ + ∠,
[ ]
But
PQR PRQ Given
PQS PRT
∠ = ∠∴ ∠ = ∠
Ans08. [ ]AOC ACO and BOD BDO given∠ = ∠ ∠ = ∠,
[ ]
[
But
AOC BOD vertically opposite angles
AOC BOD and
BOD BDO
ACO BDO
AC BD By alternate
∠ = ∠∠ = ∠∠ = ∠
⇒ ∠ = ∠∴ � int ]erior angle property
Ans09. Lines XY and MN intersect at O.
0
0
0
[ ]
,
90
90 (1)
,
180
C XON MOY vertically opposite angle
a POY
But
POY
C a
Also
POX P
∴∠ = ∠ = ∠= ∠ + ∠
∠ =
∴ ∠ = ∠ + →
∠ = − ∠0 0
0
0
180 90
90
90
OY
a b
= −=
∴ + =
0
0
0
,
: 2 : 3 [ ]
2 90
5
36 (2)
(1) (2)
36 9
But
a b Given
a
From and we get
C
=
= ×
= →
∠ = + 0
0
0
126=
67
Ans10. 0180 [ ]QPR Q R Angle sum property of∠ + ∠ + ∠ = ∆0 0 0 0
0 0
180 50 30 100
1
21
100 502
[ ]
QPR
QPT QPR
Q QPS PST Exterior angle theorem
∠ = − − =
∠ = ∠
= × =
∠ + ∠ = ∠0
0
0 0 0
0 0 0
90
90
90 50 40
50 40 10
QPS Q
x QPT QPS
=∠ = − ∠
= − == ∠ − ∠= − =
Ans11. Join BD
( )
[ ]
In ABD
b ABD BDA Exterior angle theorem
In CBD
a CBD BDC
a b CBD BDC ABD BDA
CBD ABD BDC B
∆∠ = ∠ + ∠
∆∠ = ∠ + ∠
∠ + ∠ = ∠ + ∠ + ∠ + ∠= ∠ + ∠ + ∠ + ∠( )
DA
x y
a b x y
= ∠ + ∠∠ + ∠ = ∠ + ∠
Ans12. 1
12
ABC∠ = ∠
( )
0
1 2
21
1 2 (1)2
,
180
and ACB
ABC ACB
But
ABC ACB A
∠ = ∠
∴ ∠ + ∠ = ∠ + ∠ →
∠ + + ∠ =
[ ]
0
0
0
180
1 190 (2)
2 2 (1) (2)
1 1 2 90
2
ABC ACB A
ABC ACB A
From and we get
A
∴ ∠ + = − ∠
∠ + = − ∠ →
∠ + ∠ = − ∠ (3)→
68
( )
0
0
0 0
,
1 2 180 [ ]
180 1 2
1180 90
2
But
BOC Angle of a
BOC
A
∠ + ∠ + ∠ = ∆∠ = − ∠ + ∠
= − − ∠
0 190
2A= + ∠
Ans13. PQ intersect RS at O
0 0
[ ]
4 (1)
,
75 180
QOS POR vertically opposite angles
a b
Also
a b
∴ ∠ = ∠= →
+ + =0 0
0
0
0
[ ]
180 75
105
using (1)
4 105
5 105
POQ is a straight lines
a b
b b
b
or
∴ + = −=
+ =
=
∵
0105 21
5b = =
0
0
4
4 21
84
,
2 180
sin (2)
84 2 180
a b
a
a
Again
QOR and QOS form a linear pair
a c
u g
c
∴ == ×=
∠ ∠
∴ + =
+ = 0
0 0
0
00
0 0 0
2 180 84
2 96
96 48
2, 84 , 21 48
c
c
c
Hence a b and c
= −
=
= =
= = =
69
CBSE TEST PAPER-05
CLASS - IX Mathematics (Lines and angle)
1. In figure the measure of a∠ is
(a) 030 (b) 0150(c) 015 (d) 050
[1]
2. The correct statement is-
(a) A line segment has one end point only.
(b) The ray AB is the same as the ray BA.
(c) Three points are collinear if all of them lie on a line.
(d) Two lines are coincident if they have only one point in common.
[1]
3. One angle is five times its supplement. The angles are-
(a) 0 015 ,75 (b) 0 030 ,150
(c) 0 036 ,144 (d) 0 0160 , 40
[1]
4. In figure if m n� and 1: 2 1: 2. 8 The measure of is∠ ∠ = ∠
(a) 0120
(b) 060
(c) 030
(d) 045
[1]
5. In figure if lines PQ and RS intersect at point T. Such that
040 ,PRT∠ = 0 095 75 ,RPT and TSQ find SQT∠ = ∠ = ∠ .
[2]
70
6. In figure, if 0 0, 40 50 .QT PR TQR and SPR find x and y⊥ ∠ = ∠ = [2]
7. In figure sides QP and RQ of PQR∆ are produced to points S and T respectively
if 0 0135 110 , .SPR and PQT find PRQ∠ = ∠ = ∠
[2]
8. In figure lines PQ and RS intersect each other at point O. If : 5 : 7POR ROQ∠ ∠ = .
Find all the angles.
[2]
9. In figure ray OS stands on a line POQ, ray OR and ray OT are angle bisector of . , .POS and SOQ respectively If POS x find ROT∠ ∠ ∠ = ∠
[3]
10. If a transversal intersects two lines such that the bisectors of a pair of [3]
71
corresponding angles are parallel, then prove that the two lines are parallel.
11. In figure the sides QR of PQR∆ is produced to a point S. If the bisectors of
PQR∠ and PRS∠ meet at point T. Then prove that1
2QRT QPR∠ = ∠
[3]
12. In figure PQ and RS are two mirror placed parallel to each other. An incident ray
AB striker the mirror PQ at B, the reflected ray moves along the path BC and
strikes the mirror RS at C and again reflects back along CD. Prove that AB�CD.
[3]
13. It is given that 064XYZ∠ = and XY is produced to point P. Draw a figure from the
given information. If ray YQ bisects . ZYP Find XYQ and reflex QYP∠ ∠ ∠ .
[5]
72
CBSE TEST PAPER-05
CLASS - IX Mathematics (Lines and angle)
[ANSWERS]
Ans01. (A) Ans02. (C) Ans03. (B) Ans04. (B)
Ans05. Solution,
0
0 0 0
0 0
0
0
0
PRT
P+ R+ 1=180 [ ]
95 40 1 180
1=180 135
1=45
1= 2 [vertically opposite angle]
2= 45
In TQS 2+ Q+ S=180
In
By angle sum property
∆∠ ∠ ∠
+ + ∠ =∠ −
∠∠ ∠∠ ∠∆ ∠ ∠ ∠
0 0 0
0 0
0 0
0
0
45 75 180
Q+120 180
Q=180 120
Q=60
SQT=60
Q+ ∠ + =∠ =
∠ −
∠∠
Ans06. Solution,
0 0
0
0 0 0
TQR
90 40 180 [ Angle sum property of ]
x =50
Now, Y= SPR+X
Y= 30 50 = 80
In
x
∆+ + = ∆
∴∠
∴ +
�
Ans07. Solution, 0 0
0 0
0
110 2 180 [by linear pair]
2= 180 110
2 70
+ ∠ =∠ −
∠ =0 0
0 0
1 135 180
1=180 135
1=45
∠ + =∠ −∠ �
73
0
0 0 0
0 0
0
0
1+ 2+ R=180 [By angle sum property]
45 70 180
R=180 115
R=65
PRQ=65
R
∠ ∠ ∠+ + ∠ =
∠ −
∠∠
Ans08. Solution, 0
0 0
0 0
0
180 [ linear pair of angle]
but POR: ROQ=5:7 [Given]
5 POR= 180 =75
127
Similarly ROQ= 180 =10512
Now POS= ROQ=105 [vertically opposite angle]
And SO
POR ROQ∠ + ∠ =∠ ∠
∴ ∠ ×
∠ ×
∠ ∠∠ 0Q = POR= 75 [vertically app angle]∠
Ans09. Solution,
0
0
0
tan on the line POQ
POS+ SOQ=180
But POS=X
x+ SOQ=180
SOQ=180
RayOS s ds
X
∴ ∠ ∠∠
∴ ∠∠ −
0
0
OR bisects POS,
1 1therefore ROS= =
2 2 21 1
Similarly SOT = = (180 ) 902 2 2
= 90 902 2
Now ray
xPOS x
xSOQ X
x xROT ROS SOT
∠
∠ ×∠ ⇒ × =
∠ ×∠ ⇒ × − = −
∠ = ∠ + ∠ ⇒ + − =
�
�
Ans10. Solution,
Given AD is transversal intersect two lines PQ and RS
To prove PQ�RS
Proof : BE bisects ∠ ABQ
∠ =1
(1)2
ABQ∠ →
Similarity C G bisects ∠ BCS
74
12 (2)
2But BE CG and AD is the transversal
1= 2
1 1 [By (1) and (2)]
2 2ABQ= BCS
PQ RS [ corner ponding
BCS
ABQ BCS
∴∠ = ∠ →
∴ ∠ ∠
∴ ∠ = ∠
⇒ ∠ ∠∴
�
� ∵ angle are equqal]
Ans11. Solution,
PQR
PRS= Q+ P [By Exterior angle thearem]
4+ 3= 2+ 1+ P
2 3=2 1+ P (1)
QT and Rt are bisectors of Q and PRS
In QTR
3= 1+ T (2) [By exterior angle
In ∆∠ ∠ ∠
∠ ∠ ∠ ∠ ∠∠ ∠ ∠ →
∴ ∠ ∠∆
∠ ∠ ∠ → thearem]
BY e.q. (1) and (2) we get
2[ 1+ ] =2 1+ P
2 1+2 T = 2 1+ P
1 T=
21
QTR= Hence proveel 2
T
P
QPR
∠ ∠ ∠ ∠∠ ∠ ∠ ∠
∠ ∠
∠ ∠
Ans12. Solution,
.
1= 2 (1) [Angle of incident ]
Draw MB PQand
MC RS
⊥⊥
∠ ∠ →
0
And 3= 4 (2) [is equal to angle of reflection]
MB Q= NCS=90
MB NC [ By corresponding angle property]
2= 3 (3) [ alternate interior angle]
∠ ∠ →
∠ ∠∴∴ ∠ ∠ →
∵
�
BY e.q. (1), (2) And (3)
1= 4∠ ∠
75
1+ 2= 4+ 3
ABC= BCD
AB CD [By alternate interior angles]
∠ ∠ ∠ ∠⇒ ∠ ∠
∴ �
>>
Ans13. Solution,
0 0
0 0
0 0
0
sec
1= 2
1+ 2+ 64 180 [YX is a line]
1+ 1+64 180
2 1=180 64
2 1=116
YQ bi cts ZYP∴ ∠∴ ∠ ∠
∠ ∠ ∠ =∠ ∠ =
∠ −∠
0
0 0 0
0 0
0 0
0 0
0
0
0 0
1 58
XYQ=64 58 122
2+ XYQ =180 1 2 58
2+122 180
2=150 122
2 58
Re YP=360
=360 58
QYP
QYP
flex Q QYP
∠ =∴ ∠ + =
∠ ∠ ∠ = ∠ = ∠ =∠ =
∠ −∠ = ∠ =
∠ − ∠−
0 = 302
76
CBSE TEST PAPER-01
CLASS - IX Mathematics (Congruent Triangles)
1. In fig 1.1, if AD =BC and ∠ BAD = ∠ ABC, then ∠ ACB is equal to
(A) ∠ ABD
(B) ∠ BAD
(C) ∠ BAC
(D) ∠ BDA
[1]
2. IN fig 1.2, if ABCD is a quadrilateral in which AD= CB,AB=CD, and ∠ D= ∠ B, then
∠ CAB is equal to
(A) ∠ ACD
(B) ∠ CAD
(C) ∠ ACD
(D) ∠ BAD
[1]
3. If o is the mid – point of AB and ∠ BQO = ∠ APO, then ∠ OAP is equal to
(A) ∠ QPA (B) ∠ OQB
(C) ∠ QBO (D) ∠ BOQ
[1]
4. IF AB ⊥ BC and CB = ∠ c, then the true statement is
(A) AB ≠ AC (B) AB=BC
(C) AB=AD (D) AB=AC
[1]
5. In quadrilateral ACBD, AC=AD and bisects ∠ A. show ∆ ABC ≅ ∆ ABD? [2]
6. If DA and CB are equal perpendiculars to a line segment AB. Show that CD
bisects AB.
[2]
7. L and M, two parallel lines, are intersected by Another pair of parallel lines P
and C. show that ∆ ABC ≅ ∆ CDA.
[2]
8. In fig 1.3, the bisector AD of ∆ ABC is ⊥ to the
opposite side BC at D. show that ∆ ABC is isosceles?
[2]
77
9. If ∆ ABC, the bisector of ∠ ABC and ∠ BCA intersect each other at the point o
prove that ∠ BOC =1
90 .2
A+ ∠�
[3]
10. Prove that is one angle of a triangle is equal to the sum of the other two angles,
the triangle is right angled:
[3]
11. IF fig 1.4, if PQ ⊥ PS, PQ ∏ SR, ∠ SQR = 28� and ∠ QRT = 65� , then find the values
of X and Y.
[3]
12. If in fig 1.5, AD= AE and D and E are point on BC such that BD=EC prove AB=AC. [3]
13. Prove that sum of the quadrilateral is 3600? [5]
78
CBSE TEST PAPER-01
CLASS - IX Mathematics (Statistics)
[ANSWERS]
Ans01. (D)
Ans02. (C)
Ans03. (C)
Ans04. (D)
Ans05. IN ∆ ACB and ∆ ADB,
AC=AD……… (Given)
∠ BAC= ∠ BAD……. (AB bisects ∠ A)
And AB= AB ………….. (Common)
∴∆ ABC ≅ ∆ ABD ……….. (SAS axiom)
Ans06. In ∆ AOD and ∆ BOC,
AD=BC ……….. (Given)
A= B......... (Each=90 )
AOD BOC...... (vert opp. LS)
AOD= BOC....... (AAS rule)
OA = OB ....... (C.P.C.T)
Hence CD bisects AB.
and
∠ ∠∠ ≅ ∠
∴ ∆ ∆∴
�
Ans07. and AC cuts them - (Given) L M∏ ACB= CAD......... (ALT.LS)
P Q and AC cuts them ...... (Given)
CAD= ACD...... (ALT. LS)
AC=CA .......... (common)
ABC CDA ...... (ASA rule)
∴ ∠ ∠∏
∴ ∠ ∠
∴ ∆ ≅ ∆
Ans08. In ∆ ABD and ∆ ACD
1 = 2...... (AD is the bisector of A)
And ADB= ADC=90 ..........( )
........( )
ACD ...... (ASA rule)
AB=AC ........ (C.P.C.T)
Hence ABC is isosceles.
AD BC
AD AD common
ABD
∠ ∠ ∠∠ ∠ ⊥
∴ =∆ ≅ ∆∴
∆
�
79
Ans09. BOC, we haveIn ∆
1+ 2+ BOC= 180 (1)
In ,
A+ B+ C=180
2( 1) 2( 2) 180
1 2 902
90 1+ 2=
2 substituting this value of 1+
ABC we have
A
A
A
∠ ∠ ∠ →∆
∠ ∠ ∠
⇒ ∠ + ∠ + ∠ =∠
⇒ + ∠ + ∠ =
− ∠⇒ ∠ ∠
∠ ∠
�
�
�
�
2 in (1)
90 + BOC =1802
BOC =902
So
BOC= 902
A
A
A
∠− ∠
∠∠ +
∠∠ +
� �
�
�
Ans10. =180 [ sum of three angles of o is 180 ] (1)A B C∠ + ∠ + ∠ ∆ →� �
: A+ C= B (2)
from (1) and (2)
B+ B=180
18090
2
Given that
B
∠ ∠ ∠ →
∠ ∠
⇒ ∠ = =
�
�
�
Hence ∆ ABC is right angled.
Ans11. and QR is the transversal,PQ SR∏ PQR= QRT [pair of alternate angles]
or PQS+ SQR = QRT
or x+28 =65
65 28 =37
in PQS,
SPQ+ PSQ+ PQS=180
90 180
or 90 37 =180
x
Also
Y X
Y
∴ ∠ ∠∠ ∠ ∠
∴ = −∆
∠ ∠ ∠⇒ + + =
+ +
� �
� � �
�
� �
� � �
Y=53 �
80
Ans12. ,In ADE∆ AD=AE [Given]
ADE = AED [ angles oppasrte to equal side are equal]
Now, ADE+ ADB=180 [liner pair]
Also, AED+ AEC=180 [" " ]
ADE+ ADB= AED+ AEC
But ADE= AED
N
∴ ∠ ∠∠ ∠∠ ∠
⇒ ∠ ∠ ∠ ∠∠ ∠
�
�
ow in ABD and ACE,
BD= CE
AD=AE
ADB= AEC
ABC ACE [BY SAS]
AB=AC
∆ ∆
∠ ∠∴ ∆ ≅ ∆⇒ [CPCT]
Ans13. and D to obtain two triangles ABD BCD.Join B ∆
BAD+ ABD+ BDA=180 [ is 180 ] (1)
And CBD+ BCD+ CDB=180 [ is 180 ] (2)
Adding (1) and (2),
BAD+ ABD+ BDA+ CBD + BGCD+ BCD+ CDB=360
∠ ∠ ∠ ∆ →
∠ ∠ ∠ ∆ →
∠ ∠ ∠ ∠ ∠ ∠ ∠
� �
� �
�
sum of three angles of
sum of three angles of
or BAD + ( ABD+ CBD)+ BCD+( CDB+ BDA)=360
or BAD+ ABC+ BCD+ CDA=360
i.e. A+ B+ C+ D=360
So,
Sum of quadrilateral is 360 Hence proved.
