STD: VII EVERWIN VIDHYASHRAM TERM-I MATHEMATICS

CHAPTER I - INTEGERS

Integers are a bigger collection of numbers which is formed by whole numbers and their negatives.

Properties of Integers

Property Addition Subtraction Multication Division

Closure Property

a+b=c (-2)+5=3 Holds good

a-b = c (-4)-3 = (-7) Holds good

a x b = c 3 x4 = 12 Holds good

ab = c

1 (-2) = 1

−2

Does not hold good.

Commutative Properly

a + b = b+a 3+(-7)= (-7) +3 (-4) = (-4) Holds good

a - b # b - a 3-4 # 4-3 -1 # 1 Does not hold good

a x b = b x a 3x (-4) = (-4) x3 Holds good

ab # b a

13 # 3 1 1

3 # 3

Does not hold good

Associative Property

(a+b)+c = a+(b+c) (-3+4)+6 = (-3)+(4+6) 1+6 = (-3)+10 Holds good

(a-b)-c#a-(b-c) (3-4)-6#3-(4-6) -1-6# 3-(-2) -7 # 5 Does not hold good

ax(bxc)=(axb)xc (-2)x[3x(-4)] = [(-2)x3] x(-4) -2 x-12 = -6x-4 Holds good.

(ab)c#a(

bc)

(12)3#

1(23) 1

2 x

1

3 # 1

2

3

1

6 # 1 x

3

2

1

6 #

3

2

Does not hold good

Property of Zero

a + 0 =a 3 + 0 = 0+3 =3 Holds good

a - 0 = a 4 - 0 = 4 Holds good

a x 0 =0 4 x 0 =0 Holds good

0 a = 0

But a 0= not defined

Property of one

a + 1 gives the successor of a 3+1 =4

a - 1 gives the predecessor of a (-4)-1 = -5

ax1 = a (-3)x1= -3 gives the same integer

a 1 = a

3 1 = 3 gives the same integer

Identity Property

Zero is the additive identify a+0 = a = 0+a 3+0 = 3 = 0+3

-

One is the multiplicative identify. a x 1 = a = 1xa 3x1 = 3 = 1x3

-

* Product of a +ve and a -ve integer as a negative integer. * Product of two negative integers is a positive integer. * Product of even number of -ve integers is +ve. * Product of odd number of -ve integer is -ve. * When a +ve integer is divided by a -ve integer, the quotient obtained is a -ve integer and vice-versa. * Division of a -ve integer by another -ve integer gives a +ve integer as quotient. * Under addition and multiplication, integers show a property called distributive property. That is a x (b+c) = a x b + a x c for any three integers a, b and c. Ex 1.1

1) a) Lahulspiti -80

Srinagar -20

Shimla 50

Ooty 140

Bengaluru 220

b) Hottest temp - coldest places 220 - (-80) = 22+8 = 300c c) Temp difference between Lahulspiti and Srinagar. -2(-8)=-2+8=6℃ d) Temp of Srinagar & Shimla = (-20) + 50 = 30c Temp at Shimla = 50c 30c < 50c yes temp of Srinagar & Shimla < temp at shimla. Temp of Srinagar = -20c 30c > -20c No it is not less than the temp at Srinagar. 2) Jack’s total score is 25+(-5)+(-10)+15+10 25-15+15+10=35 3) Temp on Monday =- 50c on Tuesday temp dropped = 20c

-50c - 20c = -70c on Wednesday temp rose up = 4℃ -70c +40c = -30c 4) Height above the sea level = 5000m Floating a submarine below the sea level = 1200m The vertical distance between the plane and the submarine is 5000 + 1200 = 6200m 5) Deposit amount = Rs.2000 Withdrawal amount = Rs. 1642 Balance = 2000-1642 = Rs. 358 Deposit amount is represented by +ve integer.

6) Rita moves east is represented y +ve integer. Rita moves west is represented by -ve integer. Distance from A to B = 20km Distance from B to C = 30km Distance from A to C = 20-30 = -10km

Rita is at final position from A to C is -10km. 7) Maagic square: i) Rows: 5-1- 4 =0 -5-2 +7 = 0 0+3-3=0 Columns: 5-5+0=0 -1-2+3=0 -4+7-3=0 Diagonals: 5-2-3=0 -4-2+0=-6

This is not a magic square. ii) Rows: 1-10+0=-9 -4-3-2=-9 -6+4-7=-9 Columns: 1-4-6=-9 -10-3+4=-9 0-2-7=-9 diagonals: 1-3-7=-9 0-3-6=-9

This box is a magic square. 8) Verify a-(-b) = a+b i) a=21, b=18 LHS RHS

21-(-18) 21+18 21+18 39 39 = 39

Hence verified. ii) a=118 b=125 LHS RHS

118-(-125) 118+125 118+125 243 243 = 243

iii) a =75 B=84 LHS RHS 75-(-84) 75+84

75+84 159 159 = 159

iv) a=28 b=11 LHS RHS 28-(-11) 28+11 28+11 39 39 = 39 Hence verified. a) -12 < -4 b) -15 < -2 c) -7 > -29 d) 0 < 20 e) -101 > -159 10) i)