∠ ∠ ∠ ∠ ∠ ∠∠ ∠ ∠ ∠
∠ ∠ ∠ ∠
�
�
�
�
81
CBSE TEST PAPER-02
CLASS - IX Mathematics (Congruent triangle)
1. In ∆ ABC is an isosceles triangle and ∠ B = 065 , find x.
(a) 060 (b) 070(c) 050 (d) none of these
[1]
2. If AB=AC and ∠ ACD= 0120 , find ∠ A
(a) 050 (b) 060
(c) 070 (d) none of these
[1]
3. What is the sum of the quadrilateral:-
(a) 0260 (b) 0360
(c) 0180 (d) 090
[1]
4. The sum of the triangle will be:-
(a) 0360 (b) 0270
(c) 0180 (d) 090
[1]
5. If AE=AD and BD=CE. Prove that ∆ AEB ≅ ∆ ADC [2]
6. In quadrilateral ABCD, AC=AD and AB bisects ∠ A. show that ∆ ABC ≅ ∆ ABD.
What can you say about BC and BD?
[2]
7. In ∆ ABC, the median AD is ⊥ to BC. Prove that ∆ ABC is an isosceles triangle. [2]
8. Prove that ∆ ABC is isosceles if altitude AD bisects ∠ BAC. [2]
9. In the given figure, AC=BC, ∠ DCA= ∠ ECB and ∠ DBC= ∠ EAC. Prove that ∆ DBC
and ∆ EAC are congruent and hence DC=EC.
[3]
82
10. From the following figure 2.1, prove that ∠ BAD=3 ∠ ADB. [3]
11. O is the mid-point of AB and CD. Prove that AC=BD and AC//BD. [3]
12. ABCD is a quadrilateral in which AD=BC and ∠ DAB= ∠ CBA. Prove that.
(a) ∆ ABD ≅ ∆ BAC
(b) BA=AC
(c) ∠ ABD= ∠ BAC
[3]
13. ∆ ABC is an isosceles triangle with AB=AC. AD bisects the exterior ∠ A. prove
that AD//BC.
[5]
83
CBSE TEST PAPER-02
CLASS - IX Mathematics (Congruent triangle)
[ANSWERS]
Ans01. (c)
Ans02. (b)
Ans03. (b)
Ans04. (c)
Ans05. We have,
AE=AD and CE=BD
⇒ AE+CE=AD+BD
⇒AC=AB (i)
Now, in ∆ AEB and ∆ ADC,
AE=AD [given]
∠ EAB= ∠ DAC [common]
AB=AC [from (i)]
∆ AEB ≅ ∆ ADC [by SAS]
Ans06. In ∆ ABC and ABD,
AC=AD [given]
∠ CAB= ∠ DAB [AB bisects ∠ A]
AB=AB [common]
∆ ABC ≅ ∆ ABD [SAS criterion]
∴BC=BD [CPCT]
Ans07. ,In s ABD and ACD∆
0
[ int ]
[ ]
[ 90 , ]
[ ]
[ ]
BD CD D is mid po of BC
AD AD common
ADB ADC Each AD BC
ABD ACD By SAS
AB ACE CPCT
= −=
∠ = ∠ ⊥∆ ≅ ∆∴ =
∵
84
Ans08. ,In s ABD and ACD∆0 [ 90 , ]
[ sec ]
[ ]
[ ]
[ ]
, .
ADB ADC Each AD BC
BAD CAD AD bi ts BAC
AD AD common
ABD ACB ByAAS
AB AC CPCT
Thus ABC is an isosceles triangle
∠ = ∠ ⊥∠ = ∠ ∠
=∆ ≅ ∆⇒ =
∆
Ans09. ,We have
[ ]
[ ]
( )
DCA ECB given
DCA ECD ECB ECDE adding ECD on both sides
ECA DCA i
∠ = ∠⇒ ∠ + ∠ = ∠ + ∠ ∠⇒ ∠ = ∠ →
,
[ ( )]
[ ]
Now in s DBC and EAC
DCB ECA from i
BC AC given
∆∠ = ∠
= [ ]
[ ]
[CPCT]
DBC EAC given
DBC EAC ByASA
DC EC
∠ = ∠∆ ≅ ∆
⇒ =
Ans10. Let ADC Q∠ = [ , ]
2
2 [ ]
CAD Q CA CD
Exterior ACB CAD Q Q
Q
BAC Q BA BC
⇒ ∠ = =∠ = ∠ = +
=⇒ ∠ = =
∵
∵
2
3 3 3
Hence BAD BAC CAD
Q Q
Q ADC ADB
∠ = ∠ + ∠= += = ∠ = ∠
Ans11. In s AOC and BOD∆ [ int ]
[ ]
[ int ]
[ ]
[ ]
[
AO OB O is the mid po of AB
AOC BOD vertically opposite angles
CO OD O is the mid po of CD
AOC BOD BySAS
AC BD CPCT
CAO DBO CPC
= −∠ = ∠
= −∆ ≅ ∆
=⇒ ∠ = ∠ ]
, s in sec
. . .
T
Now AC and BD are two line ter ted by a transversal AB
such that CAO DBOi e alternate angle are equal∠ = ∠
85
Ans12. ,In s ABD and BAC∆ [ ]
[ ]
AD BC given
DAB CBA given
=∠ = ∠
[ ]
( ) [ ]
( )
AB AB common
i ABD BAC SAScriterion
ii
=∆ ≅ ∆
⇒∴∵
[ ]
( ) [ ]
BD AC CPCT
iii Also ABD BAC CPCT
=⇒ ∠ = ∠
Ans13. Since AD bisects the exterior A,
EAD=1
2EAC∠
= 0 01 1180 1 90 1 ( )
2 2i − ∠ = − ∠ →
01 180 ( )EAC Linear pair ∴∠ + ∠ =
But
0
0
0
0
1 2 3 180
1 2 2 180 [ ]
2 2 180 1
1 2 90 1 (i)
2 ( )
AB AC
Hence from i and
∠ + ∠ + ∠ =⇒ ∠ + ∠ + ∠ = ∴ =⇒ ∠ = − ∠
∠ = − ∠
( )
2
/ /
ii
EAD ABC
But these are corresponding angles
AD BC
∠ = ∠ = ∠
∴
86
CBSE TEST PAPER-03
CLASS - IX Mathematics (Congruent triangle)
1. An angle is 14� more than its complement. Find its measure.
(A) 42 (B) 32
(C) 52 (D) 62
[1]
2. An angle is 4 time its complement. Find measure.
(A) 62 (B) 72
(C) 52 (D) 42
[1]
3. Find the measure of angles which is equal to its supplementary.
(A) 120� (B) 60�
(C) 45� (D) 90�
[1]
4. Which of the following pairs of angle are supplementary?
(A) 30 ,120� � (B) 45 ,135� �
(C) 120 ,30� � (D) None of these.
[1]
5. ABC is An isosceles triangle in which altitudes BE and CF are drawn to side AC
and AB respectively. Show that these altitudes are equals.
[2]
6. It AC= AE, AB=AD and BAD∠ = .EAC∠ show that BC =DE. [2]
7. Line ∠ is the bisector of an angle ∠ A and B is any point on ∠ BP and BQ are
⊥ from B to the arms of ∠ A show that :
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is A equidistant from the arms of A∠
[2]
8. In the given figure, ∆ ABC is an isosceles triangle
and ∠ B = 65� , find x.
[2]
87
9. AB is a line – segment- AX and By are equal two equal line – segments drawn on
opposite side of line AB such that AX ∏ BY. It AB and XY intersect each other at
P. prove that
(i) ∆ APX ≅ ∆ BPY,
(ii) AB and XY bisect each other at P.
[3]
10. In an isosceles ABC, with AB =AC, the bisector of ∠ B and ∠ C intersect each
other at o, join A to o. show that:
(i) OB=OC
(ii) AO bisects ∠ A.
[3]
11. Two side AB and BC and median AM of one triangle ABC are respectively equal
to side PQ and QR and median PN of ∆ PQR, show-
(I) ABM PQN∆ ≅ ∆ (ii) PQRABC∆ ≅ ∆
[3]
12. In the given figure, ABC and DBC are two triangle on the same base BC such that
AB=AC and DB=DC. Prove that ∠ ABD = ∠ ACD,
[3]
13. ∆ ABC is an isosceles triangle in which AB=AC side BA is produced to d such that
AD=AB. Show that ∠ BCD is a right angle.
[5]
88
CBSE TEST PAPER-03
CLASS - IX Mathematics (Congruent triangle)
[ANSWERS]
Ans01. (C)
Ans02. (B)
Ans03. (D)
Ans04. (B)
Ans05. ,In s ABE andACF∆
0
[ ]
90
[ ]
[ ]
[ ]
A A common
AEB AFC
AB AC given
ABE ACF AAS rule
BE CF CPCT
∠ = ∠∠ = ∠ =
=∴ ∆ ≅ ∆⇒ =
Ans06. ,In s BAC and DAE∆ [ ]
[ ]
, [ ]
AB AD given
AC AE given
Also BAD EAC given
BAC DAC EAC CAD
==
∠ = ∠∴ ∠ + ∠ = ∠ + ∠
[ ]
[ ]
BAC EAD
BAC DAE SAS criterior
BC DE CPCT
⇒ ∠ = ∠∴ ∆ ≅ ∆⇒ =
Ans07. ,In s ABP and ABQ∆
0
[ ]
90 [ ]
[ ]
( ) [ ]
( ) [ ]
BAP BAQ given
APB AQB common
AB AB common
i ABP ABQ AAS rule
ii BP BQ CPCT
∠ = ∠∠ = ∠ =
=∴ ∆ ≅ ∆
=
89
Ans08. ,In s ABC∆
0
[ ]
65
AB AC
B C Anglesoppositetoequalsidesareequal
But B
=⇒ ∠ = ∠
∠ =
0 65B C∴ ∠ = ∠ =
0
0
0
,
180
130 180
50
So
A B C
x
x
∠ + ∠ + ∠ =+ =
=
Ans09. ,In s APX and BPY∆1 2 [ ]
3 4 [ ]
[ ]
[ ]
[ ]
sec .
Alternate angle
Vertically opposite angle
AX BY given
APX BPY ByAAS
AP BP and PX PY CPCT
AB and XY bi ts each other at P
∠ = ∠∠ = ∠
=∴∆ ≅ ∆⇒ = =⇒
Ans10. ( ) In ABC,i ∆AB=AC [given]
ACB= ABC [angles oppaside to equal side]
1 1 ACB= ABC
2 2 or OCB = OBC
OB=OC [ side oppasite to equal angle]
∠ ∠
∴ ∠ ∠
∠ ∠⇒
( ) In AOC
AB = AC [ given]
ABO= ACO [ Halves of equals]
OB=OC [ proved]
AOB AOC [SAS rule]
BA
ii AOB and∆ ∆
∠ ∠
∴ ∆ ≅ ∆⇒ ∠ O= CAO [CPCT]
i.e. AO bisecls A
∠∠
90
Ans11. ( ) In s ABMand PQN,i ∆AB=PQ [ Given]
BM=QN [ Halves of equal]
AP=PN [ Given]
ABM PQN [SSS rules]∴ ∆ ⇒ ∆
( ) B= Q
Now, in s ABC and PQR,
AB=PQ [ Given]
BC=QR [Given]
B= Q [Proved]
ABC PQR [SAS ruls]
ii ⇒ ∠ ∠∆
∠ ∠∴ ∆ ≅ ∆
Ans12. ,In ABC∆AB=AC [ Given]
ABC= ACB [ angles oppaosite to equal side are equals]
Similarly in DBC, DB=DC [Given] (1)
DBC= DCB (2)
Adding (1) and (2) :
ABC+ DBC= ACB+ D
∴ ∠ ∠∆ →
∴ ∠ ∠ →
∠ ∠ ∠ ∠ CB
or ABD= ACD ∠ ∠
Ans13. [ Angles oposite to equal side]ABC ACB∠ = ∠Also, ACD= ADC [ " " " " " " " " ]
Now, BAC+ CAD=180 [ Linear Pair]
Also, CAD= ABC+ ACB [ exterior angles of ABC]
= 2 ACB
And BAC= ACD+ ADE
∠ ∠∠ ∠∠ ∠ ∠ ∆
∠∠ ∠ ∠
�
[Exterior amgles of ADC]
= 2 ACD
BAC+ CAD=2 ( ACD+ )
=2 BCD
i.e. 2 BCD=180
or BCD =90
ACB
∆∠
∴ ∠ ∠ ∠ ∠∠
∠∠
�
�
91
CBSE TEST PAPER-04
CLASS - IX Mathematics (Congruent triangle)
1. Find the measure of each exterior angle of an equilateral triangle.
(A) 110� (B) 100� (C) 120� (D) 150�
[1]
2. In an isosceles ∆ ABC, is AB=AC and 90A∠ = � , Find ∠ B.
(A) 70� (B) 80�
(C) 95� (D) 60�
[1]
3. In an ∆ ABC, is ∠ B= ∠ C= 45� , Which is the longest side.
(A) BC (B) AC
(C) CA (D) None of these.
[1]
4. In an ∆ ABC, is AB=AC and ∠ = 70� , Find ∠ A.
(A) 40� (B) 50�
(C) 45� (D) 60�
[1]
5. If ∠ E> ∠ A and ∠ C> ∠ D. prove that AD>EC. [2]
6. If PQ= PR and S is any point on side PR. Prove that RS<QS. [2]
7. Prove that MN+NO +OP+>2MO. [2]
8. Prove that MN+NO+OP>PM. [2]
9. Prove that the Angle opposite of the greatest side of a triangle is greater that [3]
92
two- third of a right angle i.e. great her than 60�
10. AD is the bisector of ∠ A of ∆ ABC, where D lies on BC. Prove that AB>BD and
AC>CD.
[3]
11. In the given figure, AB and CD are respectively the smallest and the largest side
of a quadrilateral ABCD.
Prove that ∠ A> ∠ C and ∠ B> ∠ D.
[3]
12. It the bisector of a vertical angle of a triangle also bisects the opposite side;
prove that the triangle is an isosceles triangle.
[3]
13. In the given figure, ∠ A= ∠ C and AB =BC.
Prove that ∆ ABD ≅ ∆ CBE.
[5]
93
CBSE TEST PAPER-04
CLASS - IX Mathematics (Congruent triangle)
[ANSWERS]
Ans01. (c) Ans02. (a) Ans03. (a) Ans04. (a)
Ans05. ,In ABE∆ [ ]
[ arg ] ( )
,
[ ]
( )
( ) ( )
E A Gven
AB EB Side opposite to greater angle is l er i
Similarly in BCD
C D Given
BD BC ii
Adding i and ii
AB BD EB BC
Or AD EC
∠ > ∠⇒ > →
∆∠ > ∠⇒ ∠ > ∠ →
+ > +>
Ans06. ,In PQR∆
[ ]
PQ=PR [ ]
[ ]
Now, QR< PQR
RS <QS [Side opposite to smaller angle in SRQ]
Given
PRQ PQR Angle opposite to equal side are equal
S SQR is a part of PQR
SQR PRQ OR SRQ
⇒ ∠ = ∠∠ ∠ ∠ ∠
∴∠ < ∠ ∠⇒ ∆
Ans07. ,In MON∆ [ ] (i)
Similarly in MPQ,
OP + PM > MO (ii)
Hence from (i) and (ii)
MN + NO + OP + PM > 2MO
MN NO MO Sum of any two side of is greater than third sides+ > ∆ →∆
→
Ans08. ,In MON∆ [ ] (i)
Similarly in MOQ,
MO + OP > PM (ii)
Adding (i) and (ii)
Or MN + NO + OP + MO > MO + PM
MN NO MO Sum of any two side of is greater than third sides+ > ∆ →∆
→
Ans09. ABC,In ∆AB > BC [Given]
94
0
[Angle opposite to large side is greater] (i)
,
AB > AC
C > B (ii)
Adding (i) and (ii)
2 > ( )
both sides,
3 C > ( A + B + C)
3 C > 180 [S
C A
Similarly
C A B
Adding C to
∠ > ∠ →
∴ ∠ ∠ →
∠ ∠ + ∠∠
∠ ∠ ∠ ∠
∠ 0
0
um of three angles of is 180 ]
, C > 60Or
∆∠
Ans10. ADC,In ∆3 > 2 [Exterior angles of is greater than each of the interior opposite angles]∠ ∠ ∆
2 = 1 [AD bisects A]
3 = 1
AB > BD [ Side opposite to greater angle is larger]
In ABD,
4 > 1 [Exterior opposite of is greater than each of interior opposite angle]
But,
1 2
4 >
But
∠ ∠ ∠∴ ∠ ∠⇒
∆∠ ∠ ∆
∠ = ∠∴ ∠ 2
AC > CD
[Side opposite to greater angle is larger].
∠⇒
Ans11. Join AC.
ABC,
BC > AB [AB is the smallest sides of quadrilateral ABCD]
1 > 3 [ Angle opposite to larger side is greater] (i)
In ADC,
In ∆
⇒ ∠ ∠ →∆
CD > AD [CP is the largest side of quadrilateral ABCD]
2 > 4 [Angle opposite to larger side is greater] (ii)
Adding (i) and (ii)
1 +
∠ ∠ →
∠ 2 > 3 + 4 Or A > C
Similarly, by joining BD, we can show that B > D
∠ ∠ ∠ ∠ ∠∠ ∠
95
Ans12. ABC and EDB,In s∆DC = DB [Given]
AD = ED [By construction]
ADC = EDB [Vertically opposite angle]
ADC [By SAS]
AC = EB and
[CPCT]
But DAC = BAD [ AD bisects ]
BAD =
EDB
DAC DEB
A
D
∠ ∠∴∆ ≅ ∠⇒
∠ = ∠∠ ∠ ∠
∴ ∠ ∠∵
AB = BE
But BE = AC [Proved above]
AB = AC
EB
⇒
∴
Ans13. s AOE and COD,In ∆
0 0
A = C [Given]
[Vertically opposite angle]
A +
A + 180 180
AOE COD
AOE C COD
AOE AEAEO CDO
∠ ∠∠ = ∠
∴ ∠ ∠ = ∠ + ∠
∠ ∠ + ∠⇒ − ∠ = − ∠
∵0
0
0
180 and
C + 180
(i)
Now,
180 [Linear pair]
And
O
COD CDO
AEO CDO
AEO OEB
CDO
=
∠ ∠ + ∠ =
⇒ ∠ = ∠ →
∠ + ∠ =
∠ + ∠ 0180 [Linear pair]
[Using (i)]
CEB = ADB
ODB
AEO OEB CDO ODB
OEB ODB
=⇒ ∠ + ∠ = ∠ + ∠⇒ ∠ = ∠⇒ ∠ ∠ (ii)
Now, in ABD and CBE,
A = C [Given]
ADB = CEB [From (ii)]
s
→∆
∠ ∠∠ ∠
AB = CB [Given]
ABD CBE [By AAS]∆ ≅ ∆
96
CBSE TEST PAPER-05
CLASS - IX Mathematics (Congruent triangle)
1. 0 0 a ABC, If A = 45 and B = 70 . Determine the shorteat sides of the triangles.In ∆ ∠ ∠
(a) AC (b) BC
(c) CA (d) none of these
[1]
2. In an ∆ ABC, if A∠ = 045 and B∠ = 070 , determine the longest sides of the
triangle.
(a) AC (b) CA
(c) BC (d) none of these
[1]
3. The sum of two angles of a triangle is equal to its third angle. Find the third
angles.
(a) 090 (b) 045
(c) 060 (d) 070
[1]
4. Two angles of triangles are 0 065 45and respectively. Find third angles.
(a) 090 (b) 045
(c) 060 (d) 070
[1]
5. ∆ ABC is an isosceles triangle and B∠ = 045 , find A∠ . [1]
6. ∆ ABC is an equilateral triangle and B∠ = 060 , find C∠ . [2]
7. In the figure 1.1, AB = AC and 0120 , ACD find B∠ = ∠ . [2]
97
8. In the given figure 1.1, find A∠ [2]
9. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that .B C∠ = ∠ [3]
10. AD is an altitude of an isosceles triangle ABC in which AB = AC that:
(i) AD bisects BC (ii) AD bisects A∠
[3]
11. In the given figure, PQ>PR, QS and RS are the bisectors of the s∠ Q and R
respectively. Prove that SQ>SR.
[3]
12. Prove that in a right triangle, hypotenuse is the longest (or largest) side. [3]
13. In the given figure, PR>PQ and PS is the bisector of .QPR∠ Prove
that PSR PSQ∠ > ∠ .
[5]
98
CBSE TEST PAPER-05
CLASS - IX Mathematics (Congruent triangle)
[ANSWERS]
Ans01. (b)
Ans02. (c)
Ans03. (a)
Ans04. (d)
Ans05. ABC,In ∆
0
0
0 0
AB = AC
B = C [Angle opposite to equal sides ae equal]
But B = 45
A + B + C = 180
A + 90 180
C
and
⇒ ∠ ∠∠ = ∠
∠ ∠ ∠∠ =
0 A = 90∠
Ans06. ABC,In ∆
0
0
AB = AC
B = C [Angles opposite to equal sides are equal]
But B = 60
, C = 60So
⇒ ∠ ∠∠∠
Ans07. in ABC, AB = ACSince ∆
0
B = C [Angles opposite to equal are equal]
Also, ACB + ACD = 180 [Linear pair]
⇒ ∠ ∠∠ ∠
0 0
0
ACB = 180 120
, C = B = 60and
⇒ ∠ −∠ ∠
Ans08. ABC,In ∆0 0
0 0 0
0 0
0
A + B + C = 180 [Sum of three angles of a is 180 ]
A + 60 60 180
A = 180 120
A = 60
∠ ∠ ∠ ∆∠ + =
⇒ ∠ −
∠
99
Ans09. APB and APC,In right s∆ AP = AP [Common]
Hypotenuse AB = Hypotenuse AC [Given]
APB APC [RHS rule]
B = C [CPCT]
∴ ∆ ≅ ∆⇒ ∠ ∠
Ans10. (i) In right triangle ABD and ACD,
Side AD = Side AD [common]
Hypotenuse AB = Hypotenuse AC [Given]
ABD ACD [By RHS]
BD = CD [CPCT]
Also, AD bisects BC
(ii) Also, BAD = CAD [CPCT]
i.e. AD bisects .A
∴ ∆ ≅ ∆⇒
∠ ∠∠
Ans11. Since PQ>PR
∴
> Q [Angle opposite to larger side is garden]
1 1
2 2SRQ > SQR
SQ > SR [side opposite to greater angle is larger]
R
R Q
∠ ∠
⇒ ∠ > ∠
⇒ ∠ ∠⇒
Ans12. Given a right angled triangle ABC in which 090B∠ = AC is its hypotenuse.∴
0
0
0 0 0
Now, since
B = 90
A + B + C = 180
A + C = 180 90 = 90
. . B = A + C
B > A and B > C
Hence, the side opposite to B is the hypotenuse is the longest side.
i e
∠∴ ∠ ∠ ∠
∠ ∠ −∠ ∠ ∠
⇒ ∠ ∠ ∠ ∠∠
100
Ans13.
PQR,
PR > PQ [Given]
3 > 4 [Angle opposite to larger side] (i)
Also, 6 > 1 + 3 [Exterior angle theorem] (ii)
Similarly,
In ∆
⇒ ∠ ∠ →∠ ∠ ∠ →∠5 = 2 + 4
But 2 = 1 [PS bisects QPR]
5 = 1 + 4 (iii)
(iii) from (ii)
6 - 5 = ( 1 + 3)
Subtracting
∠ ∠∠ ∠ ∠
∴ ∠ ∠ ∠ →
∠ ∠ ∠ ∠ - ( 1 + 4)
or, 6 - 5 = 3 - 4 (iv)
Now,
3 > 4
3 - 4 = 0
∠ ∠∠ ∠ ∠ ∠ →
∠ ∠⇒ ∠ ∠ (v)
From (iv) and (v)
6 - 5 > 0
6 > 5
or PSR > PSQ
→
∠ ∠∠ ∠
∠ ∠
101
CBSE TEST PAPER-01
CLASS - X Mathematics (Co-ordinate geometry)
1. The point of intersection of X and Y axes is called
(a) zero point (b) origin
(c) null point (d) none of these
(1)
2. The distance of the point (-3, -2) from x-axis is
(a) 2units (b) 3units
(c) 5units (d) 13 units
(1)
3. The distance of the point (-6, -2) from y-axis is
(a) 6units (b) 38 units
(c) 2units (d) 8units
(1)
4. The abscissa and ordinate of the point with Co-ordinates (8, 12) is
(A) abscissa 12 and ordinate (B) abscissa 8 and ordinate 12
(C) abscissa 0 and ordinate 20 (D) none of these
(1)
5. Write the name of each part of the plane formed by Vertical and horizontal
lines.
(2)
6. Write the Co-ordinates of a point which lies on the x-axis and is at a distance of
4units to the right of origin. Draw its graph.
(2)
7. Write the mirror image of the point (2, 3) and (-4, -6) with respect to x-axis. (2)
8. Write the Co-ordinates of a point which lies on y-axis and is at a distance of
3units above x-axis. Represent on the graph.
(2)
9. Locate the points (5, 0), (0, 5), (2, 5), (5, 2), (-3, 5), (-3, -5) and (6, 1) in the
Cartesian plane.
(3)
10. Take a triangle ABC with A (3, 0), B (-2, 1), C (2, 1). Find its mirror image. (3)
102
11.
In fig. write the Co-ordinates of the points and if
we join the points write the name of fig. formed.
Also write Co-ordinate of point of AC and BD.
(3)
12. In which quadrant or on which axis do each of the points (-2, 4), (2, -1), (-1, 0),
(1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian
plane.
(3)
13. See fig. and write the following
(i) The Co-ordinates of B
(ii) The Co-ordinates of C
(iii) On which axes point L lies.
(iv) The abscissa of the point D
(v) The Co-ordinates of point L
(vi) In which axes point M lies.
(vii) The ordinate of the point H
(Viii) The Co-ordinates of the point M
(ix) The point identified by the Co-ordinate (2, -4)
(x) The point identify by the Co-ordinates (-3, -5)
(5)
.
.
.
.
103
CBSE TEST PAPER-01
CLASS - X Mathematics (Co-ordinate geometry)
[ANSWERS]
Ans01. (b)
Ans02. (a)
Ans03. (a)
Ans04. (b)
Ans05. (i) Vertical line is called Y-axes.
(ii) Horizental line is called X-axes.
Ans06. (4, 0)
.
Ans07. (i) The mirror image of point (2, 3) is (2, -3) with respect to x-axis.
(ii) The mirror image of (-4, -6) is (-4,-6) with respect to x-axis.
Ans08. The Co-ordinates of the point which lies on y-axis and at a distance of 3units
above x-axis is (0, 3).
104
Ans09.
.
. . .. .
.
Ans10. Mirror image of A (3, 0), B (-2, 1) and (2, 1) are A’ (-3, 0), B’ (-2, -1), C’ (2, -1)
respectively.
. .
Ans11. (i) The Co-ordinate of point A is (0, 2), B is (2, 0), C is (-2, 0) and D is (-2, 0).
(ii) It we joined them we get square.
(iii) Co-ordinate of point of AC and BD is (0, 0).
105
Ans12. (-2, 4) lies in II quadrant.
(2, -1) lies in IV quadrant.
(-1, 0) lies on –ve x-axis.
(1, 2) lies in I quadrant.
(-3, -5) lies in III quadrant.
..
..
.
Ans13. (i) (-5, 2)
(ii) (5, -5)
(iii) Y-axis.
(iv) The abscissa of point D is 6.
(v) (0, 5)
(vi) Point M lies on X-axis.
(vii) The ordinate of point H is -3
(viii) (-3, 0)
(ix) G
(x) E
106
CBSE TEST PAPER-02
CLASS - X Mathematics (Co-ordinate Geometry)
The co-ordinate of origin in
(A) (X, o) (B) (o, y)
(C) (o, o) (D) none of thence.
[1]
2. The distance of the point (2,3) from y axis’s
(A) 2 units (B) 3 units
(C) 5 units (D) 13 units
[1]
3. The point (-2,-1) lien in
(A) 1st quadrant (B) 2nd quadrant
(C) 3rd quadrant (D) 4th quadrant
[1]
4. The point (3,0) lies in
(A) +ve x axis (B) – ve x axis
(C) + ve y axis (D) –ve y axis
[1]
5. Write abscissa and ordinate of point (-3,-4) [2]
6. State the quadrant in which each of the following points lie:
(i) (2,1) (ii) (-7,11)
(iii) (-6,-4) (iv) (-5,-5)
[2]
7. Which of the following points belongs to 2nd quadrant
(i) (2,3) (ii) (-3,2)
(iii) (2,0) (iv) (-4,2)
[2]
8. What is the name of horizontal and vertical lines drawn to determine the
position of any point in the Cartesian plane?
[2]
9. In fig of vertices find co-ordinates of ABC△ [3]
107
10. Take a quadrilateral ABCD
(A) (-5,-4) (B) (-5,2) (C) (-3,3) and (D) (-3,4) find its mirror image
with respect to y- axis.
[3]
11. Locate the points
(A) (-3, 4) (B) (3, 4) and (C) (0, 0) in a Cartesian plane write the name of figure
which is fumed by joining them.
[3]
12. Plot the following ordered pairs of number (x, y) as points in the Cartesian
plane.
[3]
13. Find some ordered pairs of the linear equation 2x+y=4 and plot them ‘how many
such ordered pairs can be found and plotted?
[5]
108
CBSE TEST PAPER-02
CLASS - X Mathematics (Co-ordinate Geometry)
[ANSWERS]
Ans01. (C) (0, 0)
Ans02. (A) 2 units
Ans03. (C) 3rd quadrant
Ans04. (A) +ve x- axis
Ans05. Abscissa -3 ordinate -4
Ans06. (2, 1) (I) Quadrant
(-7, 11) (ii) Quadrant
(-6,-4) (iii) Quadrant
(-5,-5) (iii) Quadrant
Ans07. The point (-3, 2), (-4, 2) belongs to (ii) quadrant.
Ans08. The name of horizontal line is x –axis
The name of vertical line is y – axis
Ans09. (A) (0, 0) (B) (1, 3) (c) (-2, 3)
Ans10. The mirror image of point.
(A) (-5, 4) (B) (-5, 2) (C) (-3, 3) and (D) (-3, 4) with respect to y-axis are.
A’ (+5, 4), B’ (5, 2), C’ (3, 3) and D’ (3, 4)
Ans11.
109
Ans12.
Ans13. Solution the given equation is 2x+y=4
When x=0 y=4 i.e. (0, 4)
When x=1 y=2 i.e. (1, 2)
When x=2 y=0 i.e. (2, 0)
When x=3 y = -2 i.e. (3,-2)
Some more ordered pairs satisfying the equation 2x+y=4 are given by
(4,-1) ‘(-1, 6) (-3, 10) --------
By plotting the points in the Cartesian plane.
Thus an infinite number of ordered pairs can be found and plotted.
110
CBSE TEST PAPER-03
CLASS - X Mathematics (Co-ordinate Geometry)
1. The distance of the point (3, 5) from x- axis is
(a) 3 units (b) 4 units
(c) 5 units (d) 6 units
(1)
2. The point (0, -5) lies on
(a) +ve x- axis (b) +ve y- axis
(c) –ve x- axis (d) –ve y-axis
(1)
3. The point (-2, 5) lies in
(a) 1st quadrant (b) 2nd quadrant
(c) 3rd quadrant (d) 4th quadrant
(1)
4. The distance of the point (3, 0) from x- axis is
(a) 3 units (b) 0 units
(c) 9 units (d) none of these
(1)
5. Name the points of the plane which do not belong to any of the quadrants. (2)
6. Which of the following points belong to the x- axis?
(a) (2, 0) (b) (3, 3)
(c) (0, 1) (d) (-2, 0)
(2)
7. Which of the following points belongs to 1st quadrant
(a) (3, 0) (b) (1, 2)
(c) (-3, 4) (d) (3, 4)
(2)
8. Which of the following points belongs to 3rd quadrant
(a) (1, 3) (b) (-1, -3)
(c) (0, 4) (d) (-4, -2)
(2)
9. Find Co-ordinates of vertices of rectangle ABCD (3)
111
10. Take a rectangle ABCD with A (-6, 4), B (-6, 2), C (-2, 2) and D (-2, 4). Find its
mirror image with respect to x- axis.
(3)
11. The following table gives measures (in degrees) of two acute angles of a right
triangle
Plot the point and join them.
X 10 20 30 40 50 60 70 80
Y 80 70 60 50 40 30 20 10
(3)
12. Plot each of the following points in the Cartesian Plane
(a) (3, 4) (b) (-3, -4) (c) (0, -5) (d) (2, -5) (e) (2, 0)
(3)
13. The following table given the relation between natural numbers and odd
natural numbers
X 1 2 3 4 5 6 7
Y 3 5 7 9 11 13 15
Plot the points and join them. Do you get a straight line by joining these points?
(5)
112
CBSE TEST PAPER-03
CLASS - X Mathematics (Co-ordinate Geometry)
[ANSWERS]
Ans01. (c) 5 units.
Ans02. (d) –ve y-axis.
Ans03. (b) 2nd quadrant.
Ans04. (a) 3 units.
Ans05. The points in a plane which do not belong to any one of the plane are (0, 0).
Ans06. (2, 0) and (-2, 0) belongs to x- axis.
Ans07. (1, 2) and (3, 4) belongs to 1st quadrant.
Ans08. (-1, -3) and (-4, -2) belongs to 3rd quadrant.
Ans09. The Co- ordinate of vertices of rectangle A (2, 2), B (-2, 2), C (-2, -2) and D (2, -2).