Jump Step

1 1+3=4 2 4-2=2 3 2+3=5 4 5-2=3 5 3+3=6 6 6-2=4 7 4+3=7 8 7-2=5 9 5+3=8 10 8-2=6 11 6+3=9

In 11th jump it will reach the water level.

ii)

In 5th jump it will reach back iii) a) -3+2-3+2-3+2-3+2-3+2-3+2-3+2-3+2 = -8 b) 4-2+4-2+4-2+4-2 = 8 EX 1.2 1)a) -4+(-3) = -7 12+(-5)=-7 b) 10-20=-10 c) -4+4=0 7+(-7)=0 2) a) -2-(-10)=[-2+10=8] b) (-7)+2=-5 c) (-2)-1=-2-1=-3 3) Team A Team B -40 10 10 0 0 -40 _____ _____ Total -30 -30 _____ ______ Yes, we can add integers in any order. 4) i) (-5) ii) 0 iii) (-17) iv) (-7) v) (-3)

EX 1.3 1) a) (-3) b) -225 c) 630 d)316 e) 0 f) 1320 g) 160 h) (-360) i) (-24) j) 36 2)a) LHS RHS 18x[7+(-3) [18x7]+[18x(-3)] 18x4 126+(-54) =72 =72 LHS = RHS b) LHS RHS (-21)x[(-4)+(-6)] [(-21)x(-4)]+[(-21)x(-6)] (-21) x (-10) 84+126 = 210 =210 LHS = RHS i) (-1) x a = (-a) ii) (-22) x (-1) =22 37 x(-1) = -37 0 x(-1) = 0 4) (-1)x5=-5 (-1)x3=-3 (-1)x1=-1 (-1)x(-1) = 1 (-1)x4=-4 (-1)x2=(-2) (-1)x0=0

5) a) 26x(-48)+(-48)x(-36) (-48) [26+(-36)] (-48) [-10] = 480 b) 8x53x(-125) 53x8x(-125) 53x(-1000) = -53000 c) 15x(-25)x(-4)x(-10) 15x100x(-10) 15x(-1000) = -15000 d) (-41)x102 (-41) [100+2] (-41)x100+(-41)x2 (-4100) + (-82) = (-4182) e) 625x(-35)+(-625)x65 625x(-35)+(625)x(-65) 625[(-35)+(-65)] 625(-100)= -62500 f) 7x(50-2) 7x50-7x2 350 - 14 = 336 g) (-17) x(-29) (-17) x[(-30)+1] (-17)x(-30) + (-17)x1 510+(-17) = 493 h) (-57)x(-19)+57 57x19+57x1 57(19+1) 57 (20) = 1140 6) Initial temperature = 400c Decreasing temp per hour = 50c Temp after 10 hours = 400+10(-50) = 400-500 = -100c 7) i) 4 correct ans = 4x5=20 6 Incorrect ans = 6x(-2) =-12 ___________ Total = 20+(-12)=8 ____________ ii) 5 correct answers = 5x5=25 5 Incorrect answers = 5x(-2) =-10 Total = 25+(-10) = 15

iii) 2 correct answers = 2x5=10 5 incorrect answer = 5x(-2) = -10 Total = 10+(-10)=0 8)a) Profit for 3000 bags of white cement = 3000x8=24000 loss for 5000 bags of grey cement = 5000x5 = 25000 here profit < loss

Total loss = loss-profit = 25000-24000 = 1000 Rs. b) Let x be the no of white cement bags Given, Profit = Loss

5x6400= x 8

= 56400

8 = 4000 bags.

9)a) (-3)x(-9) =27 b) 5x(-7)=-35 c) 7x(-8)=-56 d) (-11)x(-12)=132 EX 1.4 1) a) -3 b) -10 c) 4 d) -1 e) -13 f) 0 g)1 h) -1 i)1

2)a ) a(b+c)# (ab)+ (a ÷ 𝑐) a=12 b=-4 c=2 LHS RHS

12(-4+2) [12(-4)]+[122]

12(-2) -3+6 -6 # +3

b) a=(-10) b=1 c=1 LHS RHS

(-10)(1+1) (-10)1+(-10)1

-102 (-10)+(-10) -5 # -20 3) a)1 b)75 c)(-206) d) (-1) e) (-87) f) (-48) g) (-10) h) (-12)

4) i) 9(-3)=-3 ab=-3

ii) (-9)3= -3

iii) (-15)5=-3

iv) 12(-4)=-3 5) Temp decreases 20c = 1hour Temp decreases 10c =½hr

Temp decreases 180c = ½x18=9hr Total time = 12noon+9hrs=21hrs

= 9pm At 9pm the temp. would be 80c below 00c.