Ans10. The mirror image of A (-6, 4) is A’ (-6, -4) and B (-6, 2) is B’ (-6, -2), C (-2, 2) is C’
(-2, -2) and D (-2, 4) is D’ (-2, -4)
Ans11.
.. . .
. . . .
113
Ans12.
.
.
. ..
Ans13.
..
..
..
.
Yes, we get a straight line by joining them.
114
CBSE TEST PAPER-01
CLASS - X Mathematics (Heron’s formula)
1. The measure of each side of an equilateral triangle whose area is 3 cm2 is
(A) 8cm (B) 2cm (C) 4cm (D) 16cm
[1]
2. Measure of one side of an equilateral triangle is 12cm. Its area is given by
(A) 9 3 sq cm (B) 18 3 sq cm (C) 27 3 sq cm (D) 36 3 sq cm
[1]
3. Two adjacent side of a parallelogram are 74cm and 40cm one of Its diagonals is
102cm. area of the ||gram is
(A) 612 sq m (B) 1224 sq m (C) 2448 sq m (D) 4896 sq m
[1]
4. The area of rhombus is 96 sq m. If one of its diagonals is 16cm, then length of
Its side is (A) 10cm (B) 8cm (C) 6cm (D) 5cm
[1]
5. An umbrella is made by stitching 10 triangles pieces of cloth of two different
colour, each piece measuring 20cm 50cm and 50cm. How much cloth of each
colour is required for the umbrella?
[2]
6. The perimeter of a rhombus ABCD is 40cm. find the area of rhombus of Its
diagonals BD measures 12cm
[2]
7. Find area of triangle with two sides as 18cm & 10cm and the perimeter is 42cm. [2]
8. Find the area of in isosceles triangle, the measure of one of Its equals side being
‘b’ and the third side ‘a’.
[2]
9. From a point in the interior of an equilateral triangle perpendiculars drawn to
the three sides are 8cm, 10cm and 11cm respectively. Find the area of the
triangle to the nearest cm. (use 3 =1.73 )
[3]
10. A parallelogram, the length of whose side is 60m and 25m has one diagonal 65m
long. Find the area of the parallelogram.
[3]
11. A parallelogram, the measures of whose adjacent sides are 28cm and 42cm, has
one diagonals 38cm. Find Its altitude on the side 42cm.
[3]
12. Find the area of a quadrilateral ABCD in which AB=3cm, BC=4cm, CD= 4cm,
DA=5cm and AC= 5cm.
[3]
13. A field in the shape of a trapezium whose parallel side are 25m and 10m. The
non- parallel side are 14m and 13m. Find the area of the field.
[5]
115
CBSE TEST PAPER-01
CLASS - X Mathematics (Heron’s formula)
[ANSWERS]
Ans01 (b)
Ans02. (d)
Ans03. (c)
Ans04. (a)
Ans05. a=20cm, b=50cm
cloth required for each colour ∴
2 2
2 2
5 of one triangle piece
=5 44
205 4(50) (20) cm
4
=25 40 6 sq cm
=1000 6 sq cm
thus, (1000 6) sq cm cloth of each colour is required
Area
ab a
sq
= ×
× −
= × −
×
×
Ans06. 40
10cm4
AB BC CD DA cm∴ = = = =
now in ABD,
AB=10cm, BD=12cm and DA=10cm
10 12 10 S= =16cm
2 by herons pormula
area of ABD = 16(16 10) (16-12) (16-10)
= 16 6 4 6 =48sq cm
cm+ +∴
∴
−
× × ×
△
△
hom ABCD=2 area of ABD
= 2 48sq cm
=96sq cm
areaofr bus∴ ××
△
116
Ans07. Let a=18cm, b=10cm
Perimeter =42cm
42
so, C=14cm
18 10 140 S= = 21
2 2
t 21(21 18) (21-10) (21-14)
= 21 3 11 7
a b c cm
a b ccm
new area of riangles
∴ + + =
+ + + +∴ =
= −
× × ×
=21 11 cmsq
Ans08. Here 2
= units2 2
a b c a bS units
+ + +=
2 2
2 2 2 2 area of = ( )( )( )( )
2 2 2 2
2 2 = ( )( )
2 2 2 2
= 4 units4
a b a b a b a ba b c
a b b a a asq units
ab a sq
+ + + +∴ − − −
+ − ×
−
△
Ans09. Let the each side of the equilateral ∆ ABC measure be x cm.
Let OD= 11cm, OE =8cm and OF=10cm
Join OA, OB and OC
( ) ( ) ( ) ( )
1 1 111 8 10 cm
2 2 2
29= sq cm (1)
2
Now ar ABC ar OBC ar OCA ar OAB
x x x sq
x
= + +
= × + × + ×
→
△ △ △ △
But area of equilateral ∆ , the measure of whose each side of x
2
2
3 x sq cm (ii)
4from (i) and (ii)
3 29
4 2x x
= →
=
4 29 58
3 2 3
29 58 841 3 area of ABC = sq cm
2 33841 1.73
= 485 sq cm.3
x×∴ = =
×
∴ ∆ × =
× =
117
Ans10. AB=DC=60cm, BC=AD= 25m and AC=65m
Area of 11gm ABCD= Area of ∆ ABC + area of ∆ ACD
= 2 Area of ∆ ABC [∴ar ∆ ABC =ar ∆ ABD]
60 65 25 S= 75
2
area of ABC = ( )( )( )
= 75(75 60)(75 65)(75 25) sq m
=(5 3 5 2 5) sq m
=750 sqm (
now
m m
S s a s b s c
+ + =
∴ ∆ − − −
− − −× × × ×
→ II)
from (i) and (ii), we get
area of 11 ABCD =2 750=1500sq m.gm ×
Ans11. AB=DC=42cm=C
BC=AD= 28cm =b
And BD=38cm=a
Let A be the area of ∆ ABD
3 8 2 8 4 2, S = 5 4
2
A = 5 4 (5 4 3 8)(5 8 2 8)(5 4 4 2 )
= 5 4 1 6 2 6 1 2 sq c m .
= 1 4 4 3 sq cm
a rea o f A B D = 1 4 4 3 sq cm
1 ag a in a rea o f A B D =
2
n o w cm
b a se a ltitu d e
+ + =
− − −
× × ×
∴ ∆
∆ ×
1 = 4 2 , w h a re h cm is a ltitu d e
2 = 2 1 h sq c m
fro m (i) a n d (ii) ,w e g e t
2 1 h = 1 4 4 3
h sq cm× ×
∴
1 4 4 3 4 8 3 h =
2 1 7
4 8 3 th u s , req u ired a ltitu d e =
7
c m
c m
=
118
Ans12. ∴Area of quad ABC=area of ∆ ABC + area of ∆ ACD (i)
For ∆ ABC,3 4 5
62
S cm+ += =
Area of ABC= 6(6 3)(6 4)(6 5)∴ ∆ − − −
6 3 2 1 sq cm =6sq cm ( )ii= × × × →
For ∆ ACD, S=5 4 5
=7cm2
+ +
area of ACD= 7(7 5)(7 4)(7 5)
= 7 2 3 2 9.2 sq cm (iii)
from (i), (ii) and (iii), we get
area of quad ABCD =(6+9.2) =15.2 sq cm
∴ ∆ − − −
× × × = →∴
Ans13. AB=25m, CD =10m
AD=13m and BC=14
Draw ECIIAD and CP ⊥ EB
Now CE =AD=13m and
EB=AB-AE= (25-10) m= 15m
In ∆ BCE, a=15m, b=14m and C=13 m
15 14 13
2 2 =21m
area of BCE=
21(21 15)(21 14)(21 13)
= 21 6 7 8 sq m
=84 sq m (i)
1 area of BCE =
22
CP=
a b cS
Also BE CP
area
+ + + +∴ = =
∴ ∆
− − −
× × ×→
∆ × ×
of BCE 2 84 56 m= m [using(i)]
15 5
Now, area of II AECD=Base height
56 =10 m=112 sq m (ii)
5
thus area of trap ABCD=area of II of BCE
gm
gm
BE
area
∆ ×=
×
× →
+ ∆ =112 sq m+84 sq m [using(i) and (ii)]
=196 sq m
119
CBSE TEST PAPER-02
CLASS - X Mathematics (Heron’s formula)
1. The perimeter of a triangle is 60cm. If its sides are in the ratio 1:3:2, then its
smallest side is
(a) 15 (b) 5 (c) 10 (d) none of these.
(1)
2. The perimeter of a triangle is 36cm. If its sides are in the ratio 1:3:2, then its
largest side is
(a) 6 (b) 12 (c) 18 (d) none of these
(1)
3. If the perimeter of a rhombus is 20cm and one of the diagonals is 8cm. The area
of the rhombus is
(a) 24 sq cm (b) 48 sq cm (c) 50 sq cm (d) 30 sq cm
(1)
4. One of the diagonals of a rhombus is 12cm sides area is 54 sq cm. the perimeter
of the rhombus is
(a) 72 cm (b) 3 10 cm (c) 6 10 cm (d) 12 10cm
(1)
5. Find the cost of leveling the ground in the form of a triangle having its sides are
40m, 70m and 90m at Rs 8 per square meter. [use 5 = 2.24]
(2)
6. The triangular side’s walls of a flyover have been used for advertisements. The
sides of the walls are 122m, 22m and 120m. The advertisement yield on earning
of Rs 5000 per m2 per year. A company hired one of its walls for 4 months. How
much rent did it pay?
(2)
7. Find the perimeter and area of a triangle whose sides are of length 2cm, 5cm
and 5cm.
(2)
8. There is a slide in a park. One of its sides wall has been painted in some colour
with a message “KEEF THE CITY GREEN AND CLEAN”.
If the sides of the wall area 15m, 11m and 6m. Find the area painted in colour.
(2)
9. Find the base of an isosceles triangles whose area is 12cm and the length of one
of the equal side is 5cm.
(3)
10. The perimeter of a triangle is 450m and its sides are in the ratio of 13:12:5. Find (3)
120
the area of the triangle.
11. The sides of a triangle are 39cm, 42cm and 45cm. A parallelogram stands on the
greatest side of the triangle and has the same area as that of the triangle. Find
the height of the parallelogram.
(3)
12. The students of a school staged a rally for
cleanliness campaign. They walked through the
lanes in two groups. One group walked through
the lanes AB, BC and CA whiles the other group
through the lanes AC, CD and DA [fig1.1]. Then
they cleaned the area enclosed within their
lanes. If AB=9m, BC=40m, CD=15m, DA=28m and B∠ =90o, while group cleaned
more area and by how much? Find also the total area cleaned by the students.
(3)
13. The perimeter of a right triangle is 24 cm. If its hypotenuse is 10 cm, find the
other two sides. Find its area by using the formula area of a right triangle. Verify
your result by using Heron’s formula.
(5)
121
CBSE TEST PAPER-02
CLASS - X Mathematics (Heron’s formula)
[ANSWERS]
Ans01. (c)
Ans02. (c)
Ans03. (a)
Ans04. (d)
Ans05. Here S = 40 70 90
m2
+ +
= 100 m
∴ Area of a triangular ground = 100 (100 40) (100-70) (100-90) sq m−
= 100 60 30 10 sq m× × ×
= 6(10 10 5) sq m× ×
= ( )600 2.24 sq m×
= 1344 sq m
∴ Cost of leveling the ground = Rs ( )8 1344×
=Rs 10752
Ans06. The lengths of the sides of the walls are 122m, 22m and 120m.
As,
1202 + 222 = 14400 +484
= 14884
= (122)2
∴ Walls are in the form of right triangles
Area of one wall = 1
2Base height× ×
=1
120 22 sq m2
× ×
= 1320 sq m.
Rent = Rs 5000/sq m per year
∴ Rent for 4 month = Rs5000 1320 4
12
× ×
= Rs 22,00,000
122
Ans07. Here, a = 2cm, b = 5cm and c = 5cm
∴ Perimeter = a+ b+ c = (2 + 5 + 5) = 12 cm
S = semi perimeter
= 12
2 = 6 cm
using Heron’s formula,
∴ Area of triangle = ( ) (s-b) (s-c) sq cms s a−
= 6 (6-2) (6-5) (6-5) sq cm
= 24 sq cm = 4.9 sq cm
Ans08. ∴ The sides of the wall is in the triangular from with sides,
A = 15 m, b = 6 m and c = 11 m
∴ S = 15 6 11
2
+ +m
= 16 m
∴Area to be painted in colour = Area of the side wall
= ( ) (s-b) (s-c) sq cms s a−
= 16 (16 5) (16-6) (16-11) sq m−
= 4 50 sq m
= 20 2 sq m
Ans09. ∴ Area of an isosceles triangle = 2 244
ab a−
∴ 2 244
ab a− = 12
or 21004
aa− = 12
Squaring both sides, we have
( )2 2100a a− = 144×16
= 2304
or 4 2100 2300a a− + = 0
or ( ) ( )2 264 36a a− − = 0
∴ Either a2 = 64 i.e. a = ± 8
or a2 = 36 i.e. a = ± 6
∴ Required base = 8 cm or 6 cm
123
Ans10. Let the sides of the triangle be 13x, 12x and 5x
Perimeter of a triangle = 450 m
∴ 13x + 12x + 5x = 450 m
or 30x = 450
∴ x = 15
∴ The sides are 13×15, 12×15, and 5×15
I.e. 195 m, 180 m and 75 m
∴ S =2
a b c+ + =
450
2 = 225 m
∴ Area of the triangle = ( ) (s-b) (s-c) sq ms s a−
= 225 (225-195) (225-180) (225-75) sq m
= 225 30 45 150 sq m× × ×= (15×15×2×3×5) sq m
= 6750 sq m.
Ans11. To find the area of ∆ ABC
S = 45 42 39
2
+ +cm
= 63 cm
∴ Area of ∆ ABC = 63 (63-45) (63-42) (63-39) sq cm
= 63 18 21 24 sq cm× × ×= 9×7×2×3×2 sq cm
= 756 sq cm
Let h be the height of the parallogram
Now,
Area of //gm BCDE = Area of ∆ ABC
∴ h×BC = 756
or 45h = 756
h = 756
45 h = 16.8 cm
Hence, height of the //gm = 16.8 cm
Ans12. We have, right angle ∆ ABC,
AC2 =AB2 + BC2
AC2 = 92 + 402
AC2= 1680
∴ AC = 41
124
The first group has to clean the area of ∆ ABC which is right triangled
Now,
Area of ∆ ABC = 1
2×40 m ×9 m
= 180 sq m
The second group has to clean the area of ∆ ACD which has AD=28 m,
DC=15m and AC=41
Hence,
S = 28 15 41
2
+ +
= 42 m
∴ Area of ∆ ACD = 42 (42-28) (42-15) (42-41) sq m
= 42 14 27 1 sq m× × ×= 7 3 2 7 2 9 3 sq m× × × × × ×
= 126 sq m
∴ First group cleaned more = (180 – 126) sq m
= 54 sq m
∴ Total area cleaned by students = (180 + 126) sq m
= 306 sq m
Ans13. Let x and y be the two lines of the right ∠ .
∴ AB = x cm, BC = y cm and AC = 10 cm
∴ By the given condition,
Perimeter = 24 cm
Or x + y = 14 (1)
By Pythagoras theorem, 2 2x y+ = (10)2
= 100 (2)
From (1), 2 2x y+ = (14)2
Or 2 2x y+ + 2xy = 196
∴ 100 + 2xy =196 from (2)
xy = 96
2 = 48 sq cm (3)
125
Area of ∆ ABC = 1
2xy sq cm
= 1
482
× sq cm
=24 sq cm (4)
again,
(x-y) 2 = (x+y) 2 – 4xy
= (14)2 - 4×48
or x-y = ± 2
(i) When, x-y = 2 and x+y = 14, then 2x = 16
or x = 8, y = 6
(ii) When, X – y = -2 and x + y = 14, then 2x = 12
or x = 6, y = 8
Verification by using Heron’s formula:-
Sides are 6 cm, 8 cm and 10 cm
S = 24
2 = 12 cm
Area of ∆ ABC = 12 (12-6) (12-8) (12-10) sq cm
= 12 6 4 2× × × sq cm
= 24 sq cm
Which is same as found in (4)
Thus, the result is verified.
126
CBSE TEST PAPER-03
CLASS - X Mathematics (Heron’s formula)
1. The side of a triangle is 12 cm, 16 cm, and 20 cm. Its area is
(A) 100sq cm (B) 90sq cm
(C) 96sq cm (D) 120sq cm.
[1]
2. The side of an equilateral triangle is 4 3cm . Its area is.
(A) 12 3 sq cm (B) 12 6 sq cm
(C) 12 10 sq cm (D) 6 10 sq cm.
[1]
3. It the perimeter of a rhombus is 20sq cm and one of the diagonals is 8 cm. the
area of the rhombus is
(A) 40sq cm (B) 24sq cm
(C) 20sq cm (D) 13sq cm.
[1]
4. One of the diagonals of a rhombus is 12 cm and Its area is 54sq cm. the
perimeter of the rhombus is.
(A) 10 cm (B) 8 cm
(C) 6 cm (D) 12 10 cm.
[1]
5. Find the area of isosceles triangle whose side is 14 m, 12 m, 14m? [2]
6. The perimeter of a rhombus ABCD is 60 cm. find the area of the rhombus of Its
diagonal BD measures 16 cm?
[2]
7. Find the cost of leveling the ground in the from of a triangle having Its side as 70
cm, 50 cm, and 60 cm, at Rs 7 per square meter.
[2]
8. Find the area of a triangle two side of which are 18 cm, and 12 cm. and the
perimeter is 40 cm.
[2]
9. A traffic signal board indicating ‘school ahead’ is an equilateral triangle with
side ‘a’ find the area of the signal board using herons. Its perimeter is 180 cm,
what will be Its area?
[3]
127
10. An parallelogram the length of whose sides are 80m, and 40m has one diagonal
75m long. Find the area of the parallelogram?
[3]
11. The side of a triangular find is 52m, 56m, and 60m find the cost of leveling the
field Rs 18 per meter, is a space of 4cm is to be left for entry gate.
[3]
12. A floral design of a floor is made up of 16 tiles which are triangular. The side of
the triangle being 9 cm, 28 cm, and 35 cm. find the cost of polishing the tiles, at
RS 50 paisa/sq cm.
[3]
13. Radha made a picture of an aero plane with colored paper as shown in fig 1.1
find the total area of the paper wed.
[5]
128
CBSE TEST PAPER-03
CLASS - X Mathematics (Heron’s formula)
[ANSWERS]
Ans01 (C)
Ans02. (A)
Ans03. (B)
Ans04. (D)
Ans05. 14 12 14
202
S cm+ += =
Area of isosclese triangle = ( )( )( )
= 20(20 14)(20 12)(20 14) sq m
= 20 6 8 6 sq m
s s a s b s c− − −
− − −
× × ×6 = 160 =6 12.6 sq m
= 75.6 sq m
×
Ans06. side of rhombus are equal.As
8
60A B=B C =C D =D A= 15
4 A BD ,
15 151623
2
Area of AB C = 23(23 15)(23 15)(23 16)
= 23 8 8 7 = 23 7
= 8 12.7
= 101
cm
in
S cm
so
∴ =
∆+= =
∆ − − −
× × × ××
.6 sqcm
area of rhom bus =2 101.6=203.2 sq cm×
129
Ans07. 70 50 60 180
902 2
S m+ += = =
area of triangle = 90(90 70)(90 50)(90 60)
= 90 20 40 30
= 1469.7 sq m
cost of levelling the ground=RS (7 1469.7)
∴ − − −
× × ×
∴ × = 10287.9
Ans08 a=18 cm, b=12 cm and C=?Let
So a+b+c=40 cm
18+12+C=40
C=(40-30) cm =10 cm
18 12 10 S= =20 cm
2
area of triangle = 20(20 18)(20 12)(20 10)
= 20 2 8 10 sq cm
+ +∴
∴ − − −
× × × = 56.56 sq cm
Ans09. 3
units 2 2
a a a aS units
+ +=
2
3 3 3 3 A re a o f t r ia n g le = ( ) ( ) ( )
2 2 2 2
3 =
2 2 2 2
= 3 s q u n i ts4
N o w , p e r im e te r = 1 8 0 c m
1 8 0e a c h s id e = 6 0
3
A re a o f s ig n a l
a a a aa a a
a a a a
a
c m
∴ × − − −
× × ×
∴ =
∴ 23b o a rd = (6 0 ) s q c m
4
= 9 0 0 3 s q c m
Ans10. 80AB DC cm= =40 and AC=75 cm
80 40 75In ABC, S= = 97.5
2
BC AD cm= =+ +∆
130
Area of triangle = ( )( )( )
= 97.5(97.5 80)(97.5 40)(97.5 75) sq m
= 97.5 17.5 57.5 22.5 sq m
= 220746
s s a s b s c− − −
− − −
× × ×
0.94 =1485.75 sq m
= 1485.75 sq m
Area of 11 ABCD=2 Area of ACD
= 2 1485.7
= 2971.4 sq m
gm × ∆×
Ans11. the field are 52m, 56m and 60m.side of
52 56 60 S= =84m
2
Area of field = 84(84 52)(84 56)(84 60) sq m
= (7 12)(2 16(4 7)(12 2) sq m
= 7 7 12 12 2 2 4 16 sq m
+ +∴
∴ − − −
× × × ×
× × × × × × × = 1344 sqm
total cost of levelling the field = Rs5 1344
= Rs 6720
space to be left for entry gate =4 cm
space to be fenced =168-4=164
∴ ×
∴ m
cost of fencing 1 m =Rs 18
total cost of fencing the field = Rs 18 164
= Rs 2952
∴ ×
Ans12. triangalar tile, we have for Each
35 28 9 S= cm=36 cm
2
Area of Each tile = ( )( )( )
= 36(36 35)(36 28)(36 9) sq cm
= 36 6 sq cm
Area of 16 tile= 16 3
s s a s b s c
+ +
∴ − − −
− − −
∴ × 6 3 sq cm
1 cost of polishing = Rs [ 16 36 6] =Rs=288 6
2 = Rs(288 2.45) =Rs 705.60
∴ × ×
×
131
Ans13. (1) = area of isosceles triangle with a=1 cm and b= 5 cmArea
2 2 = 44
1 99 = 100 1 sq cm (approx)
4 4Area (II) = area of rectangle with
L= 6.5cm and b=1 cm
= 6.5 1 sq cm =6.5 sq cm
Area (III)
ab a−
− =
×
2
=Area of trapezium
= 3 Area of equilational with side =1 cm
3 3 1.732 5.196 =3 (1) sq cm = or sq cm
4 4 4 = 1.3 sq cm (approx)
1Area of (IV+V) =2 6
2
× ∆
×× ×
× × ×1.5 9sq cm
total area of the paper used =Area (I+II+III+IV)
= (2.5+6.5+1.3+9) sq cm
= 19.3 sq cm
sq cm =
∴
132
CBSE TEST PAPER-04
CLASS - X Mathematics (Heron’s formula)
1. The lengths of the side of a triangular park are 90m, 70m and 40m, find Its area.
(A) 1340 sq m (B) 1344 sq m (C) 1440 sq m (D) 1444 sq m
[1]
2. An equilateral triangle has a side 50cm long. Find the area of the triangles.
(A) 625 3 sq cm (B) 625 6 sq m (C) 256 6 sq m (D) 625 10 sq m
[1]
3. The area of an isosceles triangle is 12 sq cm. It one of the equal side is 5 cm, then
the length of the base is
(A) 4 cm (B) 5 cm (C) 6 cm (D) 8 cm
[1]
4. Find the area of triangle whose side is 6 cm, 10 cm and 15cm.
(A) 404.9 sq cm (B) 405.9 sq cm (C) 402.9 sq cm (D) 410sq cm [1]
5. Find the area of triangle whose side is 42m, 56m and 70m? [2]
6. Find the area of an isosceles triangle , the measure of one of Its equal side being
b and the third side a.
[2]
7. Find the area of equilateral triangle whose side is 12 cm. [2]
8. Find the area of isosceles triangle whose equal side is 6 cm, 6 cm and 8 cm. [2]
9. The measure of one side of a right triangle is 42m. It the difference in lengths of
Its hypotenuse and other side is 14 cm, find the measure of two unknown side?
[3]
10. The perimeter of a rhombus ABCD is 80 cm. find the area of rhombus if Its
diagonal BD measures 12 cm.
[3]
11. Find area of quadrilateral ABCD in which AB= 5 cm, BC= 6 cm, CD 6 cm, DA
=7cm, And AC=7 cm.
[3]
12. The triangular side walls of a flyover have been used for advertisements. The
side of the walls are 122 m, 22m And 120m. The advertisements yield a caning
of Rs 3000 per M2 per year. A company hired one of Its walls for 6 months How
much rent did It pay?
[3]
13. Shashi Kant has a vegetable garden in the shape of a rhombus. The length of
each side of garden is 35 m And Its diagonal is 42 m long. After growing the
vegetables in it. He wants to divide it in seven equal parts And look after each
part once a week. Find the area of the garden which he has to look after daily.
[3]
133
CBSE TEST PAPER-04
CLASS - X Mathematics (Heron’s formula)
[ANSWERS]
Ans01. (b)
Ans02. (a)
Ans03. (c)
Ans04. (a)
Ans05. S = 42 56 70
m2
+ + =
168 m or 84
2
∴Area of ∆ ABC = (s-a) (s-b) (s-c)s
= 84 (84-42) (84-56) (84-70) sq m
= 42×28 sq m
= 1176sq m
Ans06. S = units2
a b b+ + =
2units
2
a b+
∴ Area of triangle =2 2 2 2
2 2 2 2
a b a b a b a ba a a
+ + + + × − − −
units
= 2
units2 2 2 2
a b a a a+ × × ×
= 2 2
sq units4 4a
b a−
Ans07. S = 12 12 12
2cm
+ +
= 36
cm2
= 18 cm
∴ Area of equilateral = (s-a) (s-b) (s-c)s
= 18 (18-12) (18-12) (18-12)
= 18 6 6 6× × × = 36 3 sq cm
134
Ans08. S = 6 6 8
2cm
+ +
= 20
10 2
cm=
∴ Area of isosceles triangle = 80 (10-6) (10-6) (10-8)
= 10 4 4 2 sq cm× × × = 17.8 sq cm
Ans09. Let AB = y and AC = x and BC = 42 cm
∴ By the given condition,
x - y = 14 (i)
By Pythagoras theorem,
x2 – y2 = 1764
(x + y) (x – y) = 1764
∴ 14 (x + y) = 1764 using (ii)
∴ x + y =1764
126 (iii)14
=
Adding (ii) and (iii), we get
2x = 140
i.e. x = 70
∴ y = 126-x
y = 126 – 70
= 56
Ans10. ∴ AB = BC = CD = DA = 80
= 20 cm4
Now in ∆ ABD,
∴ S =20 20 12
2
+ +
so,
Area of ∆ ABD = 26 6 6 14 sq cm× × × = 114.4 sq cm
Area of rhombus = 2 area of ABD× ∆= 2× 114.4 sq cm
= 228.8 sq cm
135
Ans11. Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD (i)
In ∆ ABC,
S = 5 6 7
2
+ += 9 cm
∴ Area of ∆ ABC = 9 (9 5)(9 6)(9 7) sq cm− − −
= 9 4 3 2 sq cm× × ×
= 6 6 sq cm
In ∆ ACD,
S = 7 7 6
2
+ + = 10 cm
∴ Area of ∆ ACD = 10 (10 7)(10 7)(10 6) sq cm− − −
= 10 3 3 4 sq cm× × ×
= 18.9 sq cm
Area of quadrilateral ABCD = ( 6 6 + 18.9) sq cm
= (14.4 + 18.9) sq cm
= 33.3 sq cm.
Ans12. The lengths of the sides are 122m, 22m and 120m
So,
1202 + 222 = 14400 + 484 = (122)2
∴ Walls are in the form of right triangles.
Area of one wall = 1
120 222
× ×
= 1320 sq m
Rent = Rs 3000/sq m per year
∴ Rent for 6 months =3000 1320 6
12Rs
× ×
= Rs 198000
136
Ans13. Let ABCD be garden
∴ DC = 35 m . . . . . . (Given)
DB = 42 m . . . . . . (Given)
Draw CE ⊥ DB
We know that the diagonals of a rhombus bisect each other at right angles.
∴ DE = 1
2 DB =
142 or 21 m
2×
Now
CE2 = CD2 – DE2
= 352 – 212
= 784
CE = 28 m
Now area of ∆ DBC, = 1
2DB CE× ×
= 1
42 282
× ×
= 588 sq m
∴ Area of the garden ABCD = 2×588 sq m
= 1176 sq m
Area of the garden he has to look after, daily
= 1176
2 sq m
= 168 sq m
137
CBSE TEST PAPER-05
CLASS - X Mathematics (Heron’s formula)
1. If side of equilateral triangle is 25 m. Its area is
(a) 25 3 sq cm (b) 45 3 sq cm
(c) 5 3 sq cm (d) 3 sq cm
[1]
2. The perimeter of an equilateral triangle is 48 cm. Its area is
(a) 18 3 sq cm (b) 72 3 sq cm
(c) 64 3 sq cm (d) 60 3 sq cm
[1]
3. If area of isosceles triangle is 48 sq cm and length of one of its equal sides is 10
m, then what is the base?
(a) 16 cm or 12 cm (b) 12 cm or 14 cm
(c) 14 cm or 16 cm (d) 16 cm or 18 cm
[1]
4. If AB = 14 cm, BC = 13 cm, CD = 17 cm, DA = 8 cm and AC = 15 cm then area of
quadrilateral ABCD is
(a) 150 sq cm (b) 144 sq cm
(c) 142 sq cm (d) 140 sq cm
[1]
5. Find the area of an isosceles triangles, the measure of one of its equal sides
being 10 cm and the third side is 6 cm.
[2]
6. Find the area of equilateral triangle the length of one of its sides being 24 cm. [2]
7. Find the perimeter and area of a triangle whose sides are 3 cm, 4 cm and 10 cm? [2]
8. Using Heron’s formula, find area of triangle whose sides are 6 cm, 8 cm and 10
cm?
[2]
9. The perimeter of a triangle is 480 meters and its sides are in the ratio of 1:2:3.
Find the area of triangle?
[3]
10. Find the cost of leveling the ground in the form of equilateral triangle whose
side is 12 m at Rs 5 per square meter.
[3]
138
11. A kite in the shape of a square with
diagonal 32 cm and an isosceles triangle of
base 8 cm and side 6 cm each is to be made
of three different shades. How much paper
of each shade has been used in it? ( use
5 = 2.24)
[3]
12. The sides of a quadrangular field, taken in order are 29 m, 36 m and 24
respectively. The angle contained by the last two sides is a right angle. Find its
area.
[3]
13. Find the area of a triangle whose perimeter is 180 cm and two of its sides are 80
cm and 18 cm. hence calculate the altitude of the triangle taking the longest side
as base.
[3]
139
CBSE TEST PAPER-05
CLASS - X Mathematics (Heron’s formula)
[ANSWERS]
Ans01. (a)
Ans02. (c)
Ans03. (a)
Ans04. (b)
Ans05. S = 10 6 16
2 2
+ =
= 8 cm
Area if tringle = 8 (8-5) (8-5) (8-6) sq cm
= 8 3 3 2 sq cm
= 12 sq cm
∴
× × ×
Ans06. a = b = c = 24 cm
24 24 24 72 S = cm = cm
2 2 = 36 cm
+ +∴
Area of triangle = 36 (36-24) (36-24) (36-24) sq cm
= 246.12 sq cm
∴
Ans07. Perimeter = 3+4+5
= 12 cm
12 S = semiperimeter =
2 or = 6 cm
∴
6 (6-3) (6-4) (6-5) sq cm
= 6 sq cm
area of triangle =
140
Ans08.
6 8 10 24 = =
2 2 = 12 cm
S+ +
Area of triangle = 12 (12-6) (12-8) (12-10)sq cm
= 24 sq cm
∴
Ans09. Let the sides of the triangle be x, 2x, 3x Perimeter of the triangle = 480 m ∴ x + 2x + 3x = 480 m
6x = 480 m x = 80 m
∴ The sides are 80 m, 160 m, 240 m So,
80 160 240 480 = =
2 2 = 240 m
S+ +
And,
Area of triangle = (s-a) (s-b) (s-c) sq m
= 240 (240-80) (240-160) (240-240) sq m
= 0 sq cm
s∴
∴ Triangles don’t exist with the ratio 1:2:3 whose perimeter is 480 m.
Ans10. Here, sides are 12 m, 12 m, 12 m, 12 12 12
S = 2
= 18 cm
+ +∴
And,
Area of equilateral triangler = (s-a) (s-b) (s-c) sq m
= 18 (18-12) (18-12) (18-12) sq m
=
s∴
18 6 6 6 sq m
= 6 3 6 6 6 sq m
= 36 3 sq m
× × ×
× × × ×
∴ Cost of leveling ground = (5 36 1.73)Rs × × = Rs 311.4 m
141
Ans11. Let ABCD be the square and ∆ CEF be an isosceles triangles. Let the diagonals intersect each other at O. Then,
AO =1
322
× cm
= 16 cm
Area of shaded portion I = 1
16 32 2
sq cm× ×
= 256 sq cm And,
Area of portion III =
2 2
2
448
4 (6) 84
17.92 sq cm
ab a−
= × −
=
Thus, the papers of three shades required are 256 sq cm, 256 sq cm and 17.92 sq cm.
Ans12. In ∆ ADC, which are right angles? 2 2 2
2 2
AC
= (24) (7)
= 576+49=625
= AC = 625 25
AD DC∴ = +
+
∴ =
Area of quad ABCD =Area of ∆ ADC+ Area of ∆ ABC (I)→
Area of ∆ ADC= 1 1
AD CD = 24 7 2 2
sq m× × ×
= 84 sq m (II)→ For ∆ ABC,
36 25 29 S = 45
2
Area of ABC,= 45(45 36)(46 25)(46 29)
= 360 sq m (III)
M M+ +∴ =
∴ ∆ − − −→
From (I), (II) and (III) we get Area of quad ABCD = (84+360) sq m = 444 sq m
142
Ans13. Let a= 80 cm and b=18cm, perimeter = 180 cm 180= a+b+c =80+18+c
C= 82 cm
180, S= = 90 cm
2
Area of triangle = 90(90 80)(90 18)(90 82)
= 90 10 72 8 sq cm
= 720 sq
Now
∴
∴ − − −
× × × m
The longest side of triangle is 82 cm
Let h cm be length of altitude corresponding to longest side.