6) i) Total score = 20 12 correct ans = 12x(+3) =36 mark of incorrect ans = 20-36=-16

No of incorrect ans = -16(-2) = 8. 7) Initial position of nine shaft is 10m above ground level. Its final position is (-350)m

distance covered = 10-(-350) = 10+350 = 360m time taken for 6m = 1min

1m = 1

6 min

360m = 360x1

6

= 60 min = 1 hour. CH -1 INTEGER EXTRA QUESTIONS 1) Where are integers used in real-life situations? 2) What are the two important concepts of integers? 3) ‘All subtraction statements have an equivalent addition statement’ explain with two examples. 4) What will be the sign of the product if we multiply together 15 negative integers and 4 positive integers? 5) An integer divided by (-7) gives 9. Find the integer. CHAPTER-5 LINES AND ANGLES Ex-5.1 1. Complement Angles: a) 20° 90-20=70° b) 63° 90-63=27° c) 57° 90-57=33° 2. Supplement angles: i) 105° 180-105=75° ii) 87° 180-87=93° iii) 154° 180-154=26° 3. i) 65+115=180° Supplementary angle ii) 63° + 27° = 90° Complementary angle iii) 112° + 68 = 180° Supplementary angle iv) 130+50=180 Supplementary angle v) 45+45=90° Complementary angle vi) 80+10=90° Complementary angle

4. 45+45=90° Ans: 45° 5. 90+90=180° Ans: 90° 6. If ∟1 is decreased then ∟2 will increase with the same measure, so that both the angles still remain supplementary. 7. i) No, bec. sum of 2 acute angles is less than 180° ii) No, bec. sum of 2 obtuse angles is more than 180° iii) Yes, be. sum of 2 right angles is 180° 8. Let x+y=90° If x>45° Add y on both sides x+y> 45 + 𝑦 ⟹ 90° > 45 + 𝑦 ⟹ 90-45> y ⟹ 45> y ∴Its complement is less than 45° 9. i) Yes ∟1 is adjacent to ∟2 ii) No iii) Yes, they form linear pair iv) Yes v) Yes vi) ∠𝐶𝑂𝐵 ∟2 + ∟3 10. i) Vertically opp. angles ∟1 & ∟4 ∟5 & ∟2 + ∟3 ii) Linear pair ∟1 & ∟5 ∟5 & ∟4 11. ∟1 & ∟2 are not adjacent because, their vertex is not common. 12. x=55° ( Vertically opp. angles) i) 55+y=180° (Supplementary angles) y=180° − 55° =125° y=z=125° (Vertically opp. angles) ii) 40+x+25=180° (Straight angle) x=180-65 x=115° z=40° (Vertically opp. angles) y+z=180 y+40=180 y=140° 13. Fill in the blanks: i) 90° ii) 180° iii) Supplementary iv) Linear pair

v) Equal vi) Obtuse angles 14. i)∠𝐴𝑂𝐷 = ∠𝐵𝑂𝐶 ii) ∠𝐵𝑂𝐴, ∠𝐴𝑂𝐸 iii) ∠𝐵𝑂𝐸, ∠𝐸𝑂𝐷 iv)∠𝐵𝑂𝐴, ∠𝐴𝑂𝐷 ∠𝐴𝑂𝐸, ∠𝐸𝑂𝐶 ∠𝐵𝑂𝐶, ∠𝐶𝑂𝐷 ∠𝐴𝑂𝐷, ∠𝐷𝑂𝐶 v) ∠𝐵𝑂𝐴, ∠𝐴𝑂𝐸 ∠𝐴𝑂𝐸, ∠𝐸𝑂𝐷 ∠𝐸𝑂𝐷, ∠𝐶𝑂𝐷 EX-5.2 1. I) ∠1 = ∠5 Corresponding angles ii) ∠4 = ∠6 Alternate angle iii) ∠4 + ∠5 = 180° Co-Interior angle 2. i) Pair of corresponding angles: ∠4 & ∠8, ∠3 &∠7 ∠1 & ∠5,∠2 & ∠6 ii) Pairs of alternate interior angles: ∠3,∠5 & ∠2,∠8 iii) Pairs of Interior angles on the same side of the transversal ∠3,∠8 and ∠2,∠5 iv) Vertically opp. angles ∠1 & ∠3,∠2 & ∠4 ∠5 & ∠7,∠8 & ∠6 3) 125 +e=180 (Supp. angles) e=180-125=55 ∠𝑒 = ∠𝑓 = 55° (Vertically opp. angles) ∠𝑑 = 125° (Corresponding angles) ∠𝑑 = ∠𝑏 = 125° (Vertically opp. angles) ∠𝑒 = ∠𝑎 = 55° (Corresponding angle) ∠𝑎 = ∠𝑐 = 55° ( Vertically opp. angles) 4. i) y=x alternate angles 110+y=180 (Supp. angles) y=70° ∴ 𝑥 = 70 ii)

x=100° (Corresponding angle) 5. i) ∠𝐴𝐵𝐶 = 70° then ∠𝐷𝐺𝐶 = 70° corresponding angles ii) ∠𝐷𝐺𝐶 = 70° then ∠𝐷𝐸𝐹 = 70° corresponding angles 6. i) 126+44=170 𝑙 is not parallel to m bec. pairs of interior angles are not supplementary. ii) x+75=180° x=105° x & 75° are not equal S0, 𝑙 ∦ 𝑚 iii) x & 123° are alternate angles. x+57=180 x=180-57=123° 𝑙 is parallel to m. iv) 98 + 72=160° 𝑙 ∦ 𝑚 EX.12.1 1. Algebraic expressions:

a) y-z b) 1

2 𝑥 + 𝑦

c) z²(zz) d) 1

4(pq)