∴Area of triangle =1
822
hcm× ×
1720 82 cm
2720
h= cm 41
23 h= 17 cm
41
h= × ×
Hence, Area of triangle is 720 sq cm.
And length required altitude is 17231
cm.41
143
CBSE MIXED TEST PAPER-01
(Unit Test)
CLASS - IX MATHEMATICS
[Time : 1.50 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) Q. No. from 1 to 5 carries 1 mark each.
(iii) Q. No. from 6 to 9 carries 2 marks each.
(iv) Q. No. from 10 to 14 carries 3 marks each.
(v) Q. No. from 15 to 16 carries 6 marks each
1. Determine the range of the following data.
55, 36, 45, 90, 27, 25, 75, 95, 80, 15.
2. What is the mean of first five prime numbers.
3. For what value of x, the mode of following data is 7.
3,5,6,7,5,4,7,5,6,x,8,7.
4. State congruence Axiom.
5. In triangle ABC, AB = 6.5 cm, BC = 5cm, AC = 4.5 cm. Name the greatest and the smallest
angle.
6. The following data has been arranged in ascending order.
12, 14, 17, 20, 22, x, 26, 28, 32, 36.
If the media of the data is 23, find x.
7. Find mean of the following data.
Xi 5 10 15 20 35
Fi 3 5 7 7 9
144
8. In fig, OA = OB, and OC = OD
Show that
i. triangle AOD ≅ triangle BOC
ii. AD parallel BC
9. E and F are the mid points of equal sides AB and AC of ABC. Show that BF = CE.
10. The following table shows the frequency distribution of daily wages of employees of a
company.
Wages (in Rs) No. of Employess
50-59 45
60-69 12
70-19 33
80-89 36
90-99 24
Find
a. Lower limit of 3rd class interval
b. True limits of 2nd class interval
c. Class size of 4th class interval
11. Below are the marks obtained by 25 students of class IX in Maths.
20 35 60 45 75
90 50 43 80 70
72 37 60 60 40
47 89 93 15 73
51 45 87 68 29
145
Prepare a grouped frequency distribution table from the above data taking 10 as class size & one
of the class intervals be 20-30. (30 not included)
12. Find median and mode of the following data.
7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18,7.
13. AD is the altitude of an isosceles triangle ABC, in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects angel A
14. In fig. angle B < angle A and angle C < angle D.
Show that AD < BC.
15. Prove that, in an isosceles triangle , angles opposite
to equal sides are equal. Using above theorem, prove
the following result:
In fig. if AB = AC and DB = DC then show that angle
ABD = angle ACD
16. Construct a frequency polygon for the following data.
Age(in yrs) Frequency
0-5 2
5-10 4
10-15 5
15-20 7
20-25 9
25-30 9
3-35 6
35-40 5
40-45 3
45-50 1
146
CBSE MIXED TEST PAPER-02
(Unit Test)
CLASS - IX MATHEMATICS
[Time : 1.50 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
(iii) Marks allotted to each question are indicated against it.
1. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain. 4 marks
2. In figure POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between ray
OP and OR. Prove that angle ROS = ½ [angle QOS – angle POS] 4 marks
3. Prove that the sum of the angles of a triangle is 1800. 4 marks
4. In figure, the side QR of triangle PQR is produced to a point S. If the bisector of angle PQR and
PRS meet at point T, then prove that angel QTR = ½ angle QPR 4 marks.
5. State and prove ASA congruence rule. 4 marks.
147
6. In fig. AB is a line segment and P is its mid point. D and E are pints on same side of AB such that
angle BAD = angle ABE and angle EPA = angel DPB show that 4 marks
(i) triangle DAP ≅ triangle
(ii) AD = BE
7. In fig AB and CD are respectively the smallest and the longest sides of a quadrilateral AB CD
show that angle A > angle C and angle B > angle D. 4 marks
8. Prove that the diagonals of a parallelogram bisect each other. 4 marks
9. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it
is a square. 4 marks
10. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18. 4 marks
148
CBSE MIXED TEST PAPER-03
(complete test paper)
CLASS - IX MATHEMATICS
[Time : 3.00 hrs.] [M. M.: 100]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
(iii) Questions 1 to 10 carry 3 marks.
(iv) Questions 11 to 20 carry 4 marks.
(v) Questions 21 to 26 carry 5 marks.
Q1. (a) Show that 1..621621….. = 1.621 can be expressed in p/q form where p and q are integers and
0q ≠
(b) Show that 0.235353535 = 0.235 can be expressed in p/q form where p and q are integers and
0q ≠
Q2. Rationalize the denominator of 5 4 2
6 4 2
−−
Q3. Find P(0), P (1) and P(2) for each of the following polynomials
(i)P(y) = y2 – y + 1
(ii) P(t) = -t3 + 2t2 + t+ 2
Q4. Find the reminder when x3 + 3x2 + 3 x + 1 is divided by
(i) x + 1
(ii) x + Π
Q5. Solve the equation graphically 2x + 3y = 7 and 3y + 2x = 8.
149
Q6. In the fig. bisector of angle B and angle D of quadrilateral ABCD meet CD and AB produced at P
and Q respectively prove that: angle P + angle Q = ½ (angle ABC + angle ADC)
Q7. In the fig. PR>PQ and PS bisect angle QPR. Prove that angel PSR > angle PSQ
Q8. Two parallel lines I and m are intersected by transversal p. Show that the quadrilateral formed by
bisector of interior angels is a rectangle.
Q9. Prove two triangles on the same base and between same parallel lines are equal in area.
Q10. A field is in shape of trapezium those parallel sides are 25 m and 10m. The non-parallel sides are
14 m and 13 m. Find the area of trapezium.
Q11. Factorize
(a) x3 – 23x2 + 142 x – 120
(b) 8a3 + b3 + 12 a2b + 6ab2
Q12. Evaluate the following using the suitable identities.
(a) (998)3
(b) (2a – 3b)3
150
Q13. In the fig . POQ is a line. RAY OR is perpendicular
to PQ. OS is another ray lying between ray OP and OR.
Prove that angle ROS = ½ (angle QOS – angle POS)
Q14. ABCD is quadrilateral in which AD = BC and angle DAB = CBA. Prove that
(i) triangle ABD ≅ triangle BAC
(ii) BD = AC
(iii) angle ABD = BAC
Q15. (a) Show that the bisectors of angles of parallelogram form a rectangle.
(b) Show that the diagonals of a square are equal and bisect each other at right angles.
Q16. (a) Prove that the median of a triangle divides a triangle into two triangles of equal area.
(b) In the fig. side QR of triangle PQR is produced
to point to S. If the bisector of angle PQR and PRS
intersect at point P than
prove that angel QTR = ½ angle QPR.
Q17. The sides AB of a parallelogram ABCD is produced to
any point P, a line through A and parallel to CP meets
CB produced at Q and then parallelogram PQBR is
completed. Show that ar(ABCD) = ar(PBQR).
Q18. (a)In an triangle ABC,E is the mid point of AD.
Show that ar(triangle BED) = ¼ ar(triangle ABC)
(b) In the fig. D is a point on side BC of triangle ABC.
Such that AD = AC. Show that AB > AD.
Q20. In a parallelogram ABCD, E and F are the mid point of
sides AB and CD respectively. Show that the line segment
AF and EC trisect the diagonal BD.
151
Q21. (a) Prove that the um of the angles of a triangle is 1800.
(b) Find angle x and angle y in the following figures.
Q22. (a) A rhombus shaped field has green grass for 18 cows to graze. If each side of rhombus is 30 m
and longer diagonal is 48m. How much area of grass field will each cow be grazing?
(b)An umbrella is made by stitching 10 triangular pieces of cloth of two different colors, each
piece measuring 20m, 50cm, and 50cm. How much cloth of each color is required for the
umbrella?
Q23. (a) In the fig. P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(triangle APB) + ar(triangle PCD) = ½ ar(ABCD)
(ii) ar(triangle APB) + ar(APCD) = ar(triangle APD) + ar(triangle PBC)
(b) In the Fig. ar(triangle DRC) = ar(triangle DPC) and ar(triangle BDP) = ar (triangle ARC). Show
that both the quadrilaterals ABCD and DCPR are trapeziums.
152
Q24. (a)Prove that line segment joining the mid point of two sides of a triangle is parallel to the third
side and is half of it.
(b)ABCD is a trapezium in which AB parallel CD and AD = BC. Show that
(i)Angle A = Angle B
(ii) Angle C = Angle D
(iii) Triangle ABC ≅ triangle BAD
(iv) Diagonal AC = Diagonal BD.
Q25. (a) Triangle ABC and Triangle DBC are two isosceles triangles on the same base BC. Show that
triangle ABD ≅ triangle ACD
(b) AD is altitude of an isosceles tingle ABC in which AB = AC show that
(i) AD bisects BC
(ii)AD bisects A
Q26. In the following figure the sides AB and AC of a triangle ABC are produced to point E and D
respectively. If bisectors BO and CO of angle CBE and angle BCD meet at point O. Prove that
angle BOC = 90 – ½ angle BAC .
153
CBSE MIXED TEST PAPER-04
(Unit test paper)
CLASS - IX MATHEMATICS
[Time : 2.00 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
(iii) There are three Section – A, B & C in question paper
Section A : Q no. 1 to 5 carry 2 marks each. Section B : Q. no. 6 to 11 carry 3 marks each. Section C : Q no. 12 to 14 carry 4 marks each.
1. Find two rational numbers between3 4
5 5and
2. Verify whether the following are zeroes of the polynomial: ( ) ,m
P x lx m xl
= + = −
3. In which quadrant of the point (-2,4), (3,-1) lie?
4. In figure, if AC = BD, then prove that AB = CD.
154
5. In quadrilateral ABCD, AC = AD and AB bisects angle A, show that: triangle ABC≅ ABC
6. Express 0.6 in the formp
q where p and q are integers and 0.q ≠
7. Rational the denominator of1
7 6−.
8. Evaluate using identity 104 x 106.
9. Draw the graph of x + y = 7.
10. Show that in a right angled triangle, the hypotenuse is the longest side.
11. The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of quadrilateral.
12. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
13. Factories: X3 – 3x2 =-9x- 5 Or
In fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays
OP and OR. Prove that: Angle ROS = ½ (angle QOS – angle POS)
14. Prove sum of the angles of a triangle is 1800.
155
CBSE MIXED TEST PAPER-05
(Unit test paper)
CLASS - IX MATHEMATICS
[Time : 2.00 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
(iii) There are three Section – A, B C D in question paper
Section A : Q no. 1 to 5 carry 1 marks each. Section B : Q. no. 6 to 9 carry 2 marks each. Section C : Q no. 10 to 14 carry 3 marks each. Section D : Q no. 15 to 16 carry 6 marks each.
1. State the axiom used to prove AB + BC = AC in figure.
2. In train gel ABC and triangle PQR are to be made congruent in figure, Name one additional pair of
corresponding part and the congruence criteria used.
3. Find ‘x’ and ‘y’ in figure
156
4. Arrange the data given under and find its range:
8, 7, 3, 9, 6, 2, 1, 1, 4, 4, 3
5. Find the mode of : 1, 2, 3, 4, 2, 3, 2, 4, 5 and 2.
6. The following distribution is arranged in ascending order. If the median of the data is 63. Then find
the value of x : 29, 32, 48, 50, x, x + 2, 72 , 78, 84 , 95.
Or
The mean of the date given under is 9 find ‘x’
X i 4 6 9 10 15
f i 5 10 10 x 8
7. In fig. x + y = w + z, then show that AOB is a line
8. Prove that every line segment has one and only one mid point. Or
Prove that an equilateral triangle can be constructed on any give line segment .
9. In fig. PR > PQ and PS bisects angle PPR . Prove that angle PSR > angle PSQ.
10. Prove that the sum of three angles of a triangle is 1800.
11. Twenty student of class IX were asked about the number of hours they watched TV programmes in
the previous week. The result was found as follows:
2 13 11 9 17
4 5 12 2 7
5 14 12 6 10
10 18 5 4 9
Make a grouped frequency table for this data taking class width 4 and one of the intervals as 4-8.
12. A random survey of number of persons of various age group present in a park was found as
follows:
157
Age group: 1-10 11-20 21-30 31-40 41-50
No. of persons 12 29 33 19 7
Answer the following (i) class size of the data (ii) class mark of 21-30. (iii) true class limit of the
class with maximum frequency.
13. ABCD is a quadrilateral in which AD = BC and angle DAB = angle CBA. Prove that BD = AC.
14. If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are
parallel, then prove that the two lines are parallel.
Or
In fig. show AB parallel EF.
15. In a city the weekly observation made in a study on the cost of living index are given in table:
Cost of living index 140 – 150 150-160 160-170 170-180 180-190 190-200 Total
No. of weeks 7 10 14 12 6 3 52
Draw histogram and frequency polygon of this data.
16. Prove that the angle opposite to equal sides of isosceles triangle is equal. Use it to show angle
ABD = angle ACD when AB = AC and BD = CD
Or
Prove that: Two triangles are congruent if two angels and the included side of one triangle are
equal to two angles and the included side of another triangle.
158
CBSE MIXED TEST PAPER-06
(Unit test paper)
CLASS - IX MATHEMATICS
[Time : 1.5 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
Section – A
1. Give an example such that sum of two irrational numbers is a rational number. 1-mark
2. Write the value of up to three decimal places. 1-mark
3. Write the degree of polynomial 2 – y2 – y3 + 2y8 1-mark
4. Represent. 1.25 on the number line. 1-mark
5. Why (x, y) (y, x) 1-mark
Section - B
6. What do you mean by ordered pairs 2-marks
7. Expand (1 + X)3 2-marks
8.
If x + y + z = 0 then write the value of x
x + y3 + z3 – 3xyz.
2-marks
159
9. Write the factors of 12 ky2 2-marks
Section - C
10. Represent on the number line. 3-marks
11. Show that can be expressed in the form. 3-marks
12. Visualize on the number line upto 4 decimal places. 3-marks
13. Renationalize the denominators of
3-marks
14.
Define zero of a polynomial. Also find zeros of polynomial
P(x) = x2 – 5x + 6
3-marks
Section - D
15.
Draw the graph of
X + 2y = 5 and
2x + 3y = 12 on the same graph.
Using graph Check whether x = 1, y = 2 is the solution of x + 2y = 5
6-marks
16.
State remainder theorem and factor theorem.
Factorize
X3 – 23x2 + 142x – 120
Using factor theorem.
6-marks
160
CBSE MIXED TEST PAPER-07
(Unit test paper)
CLASS - IX MATHEMATICS
[Time : 1.5 hrs.] [M. M.: 40]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
Section - A
1. Insert two rational and two irrational numbers between .
2-marks
2. Represent on same number line. 2-marks
3. (i) Rationalise
(ii) Find
2-marks
4. Find the value of polynomial 5x – 4x2 + 3 at x = -2 2-marks
5. Factorise 3x2 – x – 4
Or
Factorise 27y3 + 125 x3
2-marks
6. Evaluate using identities 97 x 105. 2-marks
7. Write two solutions of 2x + 3y = 4. 2-marks
161
8. Express in form .
2-marks
Section - B
9. Find remainder, using remainder theorem, which x3 + 3x2 + 3x + 1 is divided by
.
3-marks
10. If x-4 is a factor of 5x3 – 7x2 – kx – 28 find value of k. 3-marks
11. Plot points P (0,0), Q (5,6), R (7,3) S(2,3) on cartesian plane and identify the
figure PQRS.
3-marks
12. Give geometrical representation of 2x + 9 = 0 as an equation
(i) In one variable
(ii) In two variable
3-marks
Section - C
13. Using factor theorem factorise x3 – 3x2 – 9x – 5 4-marks
14. Draw graph of x + 3y = 6 and find points where it meets
(i) x-axis
(ii) y-axis
4-marks
15. Evaluate 4-marks
162
CBSE MIXED TEST PAPER-08
(Unit test paper)
CLASS - IX MATHEMATICS
[Time : 3.0 hrs.] [M. M.: ]
General Instructions:-
(i) All questions are compulsory.
(ii) There is no overall choice. However internal choice has been provided.
1. Write the value of
1-marks
2. Find the product of 1-marks
3. Find the value of 5x3-4x2+3 at x = 2 1-marks
4. The cost of a note book is twice the cost of a pen. Write a linear equation in two
variables for the statement.
1-marks
5. Express 5 = 2x in the form of ax + by=c = 0 and indicate the value of a, b and c. 1-marks
6. Factorize: - x2 – y2 1-marks
7. Check whether (4, 0) is a solution of equation x – 2y = 4 or not? 1-marks
8. Explain parallel and perpendicular line with diagram. 1-marks
9. Represent (625)-4 as positive number in lowest form. 1-marks
10. Check whether is form of p(x) = lx + m or not?
1-marks
11. and are two isosceles triangle on the same base BC, show
that angle ABC = angle ACD
2-marks
12. In figure, if AC = BD then prove that AB = CD 2-marks
163
13. Represent or on a number line. 2-marks
14. Find the value of (-15)3 + (8)3 + (7)3 without actual multiplication 2-marks
15. In the given figure, AB parallel DE, angle BAC = 350, angle CDE = 530 E, other find
angle DCE = ?
2-marks
16. (1)Find four rational number between 3 and 4.
(2)Evaluate 0.265 (express in p/q form).
3-marks
17. Find a and B if
3-marks
18. Find the value of k if x-1 is a factor of 4x3 + 3x2 – 4x + x. 3-marks
19. AB and CD are respectively the smallest and largest side of quadrilateral ABCD
than show that <A> angle C and <B> angle D <A> (OR)
If the bisectors of angle B and angle C of a triangle ABC meet at o the prove that
BOC = 900 + angle BAC
3-marks
20. In figure, AC = AE, AB = AD and angle BAD = angle EAC that show that BC = DE 3-marks
21. Write all the postulates of Euclid in geometry. 3-marks
22. Give the geometrical representation of y = 3 in one variable and two variable on
your answer sheet.
3-marks
23. Triangle ABC is an isosules triangle in which AB = AC side BA is produced to D 3-marks
164
such that AD = AB then show that angel BCD = 900
24. Factories (i) 3x2 – x – 4 (ii) 4y2 – 4y + 1 3-marks
25. Evaluate the factors of (i) 27y3 + 125Z3 (ii) 8x3 + y3 + 27Z3 – 18xy2 using
suitable identify:
3-marks
26. “The sum of any two side of a triangle is greater than the third side” prove it. 5-marks
27. The taxi fare in a city is comprises of the first kilometer fare an is 8 and far the
subsequent distance it is Rs5 per km. write a linear equation for it taxing the
distance covred as x km and total fare a Rs y. Also draw its graph by taking at
least four solution of it.
5-marks
28. Factories p(x) = x3 + 13x2 + 32x + 20 completely using the factor theorem. 5-marks
29. (i)AT line XY and MN intersect at o and angle POY = 900 with a: b = 2:3 than find
the valued a, b, and c.
(ii)Find the remainder when x3 – 9x2 + 6x – a is divided by (x – a)
(iii)Factories 4x2 - y2 + Z2 – 4xy – 2yZ + 4xZ using identify.
5-marks
30. “If a traversal intersects two parallel line than each pair of alternate introvert
angle is equal” Prove it. Hence evaluate x in the figure if AB||CD and angle AMO =
50o, angel CNO = 45o
5-marks
165
CBSE MIXED TEST PAPER-09
(Unit Test)
CLASS - IX MATHEMATICS
[Time : 3.00 hrs.] [M. M.: ]
General Instructions:-
1. All questions are compulsory.
2. Q. No. 1 to 10 carry 1 mark each. Q.No. 11 to 15 carry two marks, each. Q.No. 16 to 25 carry 3
marks and Q. No. 26 to 30 carry 6 marks each.
SECTION A
1. Find two irrational numbers between 2 and 3.
2. Evaluate (99)2 using some identity.
3. Express1
7 into decimal form.
4. Simplify:
51 2516
−
5. Simplify: 2 4x y−
6. Plot the points A(-2,3) and B(2,-3) in Cartesian Coordinate system.
7. How many lines can pass through two points?
8. Find the degree measure of angle A in right angled isosceles triangle ABC, right angle ed. At B.
9. If l 1 and l2 are parallel, then find x.
166
10. A triangle and a �gm have common base and lie between same parallel lines. What is the
relationship between their areas?
SECTION B
11. Express 2.765 into p/q form of a rational number.
12. Rationalize the denominator of :1
5 3 2−
13. Factories: 2x2 – 3x + 1.
14. Using factor theorem, find whether x – 2 is a factor of 2x4 – 5x3 + 2x2 x + 2?
15. Prove that any exterior angle of a triangle is equal to the sum of its interior opposite angles.
SECTION C
16. Using remainder theorem, find the remainder when x3 – 6x2 + 2x – 4 is divided by x + 1.
Check your answer by actual division.
17. Represents 6 on the number line.
18. Factories: 64x3 – 27y3 + 8z3 + 72 xyz.
19. Two parallel lines are intersected by a transversal. Prove that the bisector of any pair of alternate
interior angles are also parallel.
20. Prove that he diagonals of rhombus are perpendicular to each other.
21. If the altitude from two vertices of a triangle are equal, then the triangle is isosceles. Prove.
22. Prove that the sum of any two sides of a triangle is always greater than the third side.
23. P and Q are any two points lying on the sides DC and AD respectively of �gm ABCD. Show
that ( ) ( )ar APB ar BQC∆ = ∆
24. Construct a frequency polygon for the following data:
Age: (in year) 0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16
Frequency 2 4 6 8 10 6 5 3
25. Find the mean of the following distribution:
Wt.(in Kg) 67 70 72 73 75
No. of students 4 3 2 2 1
167
SECTION D
26. Prove that the triangles on the same base and between same parallel lines are equal in area.
27. If a transversal makes equal intercepts on three parallel lines then prove that they make equal
intercepts on any other transversal.
28. Prove that the four triangles formed by joining the midpoints of the sides of a triangle are
congruent.
29. In fig., PS is the bisector of QPR∠ and PT ⊥ QR.
Show that 1
( )2
TPS Q R∠ = ∠ − ∠
30. Factories: (a+b)3 + (b + c)3 (c + a)3
-3(a+b) (b+c) (c +a)
168
CBSE MIXED TEST PAPER-10
(Second Unit Test)
CLASS - IX MATHEMATICS
[Time : 3 hrs.] [M. M.: 80]
General Instructions:-
1. All questions are compulsory.
2. The question paper consists of thirty questions into 4 sections. A, B, C and D. Section-A
comprises of ten questions of 1 mark each, Section B comprises of five questions of 2
marks each, Section C comprises of ten questions of 3 marks each and Section D comprises
of five questions of 6 marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exact
requirement of the questions.
4. There is no overall choice. However, internal choice has been provided in one question of
2 marks each, three questions of 3 marks and two questions of 6 marks each. You have to
attempt only one of the alternative in all such questions.
5. Use of calculator is not permitted. However you may ask for mathematical tables.
6. In questions on construction (if any), drawing should be neat and exactly as per the given
measurements.
7. Cancel the previous question, if attempted again, for any reason.
8. Use two graph paper one for Q. No. 17 & 18 and another for Q.No. 28 & 30.
Section A
1. Rationalize the denominator of1
.7 6−
2. Find the zero of the polynomial p(x) = 2x + 5.3. What are the coordinates of the origin?
4. Write 4 = 5x – 3y in the form of ax + by +c = 0 and indicate the value of a, b and c.
5. Find the value of x and y in the given figure if AB parallel CD.
169
6. Write the statement of SAS congruence rule.
7. Write the condition for a quadrilateral to be parallelogram.
8. If two figures A and B are congruent then write a relation between the area for the figure A
and B.
9. In figure, angle ABC = 700, angle ACB = 300 Find angle BDC.
10. Find the mode of the following marks (out of 10) obtained by 20 students:
4,6,5,9,3,2,7,7,6,5,4,9,10,10,3,4,7,6,9,9
Section B
11. x3 – ax2 +6x – a is divided by (x-a).
12. In figure line XY and MN intersect at O. If angle POY = 900 and a : b = 2 : 3 find c.
13. In figure AD and BC are equal perpendicular to a line segment AB. Show that CD bisect AB.
170
14. In figure, ABCD is a parallelogram, AE perpendicular DC and CF perpendicular AD, if AB =
16 cm, AE = 8 cm and CF = 10 cm, find AD.
15. In figure, angle PQR = 1000, where P, Q, R are points on a circle with centre O, find angle
OPR.
Or
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the
chord at a point on the minor arc and also at point on the major arc.
Section C
16. Represent 9.3 on the number line.17. Locate the points (5,0), (0,5), (2,5), (-3,5), (-3, -5) and (5, - 3) in the Cartesian plane.18. Draw the graph of the linear equation in two variables. X + y = 419. In figure if AC = BD then prove that AB = CD.
20. In figure, the sides AB and AC of triangle ABC are produced to points E and D respectively. Ifthe bisector BO and CO of angle CBE and angle BCD respectively meet at point O, then prove
that 0 190
2BOC BAC∠ = − ∠
171
Or
Given a triangle ABC, if the bisector BO and CO of angle ABC and angle ACB respectively
meet at point O, then prove that angle BOC = 0 190
2+
21. Show that in a right angled triangle, the hypotenuse is the longest side.
22. ABCD is a quadrilateral in which P, Q, R and S are mid points of the sides AB, BC and DA.
AC is a diagonal, show that:
(i) SR �AC and SR = 1
2AC
(ii) PQ = SR
(iii) PQSR is a parallelogram.
Or
ABC is a triangle right angled at C. A line through the mid point M of the hypotenuse AB and
parallel to BC intersect AC at D. Show that
(i) D is the mid point of AC
(ii) MD AC⊥
(iii) 1
2CM MA AB= =
172
23. In figure E is any point on median AD of a triangle ABC. Show that ( ) ( )ar ABe ar ACE=△ △
24. Prove that : Equal chords of a circle (or of congruent circles) are equidistant from the centre(centers). Or Prove that: Chords equidistant from the centre of a circle are equal in length.
25. The following observations have been arranged in ascending order. If the median of data is 63.Find the value of x. 29, 32, 48, 50, x, x+2, 72, 78, 84, 95.
Section D
26. Express 1.27 in the form ofp
q, where p and q are integers and 0.q ≠
27. Factories : x3 + 13x2 + 32x + 20.28. The taxi fare in a city is as follows: For the first kilometer, the fare is Rs. 8 and for the
subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs. y. Write a linear equation for this information and draw its graph.Or Give the geometric representation of y = 3 as an equation. (i) In one variable (ii) In two variables.
29. State and prove that the mid-point theorem.Or
30. Prove that: The line drawn through the mid-point of one side of a triangle, parallel to anotherside bisect the third side.
31. The following table gives the number of marks obtained by 50 students.
Draw a histogram and a frequency polygon for the above data on the same graph.
Marks No. of students 0-20 20-40 40-60 60-80 80-100
4 11 23 8 4
Total 50
173
CBSE MIXED TEST PAPER-11
(Unit Test)
CLASS - IX MATHEMATICS
[Time : 3 hrs.] [M. M.: ]
General Instructions:-
1. All questions are compulsory.
2. The question paper consists of 30 questions into 4 sections.
3. Section A comprises of 10 questions of 1 mark each.
Section B comprises of 5 questions of 2 mark each.
Section C comprises of 10 questions of 3 mark each.
Section D comprises of 5 questions of 6 mark each.
4. There is no over all choice. However internal choice has been provided ‘1’ question of 2
marks , 3 questions of 3 marks each and 2 questions of 6 marks each.
Section A
1. Rationalize the denominator of1
7.
2. Find the value of 1
3(125) .
3. Write the coefficient of x2 in -7x4 + 8x3 -3x2 + x + 1.
4. Find the value of the polynomial 5x – 4x2 + 3 at x = -1.
5. Write the name of point where x axis and y axis intersect.
6. What is the name of horizontal and the vertical lines drawn in the cartesion plane.
7. Coordinates of a point P(-3,-7) write the value of abscissa and ordinate.
8. Express 3x + 9 = 0 as an equation in two variables.
9. 0, , 60 ,ABC AB AC b find A= ∠ = ∠△
10. The points scored by a term in a series of matches are as follows. 19, 3, 18, 17, 7, 5, 3, 4. Findthe median of the points scored by the team.
Section B
11. Express 0.47 in form ofp
q
12. Show how 5 can be represented on the number line.13. In which quadrant or on which axis do each of the point lie? (-3,7), (0,4), (-8,0),(3,4)
14. ABC is a triangle in which BC is produced to DCA, is produced to E0 0105 , 126 ,DCA BAE Find ABC∠ = ∠ = ∠
174
Or
Calculate the value of x in the following fig.
15. Find the mode of 15, 16, 0, 3, 2, 13, 14, 15, 9, 15, 13, 12, 15, 7, 15 by arranging in ascending
order.
Section C
16. Rationalize the denominator of1
7 6−17. By division method find the remainder when x3 + 3x2 + 3x + 1 is divided by 2x + 5.18. Use the factor theorem to determine whether g(x) is a factor of p(x).
p(x)=x3 – 4x2 + x + 6, g(x) = x-3 Or Find the value of K, if x-1 is a factor of p(x) = kx2 – 3x + k.
19. Give the equations of three lines passing through (2,14).20. If the points (3,4) lies on the graph of the equation 3y = ax + 7, then find the value of a?
Or Check which of the following are solutions of the equation x-2y = 4 and which are not (i) (0,2) (ii) (2,0)
(iii) ( 2,4 2)
21. Locate the points (5,8), (0,5), (2,5), (5,2), (-3,0), (8,0) in the Cartesian plane.
22. Evaluate:
1
4625
81
−
23. ABC and DBC are two isosceles triangles on the same base BC. Show that angle ABD = angleACD
175
24. In above fig E is any point on median AD of a triangle ABC. Show that ( )ar ABe ar AEC=△
Or
In triangle ABC, E is the mid-point of median AD show that 1
4ar BED arABC=△
25. Thirty children were asked about the number of hours they played in the previous week . Theresults were found as follows. 1,6,2,3,5,13,5,8,4,5 10,3,8,5,2,8,15,1,13,6 3,2,4,12,6,7,9,7,14,13 Make a grouped frequency distribution table for this data, taking one of the class intervals as 5-10.
Section D
26. The length of 40 leaves of a plant are measured correct to one millimeter and the obtained datais represented in the following table.
176
Length (in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2
27. A taxi fare in a city is as follows: For the first kilometer, the fare is Rs 8 and for the subsequentdistance it is Rs 5 per km. Taking a distance covered as x km and total free as Rs 4 write a linear equation for this information and draw its graph.
28. Prove that a diagonal of a parallelogram divides it into two congruent triangles. By using abovetheorem. �gm ABCD, diagonal AC bisects angle A, Prove that it also bisects angle C. Or Prove that parallelogram on the same base and between the same parallels are equal in area By sing above using theorem. Find the area of �gm. ABEF is 50 squcm. If both are on the same base AB and between the same parallels FC � AB.
29. Show that the quadrilateral formed by joining the mid – points of the consecutive sides of arectangle is a rhombus.
In above fig AB and Cd are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ( ) ,
( )
i A C
ii B D
∠ > ∠∠ > ∠
30. a. Prove that the sum of the three angles of a triangle is 1800. b. The angles opposite to equal sides are equal prove it.
177
CBSE MIXED TEST PAPER-12
(Half Yearly Test)
CLASS - IX MATHEMATICS
[Time : 3 hrs.] [M. M.: ]
General Instructions:-
1. All questions are compulsory.
2. Questions are divided into four sections A, B , C and D. Marks are indicated in brackets.
Section A
1.What property q must satisfy such thatp
q is terminating. 1mark
2. Why 3 2t t+ is not a polynomial? 1mark
3. Write the co-ordinates of a point which lies 4 units above x-axis and lies on the y-axis.1mark4. Give a linear equation passing through (2, 11). 1mark5. Find angle ABC in the given figure. 1mark
6. Find the angle ∠ AOC in the figure. 1mark
7. Write two conditions that a triangle can be formed. 1mark8. In figure, identify one triangle and a quadrilateral on the same base and between the same paralles. 1mark.
AB CD�
9. Factorize : x3 + 8y3.
178
10. Find the mean of first four prime numbers. 1 mark
Section B
11. Represent 3 on number line. 2 marks
12. Check whether x = 1
3− and x =
2
3are zero of polynomial. 2 marks
p(x) = 3x2 – 1 13. Find the value of k if X = 2 and y = -1 is a solution of equation 2x + 3y = k. 2 marks14. ABCD is a parallelogramAE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm find AD. 2 marks 15. In an equilateral triangle prove that median is ⊥ to the opposite side. 2 marks.
Section – C
16. Simplify:5 2 6 2
7 3 2 7 3 2+
− + 3 marks
17. Divide: 3x4 – 4x3 – 3x – 1 by x – 1. 3 marksOr Factories: 6x2 + 5x – 6 18. Locate the points : (5, 0), (2, 5), (-3, 5) and (-3, -5). Also write its quadrant. 3 marks.
19. In the given figure, prove that 1
.2
QTR QPR∠ = ∠
Given QR of triangle PQR is produced to a point S. If the bisector of angle PQR and angle PRS meet at point T. 3 marks 20. In fig. POQ is a lies Ray, OR is perpendicular to PQ. OS is another ray lying between rays OP and
OR. Prove that 1
( )2
ROS QOS POS∠ = ∠ − ∠ 3 marks
179
21. Verify:
x3 + y3 + z3 – 3xyz = 1
2(x + y + z) x [(x – y)2 + (y-z)2 + (z – x)2]. 3 marks.
22. In triangle ABC, AB = AC. D and E are points on BC such that BE = CD. Show that AD = AE. 3marks 23. Show that the bisectors of angles of a parallelogram form a rectangle. 3 marks.24. Show that a median of triangle divides it into two triangles of equal areas. 3 marks25. Marks obtained by 10 students are given below:10, 20, 36, 92, 95, 40, 50, 60, 70, 10. Find its Median and mode. Or Median of data 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 arranged in increasing order is 63. Find x. 3 marks.
Section – D
26. State factor theorem. Using factor theorem factories.x3 – 23x2 + 142x -120. 6 marks 27. Draw graph of the linear equations x + y = 5 and 2x + 3y = 7. Also find its point of intersection. 6marks.