e) x²+y² f) 5+3mn g) 10-yz h) ab-(a+b) 2.i) Terms & their factors: a) x-3 Terms x -3 factors 1 x

b) 1+x+x² Terms 1 x x² Factors 1 x x x c) y-y³ y y³ 1 y y y y d) 5xy²+7x²y 5xy² 7x²y 5 x y y 7 x x y e) -ab+2b²-3a² -ab 2b² -3a² -1 a b 2 b b -3 a a ii) a) -4x+5 T=-4x, 5 F=-4, x; 5 b) -4x+5y T=-4x, 5y F=-4, x; 5, y c) 5y+3y² T=5y, 3y² F=5, y; 3, y, y d) xy+2x²y² T=xy, 2x²y² f=x, y, 1; 2, x, x, y, y e) pq+q T=pq, q F=p, q; q. 1 f) 1.2ab-2.4b+3.6a T=1.2ab, -2.4b, 3.6a F=1.2, a, b; -2.4, b; 3.6, a

g) 3

4𝑥 +

1

4

T=3

4𝑥,

1

4 F=

3

4, 𝑥;

1

4

h) 0.1p²+ 0.29q² T=o.1p², 0.2q² F=0.1, p, p; 0.2, q, p

3. Numerical Co-efficients.

Term N. Co-efficient

i 5-3t² -3t² -3

ii 1+t+t²+t³ t t² t³

1 1 1

iii x+2xy+3y x

2xy 3y

1 2 3

iv 100m+1000n 100m 1000n

100 1000

v -p²q²+7pq -p²q² 7pq

-1 7

vi 1.2a+0.8b 1.2a 0.8b

1.2 0.8b

vii 3.14r² 3.14r² 3.14

viii 2(l+b)=2l+2b 2l 2b

2 2

ix 0.1y+0.01y² 0.1y

0.01y² 0.1

0.01 4. a) Terms contain x and coefficient of x

Terms contain x Coefficient of x

i) y²x+y Y²x y²

ii 13y²-8yx -8yx -8y

iii x+y+2 x 1

iv 5+z+zx zx z

v 1+x+xy x

xy 1 y

vi 12xy²+25 12xy² 12y²

vii 7x+xy² 7x xy²

7 y²

b)

Terms with y² Coefficient of y²

i 8-xy² -xy² -x

ii 5y²+7x 5y² 5

iii 2x²y-15xy²+7y² -15xy²

7y² -15x

7 5.

Monomials Binomials Trinomials

ii) y² (i) 4y-7z (iii) x+y-xy

(iv) 100 (vi) 5-3t (v) ab-a-b

(viii) 7mn (x) a²+b² (ix) z²-3z+8

(xi) z²+z (xii) 1+x+x²

(vii) 4p²q-4pq²

6. i) Like terms ii) Like terms iii) Unlike terms iv) Like terms v) Unlike terms vi) Unlike terms 7. a) Like Terms -xy², 2xy² -4yx², 20x²y -8x², -11x², -6x² 7y, y 100x, 3x -11yx, 2xy b) Like Terms 10pq, -7pq, 78pq 7p, 2405p 8q, -100q -p²q², 12p²q² -12, 41 -5p², 701p² 13p²q, qp² EX-12.2 1. Simplify: i) 21b+7b-20b-32=28b-20b-32=20b-32 ii) -z²+13z²+7z³-5z-15z=12z²+7z³-20z iii) p-(p-q)-q-(q-p)= p-p+q-q-q+p= p-q iv) 3a-2b-ab-a-b-ab+3ab+b-a 3a-a-a-2b-b+b-ab-ab+3ab= a-2b+ab v) 5x²y-5x²+3yx²-3y²+x²-y²+8xy²-3y² 5x²y+3x²y-5x²+x²-3y²-y²-3y²+8xy² 8x²y-4x²-7y²+8xy²

vi) (3y²+5y-4)-(8y-y²-4) 3y²+5y-4-8y+y²+4 = 4y²-3y 2. Add i) 3mn-5mn+8mn-4mn=2mn ii) t-8tz, 3tz-z, z-t= t-t-8tz+3tz-z+z=-5tz iii) -7mn+12mn+9mn-2mn+5+2-8-3 =12mn-4 iv) a+a-a+b+b-b-3+3+3=a+b+3 v) 14x-7x+10y-10y-12xy+8xy+4xy-13+18= 7x+5 vi) 5m-4m+2m-7n+3n-3mn+2-5 =3m-4n-3mn-3 vii) 4x²y+(-3xy²)+(-5xy²)+5x²y=9x²y-8xy² viii) 3p²q²-10p²q²+7p²q²-4pq+9pq+5+15 =5pq+20 ix) ab-ab-4a+4a+4b-4b =o x) x²-x²-x²-y²+y²-y²-1-1+1 =−𝑥2 − 𝑦2 − 1 3. Subtract: i)y²-(-5y²)=y²+5y²=6y² ii) -12xy-6xy=-18xy iii) a+b-a+b=2b iv) 5b-ab-ab+5a =5a+5b-2ab v) 4m²-3mn+8+m²-5mn =5m²-8mn+8 vi) 5x-10+x²-10x+5 = x²-5x-5 vii) 3ab-2a²-2b²-5a²+7ab-5b² =10ab-7a²-7b² viii) 5p²+3q²-pq-4pq+5q²+3p² =8p²+8q²-5pq 4. a) Let p should be added x²+xy+y²+p=2x²+3xy p=2x²+3xy-x²-xy-y² p=x²+2xy-y² b) Let p should be subtracted 2a+8b+10-p=-3a+7b+16 2a+8b+10+3a-7b-16=p 5a+b-6=p 5. Let p be taken away 3x²-4y²+5xy+20-p=-x²-y²+6xy+20 3x²-4y²+5xy+20+x²+y²-6xy-20=p 4x²-3y²-xy=p

6. a) 3x-y+11-y-11-3x+y+11 -y+11 b) 4+3x+5-4x+2x²-(3x²-5x-x²+2x+5) -x+2x²+9-2x²+3x-5 2x+4 EX-12.3 1. m=2 i) m-2=2-2=0 ii) 3m-5=6-5=1 iii) 9-5m=9-10=-1 iv) 3m²-2m-7=12-4-7=1

v) 5𝑚

2− 4 = 5 − 4 = 1

2. p=-2 i) 4p+7=-8+7=-1 ii) -3p²+4p+7=-12-8+7=-13 iii) -2p²-3p²+4p+7 16-12-8+7 3 3. x=-1 i) 2x-7=-2-7=-9 ii)( -x+2=1+2=3 iii) x²+2x+1=1-2+1=0 iv) 2x²-x-2=2+1-2=1 4. a=2, b=-2 i) a²+b²=4+4=8 ii) a²+ab+b²=4-4+4=4 iii) a²-b²=4-4=0 5. a=0, b=−1 i) 2a+2b=0-2=-2 ii) 2a²+b²+1=0+1+1=2 iii) 2a²b+2ab²+ab 0+0+0=0 iv) a²+ab+2=0+0+2=2 6. x=2 i) x+7+4(x-5)=2+7+4(-3) = 9-12=-3 ii) 3(x+2)+5x-7 12+10-7=15 iii) 6x+5(x-2) 12+0=12 iv) 4(2x-1)+3x+11 4(3)+6+11=12+6+11=29 7. x=3, a=-1, b=-2 i) 3x-5-x+9 9-5-3+9=10

ii) 2-8x+4x+4 2-24+12+4=-6 iii) 3a+5-8a+1 -3+5+8+1=11 iv) 10-3b-4-5b 10+6-4+10=22 v) 2a-2b-4-5+a -2+2-4-5-1=-10 8. i) z=10 z³-3(z-10) 1000-0=1000 ii) p=-10 p²-2p-100 100+20-100=20 9. When x=0 2x²+x-a=5 0+0-a=5 a=-5 10. a=5, b=-3 2(a²+ab)+3-ab 2(10)+3+15=20+3+15 =38 EX-12.4 1. a) i) For 5 digits: Pattern (5n+1) =5(5)+1=26 ii) For 10 digits: 5(10)+1=51 iii) For 100 digits: 5(100)+1=501 b) i) For 5 digits: Pattern (3n+1) 3(5)+1=16 For 10 digits: 3(10)+1=31 For 100 digits: 3(100)+1=301 c) i) For 5 digits: Pattern (5n+2) 5(5)+2=27 ii) For 10 digits: 5(10)+2=52 iii) For 100 digits: 5(100)+2=502 2. Term

Expression 5th 10th 100th

2n-1 9 19 199

3n+2 17 32 302

4n+1 21 41 401

7n+20 55 90 720

n²+1 26 101 10001

5th →2n-1=2(5)-1=10-1=9 10th →7n+20=7(10)+20=90 100th → n²+1=100+1=10000+1=10001 CHAPTER-11 PERIMETER AND AREA EX.11.1 1. l= 500m b=300m i) Area=𝑙𝑏 = 500300 = 150000𝑚2 ii) Cost for 1m²=10,000 Rs ∴ 150000m²=15000010,000 =150,00,00,000 Rs 2. Perimeter of square = 320m 4a=320

a= 320

4 = 80

Area=a²=8080=6400m² 3. Area of Rectangle=440m² 𝑙 = 22𝑚 b= ? 𝑙𝑏 = 440

b=440

22

b=20m Perimeter=2(l+b) =2(22+20) =2(42)=84m² 4. Perimeter of rectangular=100cm 𝑙=35cm b=? 2(𝑙 + 𝑏) = 100 2(35+b)=100 35+b=50 b=15cm Area=𝑙𝑏 = 3515 = 525𝑐𝑚² 5. Area of square=Area of rectangle a²=𝑙𝑏 6060=90b

3600

90= 𝑏 40m=b

6. 𝑙 = 40 𝑐𝑚 b=22cm Perimeter of Rectangle=2(𝑙 + 𝑏) =2(62) =124 cm Perimeter of Square=124 cm 4a=124

a= 124

431 𝑐𝑚

Area of Rectangle = 𝑙𝑏 = 40 22 =880 cm² Area of Square = 31 31=961 cm² Square encloses more area than Rectangle. 7. Perimeter of Rectangle=130 cm b= 30 cm 𝑙 =? 2(𝑙 + 𝑏) = 130 𝑙 + 30 = 65 ⟹ 𝑙 = 35 𝑐𝑚 Area= 𝑙𝑏 =3530 =1050 cm² 8. Door: 𝑙 = 2 𝑚, 𝑏 = 1 𝑚 Area of door = 𝑙𝑏 =2 m² Wall: 𝑙=4.5m b=3.6m Area of wall=4.53.6 =16.2 m² Area of shaded part is Area of wall - Area of door