28. State and prove angle sum property of a triangle. 6 marks
Using above statement find the value of x and y in figure. Or State and prove ASA congruence rule. Using ASA prove AOB DOC≅△ △ in fig.
Given angle A = angle D = 900. OA = OD
180
29. In triangle ABC, D, E and F are respectively the mid points of sides AB, BC and CA respectively.Show that triangle ABC is divided into four congruent triangles by joining D, E and F and
1( ) ( )
4ar DEF ar ABC=△ △
Or In fig. Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD, If AB = CD, then show that
( ) ( ) ( )
( ) ( ) ( )
i ar DOC ar AOB
ii ar DCB ar ACB and
==
△ △
△ △
(iii) ABCD is a parallelogram.
30. 100 surname were randomly picked up and information arranged is alphabet as given below intabular form:
No. of letters No. of surname 1-4 4-6 6-8 8-12 12-20
6 30 44 16 4
(i)Draw a histogram and frequency polygon for above information. (ii) Write the interval in which max. number of surnames lies. 6 marks.
181
CBSE MIXED TEST PAPER-13
(Half Yearly Test)
CLASS - IX MATHEMATICS
[Time : 3 hrs.] [M. M.: 80]
General Instructions:-
1. All questions are compulsory. There are four sections A, B , C and D.
Section A: Q 1 to 10 carry 1 mark each.
Section B: Q 11 to 15 carry 2 marks each.
Section C: Q 16 to 25 carry 3 marks each.
Section D: Q 26 to 30 carry 6 marks each.
2. Internal choices have been provided in some questions.
Section A
1. Write the Zero of polynomial p(x) = 2x + 5.
2. Write the equation of x-axis
3. Factorize x2 – 81y2
4. If AB = AC and angle ACD= 1200. Find angle A.
5. Find range of data 70, 65, 71, 36, 55, 61, 62, 41, 40, 39, 35.
6. Is ( 3 4)( 3 4)+ − a rational or irrational number?
7. If B lies between A and C and AC = 15cm, BC= 9cm. Find AB2.8. Write an irrational number between 1.3, and 1.4.9. Find value of k if x=2, y = 1 is a solution by 2x + 3y = k.
10. Find the value of 3
416
182
Section B
11. Factorize: 2 72
16x x− +
Or 2 5 3 30x x− +
12. Find the value of a if (2x-4) is a factor of 2x3 + ax2 + 11 x + a + 2.13. If AB CD� , find value of x and y.
14. Find two rational numbers between5 1
7 11and
15. ABCD is a parallelogram, AE perpendicular DC and CF perpendicular AD. If AB = 16 cm, AE= 8cm, CF = 10 cm. Find AD.
Section C
16. Represent 2, 3 5and on same number line.17. (i) In fig find x
183
(ii)
0
,
126
.
AB CD EF CD
GED
Find AGEand GEF
⊥
∠ =∠ ∠
�
18. ABCD is a rectangle, P, Q, R, S are mid points of sides AB, BC, CD and DA respectively.Show that the quadrilateral PQRS is a rhombus.
19. Give geometric representation of y = 4 as an equation (i) in one variable (ii) in two variable.
20. AB and CD are respectively the smallest and longest sides of Quadrilateral ABCD show that( )
( )
i A C
ii B D
∠ > ∠∠ > ∠
21. Calculate the mean and mode of followingx : 10 30 50 70 89 f: 7 8 10 15 10
22. Rationalize:5
7 5−23. Line l is bisector of angle A and B is any point on i. BP and BQ are perpendicular from B on
arms of angle A. Show that
(i)APB AQB≅△ △
(ii)B is equidistant from arms of angle A.
24. The median of following observation arranged in ascending order is 25. Find x11, 13, 15, 19, x+2, x+4, 30, 35, 39, 46. Or If mean of 10, 12, 18, 13, P and 17, is 15. Find value of P.
184
25. ABC and ABD are two triangles on same base. AB. If lines segment CD is bisected by AB at O.Show that
( ) ( )ar ABC ar ABD=△ △
Section D
26. Prove that parallelogram on same base and between same parallel lines are equal in area.27. Draw the graph of equation 2x + y = 5.
From the graph find two more solution from the graph find y, if x = 3. 28. Factorize x3 – 23 x2 + 142x + 120.29. The time taken in seconds to solve a problem for 25 pupils is as follows.
16,20,26,27,28,30,33,37,38 46,46,48,49,50,53,58,59,60 40,64,42,52,43,20,46
30. (i)Construct a frequency distribution for the above data using class interval of 10 seconds.31. (ii)Draw a histogram and frequency polygon to represent the frequency distribution32. (i)Prove that sum of angles of a triangle is 1800.33. (ii)In fig
0 0, 70 , 100
min &
PQ RS PAB ACS
Deter e ABC BAC
∠ = ∠ =∠ ∠
�
185
Sample Paper-I Half Yearly Examination
Subject: Mathematics Class: IX
Time Allowed: 2½Hours Max: Marks:
General Instructions: (1). All questions are compulsory. (2). Section- A Contains 10 Questions of 1 marks each. (3). Section- B Contains 05 Questions of 2 marks each. (4). Section -C Contains 10 Questions of 3 marks each. (5). Section- D Contains 05 Questions of 6 marks each.
SECTION –A
Q1. Find two rational numbers between 3 and 4.
Q2. Write 3/13 in decimal form.
Q3. Find the remainder when x³+3x²+3x+1 is divided by x+1.
Q4. Find the product by using suitable identity:- (x+4) (x-10)
Q5. If ∆ABC=∆PQR, <B=55˚ and <C= 95˚, find < P
Q6. Check whether x=-1 and x=2 are the Zeros of the polynomial p(x) = (x+1) (x-2)
Q7. The abscissa of a point on the y-axis is ………………….
Q8. Find: (125) ⅓
Q9. In ∆ABC, AB=BC=5cm and angle A=55˚, find angle B
Q10. Write a linear equation in two variables for the statement:- Cost of a notebook is twice the cost of a pen.
186
SECTION-B
Q11. Draw the graph of following linear equation in two variables. y = 2x+4
Q12. Find the value of k, if x-1 is a factor of 2x²+kx+√2 __
Q13. Show that1.27 can be expressed in the form p/q, where p and q are integers and q ≠ 0.
Q14. Show that the sum of the Angles of a triangle is 180˚.
Q15. Rationalize the denominator of : _ 1____ √7- √6
SECTION-C
Q16. If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD, Show that ar (EFGH) = ½ ar(ABCD)
Q17. Represent √3 on number line
Q18. Factorise: 3√3x³+2√2y³
Q19. Expand :- (-2x+3y+2z) ² by using a suitable identity.
Q20. Plot the points (4, 0), (2, 5) and (4,-3) in the Cartesian plane by using a graph paper.
Q21. Factorise:- 8x³+27y³-z³+18xyz
187
Q22. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.
A
B C
F
E
Show that i) ∆ABE= ∆ACFii ) AB = AC
Q23. In given figure if lines PQ and RS intersect at point T, such that <PRT =40˚, <RPT=95˚and <TSQ = 75˚, find <SQT.
Q24. Divide the polynomial 3x³-4x²-3x-1 by x-1
Q25. Factorise:- 27y³+125z³ by using the identity a³+b³= (a+b) (a²-ab+b²)
P
R S
Q
T 75˚
95˚
40̊
188
SECTION-D
Q26. ∆ABC and ∆DBC are two isosceles triangles on the same base BC. Show that < ABD = <ACD.
Q27. Factorise: x³+13x²+32x +20 Q28. Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two sides and the included side of other triangle. Q29. Verify that: x³+y³+z³-3xyz = ½(x+y+z) [(x-y)² + (y-z)² + (z-x)²] Q30. Show that the diagonals of a rhombus are perpendicular to each other. …………………………………………
A
B C
D
189
Sample Paper-II Half Yearly Examination
Subject: Mathematics Class: IX
Time Allowed: 2½Hours Max: Marks: General Instructions: (1). All questions are compulsory. (2). Section- A Contains 10 Questions of 1 marks each. (3). Section- B Contains 05 Questions of 2 marks each. (4). Section- C Contains 10 Questions of 3 marks each. (5). Section- D Contains 05 Questions of 6 marks each.
SECTION –A
Q1. Determine whether x-3 is a factor of x³-4x²+x+6.
Q2. Simplify: (√ 5-√2) (√5+√2)
Q3. Find the zero of the polynomial p(x)= 2x+5
Q4. Find the value of ‘k’ if x=2, y= 1 is a solution of the equation 2x+3y=k.
Q5. Find the remainder when x³-ax²+6x-a is divided by x-a
Q6. Find the product by using suitable property: (3-2x)(3+2x)
Q7. Evaluate by using suitable property: (99)³
Q8. In the given figure if AC=BD, then prove that AB = CD.
A
BC
D
Q9. Simplify: 7½. 8½
Q10. If the point (3, 4) lies on the graph of equation 3y=ax+7, then find the value of ‘a’
190
SECTION-B
Q11. Draw the graph of following linear equation in two variables. y = 2x-4 Q12. Find the value of k, if x-1 is a factor of p(x) = kx²-√2x+1 ___ Q13. Show that 0.235 can be expressed in the form p/q, where p and q are integers and q ≠ 0. Q14. In the given figure ABCD is a parallelogram, AE┴ DC and CF ┴ AD. If AB=16cm, AE=8cm and CF= 10cm. Find AD
A B
D C
F
E
Q15. A diagonal of a parallelogram divides it into two congruent triangles.
SECTION-C
Q16. Represent √5 on number line Q17. Factorise: 64m³ -343n³ Q18. Factorise:- 8x³+y³-27z³+18xyz Q19. Visualise 3.765 on the number line, using successive magnification. Q20. Give two different irrational numbers between 5/7 and 9/11. Q21. In a triangle ABC, E is the mid point of median AD. Show that ar( BED)= ¼ ar(ABC)
191
Q22. Solve the equation 2x+1=x-3, and represent the solutions on (i) The Number line (ii) The Cartesian plane.
Q23. Prove that the angles opposite to equal sides of an isosceles triangle are equal.
Q24. If x+y+z=0, Show that x ³ + y³+ z³ =3xyz
Q25. Rationalise the denominator of the: ___ 1_____ 7-3 √2
SECTION-D
Q26. Expand by using suitable identity: (-2x+5y-3z)²
Q27. In given, figure diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD if AB = CD, then show that
(i) ar (DOC)= ar(AOB) (ii) ar (DCB= ar(ACD) (iii) DA║ CB or ABCD is a parallelogram.
D A
C B
O
192
Q28. Factorise: 2x² + y ²+ 8z²-2 √2 xy + 4√ 2yz – 8xz Q29.Factorise: x³-23x²+142x-120 Q30. In the figure, PR >PQ and PS bisects < QPR. Prove that angle PSR > angle PSQ.
************************
P
Q R S
193
CBSE MIXED TEST PAPER-19
UNIT TEST
CLASS - IX MATHEMATICS
[Time : 11/2hrs.] [M. M.: ]
GENERAL INSTRUCTIONS:-
1. Q. No. from 1 to 5 one marks each.
2. Q. No. from 6 to 10 two marks each.
3. Q. No. from 11 to 15 three marks each.
4. Q. No. from 1 to17 five marks each.
Q1. In the give x||y, z ⊥ y, then find P.
Q2. Find mean of firs five prime numbers.
Q3. Find y, if A, O, b are collinear
Q4. Find mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Q5. The ratio of angles of Quadrilateral are 1:1:1:1 the quadrilateral is ………………
Q6. The mean of 16 numbers is 8. If 2 is added to every number, what will be new mean.
194
Q7. ABC is a triangle in which BE and CF are altitude to sides AC and AB are equal Prove that
∆ ABC ≅ ∆ ACF
Q8. The following data is arrangement in ascending order
19, 20, 36, 46, 50, x – 2, x + 4, 59, 70, 80, 83, 88
If median is 57. Find x
Q9. D is point on side BC of ∆ ABC such that AD = AC show that AB >AD
Q10. If PQ||ST, ∠ PQR = 110o and ∠ RST = 130o Find ∠ QRS
Q11. In the fig BO and CO are bisectors o interior angles ∠ B and ∠ C intersecting at O.
Show that ∠ BOC = 90o +1
2 ∠ BAC
Q12. In a parallelogram ABCD E and F are mid-points of sides AB and CD respectively show that
line segment AF and EC trisect diagonal BD.
195
Q13. Find value of P in the mean of following data 40.4
Q14. ∠ X = 62o, ∠ XYZ = 54o, if YO and ZO are bisectors of ∠ XYZ and ∠ XZY. Find ∠ OZY and
∠ YOZ
Q15. Prove the sum of angles of a triangle is 180o.
Q16. The distribution of weight (in kg) of 100 people given below:
Weight (in kg) Frequency
40-45
45-50
50-55
55-60
60-65
65-70
70-75
13
25
28
15
12
5
2
100
Construct a histrogram and frequency polygon for the above data
Q17. Prove that two triangles are congruent if two angles and an included side of one triangle are
equal to two angles and an included side of other triangle.
Variable (x) 10 20 30 40 50 60 70
Frequency (f) 3 8 12 5 P 7 5
196
CBSE MIXED TEST PAPER-20
HALF YEARLY EXAMINATION
CLASS - IX MATHEMATICS
[Time : 3hrs.] [M. M.: 100]
Q1. Write True of False: [10]
(a) Every natural is whole number.
(b) Every integer is a whole number.
(c) Every rational number is whole number.
(d) Every irrational number is a real number.
(e) Every point on the number line is the form m where m is a natural number.
(f) Every whole number is natural number.
(g) Every integer is a rational number.
(h) Every rational number is an integer.
(i) A parallelogram is a trapezium.
(j) A kite is a parallelogram.
Q2. Rationalise the denominator of 1
.7 3 2+
[3]
Q3. Express as a rational number: 1.27 [3]
Q4. Find p(0), p(1) and p(2) for the polynomial: [3]
p(t) = 2 + t + 2t2 – t3
Q5. Factories: 12x2 – 7x + 1 [3]
Q6. Show that the sum of the angles of a quadrilateral is 360o. [3]
Q7. Find the remainder when x3 + 3x2 + 3x + 1 is divided by x – ½. [3]
Q8. In fig. lines PQ and RS intersect each other at point O if ∠ POR: ∠ ROQ o= 5:7. Find all the
angles. [3]
197
Q9. The following observations have been arranged in ascending order. If the median of the data
is 63, find the value of x. [3]
29, 32, 48, 50, x, x+2, 72, 78, 84, 95
Q10. Fill in the blanks: [10]
(a) (x + y + z) 2 = ………..+…………+……………. +…………..+……………. +
(b) x3 + y3 + z3 – 3xyz = (………..+…………+…………)(………+…………-………….-………..-…………..)
(c) The sides opposite to equal angles of a triangle are ………………….
(d) If two sides of a triangle are unequal, the angle opposite to the longer side is ……………..
(e) If each pair of opposite sides of a quadrilateral is equal, then it is a ………………….
(f) The line segment joining ht mid points of two sides of a triangle is……….. To the ……………
(g) Parallelogram on the same bases and between the same parallels are ……………….
(h) A quadrilateral is a parallelogram if a pair of ……………… is equal and……………..
(i) In any triangle the side opposite to the larger angle is ……………..
(j) Lines which are parallel to a given line are…………… to ………………
Q11. In fig. the side YZ of ∆XYZ is produced to a point P. if the bisectors of ∠ XYZ and ∠ XZP meet
at point Q. then prove that ∠ YQZ = ½ ∠ YXZ. [4]
Q12. In fig. PR>PQ and PS bisects ∠ QPR prove that ∠ PSR> ∠ PSQ.
198
Q13. In fig. if a transversal intersects two lines such that the bisectors of a pair of corresponding
angles are parallel, and then prove that two liens PQ and RS are parallel.
Q14. Prove that parallelogram on the same base and between the same parallels are equal in area.
Q15. The following table gives the distribution of students of two sections according to the marks
obtained by them: [6]
Q16. 100 surnames were randomly picked up from a local telephone directory and a frequency
distribution of the numbers of letter in the English alphabet in the surnames was found as
follows:
(a) Draw a histogram to depict the given information.
(b) Write the class interval in which maximum numbers of surnames lie.
Q17. Show that a median of a triangle divides its who triangles of equal area. [6]
SECTION – A SECTION – B
MARKS FREQUENCY MARKS FREQUENCY
0-10
10-20
20-30
30-40
40-50
3
9
17
12
9
0-10
10-20
20-30
30-40
40-50
5
19
15
10
1
Number of letters Number of surnames
1-4
4-6
6-8
8-12
12-20
4
30
44
16
4
199
Q18. In fig. AP//BQ//CR
Prove that ar(ACQ) = ar(PBR)
Q19. In ∆ABC and ∆DEF
AB = DE, AB//DE, BC = EF and BC//EF vertices A, B and C are joined to vertices D, E and F
respectively show that “
(1) Quadrilateral ABCD is a parallelogram
(2) Quadrilateral BEFC is a parallelogram
(3) AD//CF and AD = CF.
(4) Quadrilateral ACFD is parallelogram
(5) AC = DF
(6) ∆AB≅C ∆DEF
Q20. In fig. ABC is an isosceles triangle with AB = AC. D and E are points on BC such that BE = CD
Show that AD = AE
200