16.2 m² - 2m² =14.2 m² Cost of White Wash: Rs: 20 per m² ∴ For 14.2 m²=14.2 20 =284 Rs Ex-11.2 1. a) Area of Parallelogram=base height=74=28 cm² b) Area of Parallelogram= base height = 5 3=15 cm² c) Area of Parallelogram= base height= 2.5 3.5 125 75 8. 75 cm² d) Area of Parallelogram= base height=5x4.8 =24.0 cm² e) Area of Parallelogram= base height==2.4.4=8.8 cm²

2. a) Area of Triangle=1

2𝑏𝑎𝑠𝑒𝑕𝑒𝑖𝑔𝑕𝑡

=1

243 =

12

2= 6 𝑐𝑚2

b) Area of Triangle=1

2𝑏𝑎𝑠𝑒𝑕𝑒𝑖𝑔𝑕𝑡=

1

253.2 = 8.0 𝑐𝑚2

c) Area of Triangle=1

2𝑏𝑎𝑠𝑒𝑕𝑒𝑖𝑔𝑕𝑡=

1

234 = 6 𝑐𝑚2

d) Area of Triangle=1

2𝑏𝑎𝑠𝑒𝑕𝑒𝑖𝑔𝑕𝑡=

1

232 = 3 𝑐𝑚2

3. a) Area of Parallelogram= bh 246=20h

246

20= 𝑕

h=12.3 cm b) Area of Parallelogram= bh 154.5=b15

154.5

15= 𝑏

∴ 𝑏 = 10.3 𝑐𝑚 c) Area of Parallelogram= bh 48. 72=b8.4

48.72

804= 𝑏

5.8=b ∴ 𝑏 = 5.8 𝑐𝑚 d) Area of Parallelogram= bh 16. 38= 15.6h

16.38

15.6= 𝑕

h=1.05 cm

4. a) Area of Triangle=1

2bh

87=1

2b15

872=15b

872

15= 𝑏

174

15= 𝑏

b=11.6 cm

b) Area of Triangle=1

2bh

1256= 1

2b31.4

12562=31.4b

12562

3.14= 𝑏

2512

31.4= 𝑏 b=80 mm

c) Area of Triangle=1

2bh

170.5=1

222h

170.52 = 22𝑕

170.52

22= 𝑕

314

22= 𝑕 ∴ 𝑕 = 15.5 𝑐𝑚

5. a) Area of Parallelogram= bh = SRQM =127.6 Area of Parallelogram=91.2 cm² b) Area of Parallelogram= bh =PSQN 91.2=8QN

91.2

8= 𝑄𝑁 ⟹ 11.4 𝑐𝑚

6. Area of Parallelogram= bh 1470=ABDL 1470=35DL

1470

35= 𝐷𝐿

∴ 𝐷𝐿 = 42𝑐𝑚 Area of Parallelogram= bh 1470=ADBM 1470=49BM

1470

49= 𝐵𝑀

∴ 𝐵𝑀 = 30𝑐𝑚

7. Area of Triangle=1

2bh

=1

2512 =

60

2= 30𝑐𝑚2

=30 cm² To find the length of AD

Area of Triangle=1

2bh

30=1

213𝐴𝐷

60=13AD

∴ 𝐴𝐷 =60

13 𝑐𝑚

8.

Area of Triangle=1

2bh

=1

269=27 cm²

To find height from C to AB

Area of Triangle=1

2bh

27=1

27.5𝐶𝐸

54=7.5CE

CE=54

705= 7.2 𝑐𝑚

∴ 𝐶𝐸 = 7.2 𝑐𝑚 EX-11.3 1. Circumference of the Circle=2𝜋𝑟 a) r=14cm

=222

714 = 88 cm

b) 28 mm

=222

728

=176 mm c) 21 cm

=222

721

=132 cm 2. Area of Circle 𝜋𝑟2 a) r=14mm

=22

71414

= 4414 =616 mm² b) d=49 m

r=49

2cm

=22

7

49

2

49

2

=7749

2= 1886.5 𝑚2

c) r= 5cm

= 22

755 = 78.57 𝑐𝑚2

(or) 3.1455=78.5 cm² 3. Circumference=154 m 2𝜋𝑟 = 154

r=1541

2

7

22

r=49

2𝑚

Area= 𝜋𝑟2 =22

7

49

2

49

2

=7749

2

=3773

2 𝑚2

=1886.5 m²

4. d=21m ⟹ 𝑟 =21

2𝑚

Fence= Circumference

2𝜋𝑟 = 222

7

21

2= 66 𝑚

For 2 rounds=266 =132 m Cost: Rs. 4per m For 132 m=1324 =Rs. 528 5. Circular Sheet r=4 cm

Area 𝜋𝑟2 =22

744

= 352

7= 50.24𝑐𝑚2

Circle r= 3cm

22

733 =

198

7 𝑐𝑚² = 28.26

Remaining Sheet=352

7−

198

7=

154

7= 22 𝑐𝑚2

(or) 50.24-28.26=21. 98 cm² 6. Diameter of Circular table cover=1.5 m

r=1.5

2𝑚

Perimeter=2𝜋𝑟

=222

7

1.5

2

=23.141.5

2

=3.141.5 =4. 71m Required lace length=4.71m cost⟹ 𝑅𝑠. 15/𝑚 For 4. 71 m=154. 71=70. 65 Rs. 7. d=10cm r=5 cm Perimeter of full circle=2𝜋𝑟

∴Perimeter of semi circle=2𝜋𝑟

2= 𝜋𝑟

=3. 145 =15. 70 cm Perimeter of the given Shape=d+ Perimeter of Semi circle =(5+15.7) cm=20.7cm 8. d=1.6 m

r=1.6

2𝑚

Area of top=𝜋𝑟2 =3.140.80.8 =2.0096m²

Polishing: Rs 15/m² ∴ For2.0096 m² = 2.009615 = Rs 30.144 9. Length of the wire= Circumference of the circle 44=2𝜋𝑟

447

222= 𝑟

7=r r=7 cm

Area = 𝜋𝑟2=22

777 = 154 𝑐𝑚2

Perimeter of square=Length of the wire 4a = 44 a =11 cm Area=1111=121 cm² ∴ Area of circle encloses more the square. 10. Circle: r=14 cm

Area= 22

71414

=616 cm ² Two small circles: r=3.5 cm

Area of 2 circles=22

73.53.52

=77 cm² Rectangle: 𝑙 = 3 𝑐𝑚 b= 1 cm Area=𝑙𝑏 = 31 = 3 𝑐𝑚2 Remaining Area =616-(77+3) =616-80=536 cm² 11. Aluminium sheet: a= 6 cm Area= 66=36 cm² Circle: r=2 cm Area= 𝜋𝑟2 = 3.1422 =12. 56 cm² Remaining Sheet: 36-12. 56=23.44 cm² 12. Circumference of a circle=31.4 cm 2𝜋𝑟 = 31.4

r=31.41

23.14

r=31.4

23.14

10

10 r=

31.4

231.410 r= 5 cm

Area= 𝜋𝑟2 = 3.1455 Area=78.5 cm² 13. Flower bed d=66 m r=33 m Area=3.143333 =3419.46 m² Width of path= 4m

∴Diameter of outer circle= 4+66+4=74 m r=74

2= 37𝑚

Area= 3.143737 =4298.66,² Area of the path= Area of outer circle-Area of Flower bed =4298.66-3419.46 =879.2 m² 14. Flower Garden Area= 314 m² Sprinkler at the centre r=12 m Area of sprinkler=3.141212 =452.16 m² Yes, the sprinkler can water the entire garden. 15. Inner Circle: r=19-10=9 m Circumference=2𝜋𝑟 = 23.149 =569.52m Outer Circle: r=19 m Circumference=2𝜋𝑟 =23.1419 = 119.32 m 16. Radius of Wheel=28 cm Circumference of a circle= 1 full rotation

Circumference of a circle=222

728

=176 cm

No. of rotation=𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

1 𝑓𝑢𝑙𝑙 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛

= 352 𝑚

176 𝑐𝑚=

35200

176

= 200 rotations. 17. Length of minute hand= Radius of clock r= 15 cm Tip of min. hand move in 1 hour= Circumference of clock =23.1415 =94.2 cm

EX-11.4 1. Inner Rectangle: Area: 𝑙𝑏 = 9075 = 6750 𝑐𝑚2 Outer Rectangle: Area:= 10085 =8500 m² Area of path: 8500-6750=1750 m² Area of Garden= Area of Inner Rectangle Garden in hectare 10,000 m²= 1 hectare

1 m²=1

10,000 𝑕𝑒𝑐

∴6750 m²=6750

10000𝑕𝑒𝑐

=0.675 hectare =0.675 hec 2. Rectangular Park 𝑙 = 125 𝑚 b= 65m Area=𝑙𝑏 = 12565 = 8125 𝑚2 Wide of path= 3 cm Outer of Rectangle: 𝑙 = 3 + 125 + 3 = 131m b=3+65+3=71 m Area= 𝑙𝑏 = 13171 = 9301 𝑚2 Area of path =9301-8125 =1176 m² 3. Cardboard 𝑙 = 8 𝑐𝑚 b= 5cm Area= 85 = 40 cm² Picture 𝑙 = 8 − 3 = 5𝑐𝑚 b=5−3= 2 cm Area= 52=10 cm² Area of margin: = 40−10 = 30 𝑐𝑚²

4. Room: 𝑙 = 5.5 𝑚, 𝑏 = 4 𝑚 Area=5.54=22.0 m² Width=2.25 m Verandah: 𝑙 = 2.25 + 5.5 + 2.25 = 10𝑚 b=2.25+4+2.25=8.5 m Area of outer Rectangle=108.5 =85 =85m² i) Area of Verandah: = 95-22=63 m² ii) Cementing Verandah: Cost=Rs. 200 /m² ∴For 63 m²=63200 =12600 Rs 5. Square Garden a=30 m Area= 900 m² Width= 1m Square Garden without path a=30−2 = 28 𝑚 Area=2828 =784 m² i) Area of path: 900-784=116 m² ii) Cost: Rs. 40 /m² ∴For 784 m²=78440 =Rs 31, 360 6. PQ= 10 m PS=300 EH=10 m EF= 700m KL=10m KN=10 m Area of road=Area of PQRS + Area of EFGH − Area of KLMN =3000+ 700−100 =9,900 m²

Area of hectare:

1 m²= 1

10,000 𝑕𝑒𝑐

9900m²=9900

10000

=0.99 hec Area of Park: 700 300=210000 m² Area of park excluding cross roads: 210000-9900 =200100 m² =20.01 hectare 7.

PQ= 3 m PS= 60 m EH= 3 m EF=90 m KL= 3 m KN= 3M i) Area of road=Area of PQRS + Area of EFGH - Area of KLMN =180+270−9 =441 m² ii) Cost of constructing road: Rs 110 per m² ∴ For 441 m²=441110 =48, 510 Rs 8. Radius of pipe= 4cm Wrapping around circle=2𝜋𝑟 =23.144 =25.12 cm Wrapping around square= 4a =44=16 cm Remaining Cord=Cord Wrapped on Pipe − Square =25.12-16 =9.12 cm She has left 9.12 cm cord. 9. 𝑙 = 10 𝑚 r=3 m b= 5 m i) Area of whold land=𝑙𝑏 =105=50 m²

ii) Area of flower bed=𝜋𝑟2 =3.1422 =12. 56 m² iii) Area of lawn excluding flower bed 50−12.56 = 37.44 𝑚2 iv) Circumference of flower bed=2𝜋𝑟 =23.142 =12.56 m 10. i) 𝑙 = 18 𝑐𝑚 b=10 cm Area of Rectangle=180 cm²

Area of Triangle I=1

2𝑏𝑕

=1

2106

=30 cm²

Area of Triangle II=1

2810

= 40 cm² Area of Shaded part =180− 30 + 40 =110 cm² ii) Area of square =2020=400 cm²

Area of Triangle I=1

21010

= 50 cm²

Area of Triangle II=1

21020

= 100 cm²

Area of Triangle III=1

21020

=100 cm² Area of Shaded part =400− 50 + 100 + 100 = 150 cm² 11. Area of quadrilateral ABCD=Area of Triangle ABC+ Area of Triangle ACD

=1

2223 +

1

2223

= 33+ 33 =66 cm² CHAPTER 6 THE TRIANLGES AND ITS PROPERTIES

EX 6.1

1. 𝑃𝑀 is altitude PD is median OM≠ 𝑀𝑅

2. a) B A E C b) c) Y R Q P X Z L 3. Isosceles triangles means any two sides are same. Take∆ 𝐴𝐵𝐶 & draw the median when AB=AC. AL is the median and altitude of the given triangle. A B L C EX 6.2 1. An exterior angle of a triangle is equal to the sum of its interior opposite angles. i) 50+70=x=120° ii) 65+45=x=110° iii) 30+40=x=70° iv) x=60+60=120° v) x=50+50=100° vi) x=30+60=90° 2.i) x+50=115 x=115-50=65° ii) x+70=100 x=100-70=30° iii) x+90=125 x=125-90=35° iv) x+60=120 x=120-60=60° v) x+30=80 x=80-30=50° vi) x+35=75° x=75-35=40° EX 6.3 1. The total measure of the 3 angles of a Triangle is 180°. i) 50+60+x=180 x=180-110=70° ii) 90+30+x=180 x=180-120=60°

iii) 30+110+x=180 x=180-140=40° iv) 50+x+x=180 2x=180-50

x=130

2 =65°

v) x+x+x=180° 3x=180

x=180

3 =60°

vi) x+2x+90=180 3x=180-90 3x=90

x=90

3=30°

2.i) 50+x=120 x=120-50 x=70° 50+70+y=180 y=180-120 y=60° [OR] y+120=180° (Supp. angle) y=60° 50+60+x=180 x=180-110 x=70° ii) y=80°( Vertically opp. angles) 50+y+x=180° 50+80+x=180 x=180-130=50° iii) x=50+60° x=110° x+y=180° (Supplementary angles) 110+y=180 y=180-110=70° iv) x=60° (Verti. opp. angles) 30+x+y=180° 30+60+y=180° y=180-90=90° v) y=90° (Ver. opp. angles) x+x+90=180° 2x=180-90=90°

x=90

2= 45°

vi) x=y (Ver. opp. angles) x+x+x=180° 3x=180°

x=180

3= 60°

Ex 6.4 1. i) 2 cm, 3cm, 5 cm 2+3=5 (NO) 3+5>2 (No) 2+5>3 (No) It is not possible. ii) 3+7>7 Yes 6+7>3 Yes 7+3>6 Yes It is possible iii) 6+3>2 No 3+2< 6 Yes 6+2>6 No It is not possible 2. i) Yes ii) Yes iii) Yes 3. In ∆ 𝐴𝐵𝑀, 𝐴𝐵 + 𝐵𝑀 > 𝐴𝑀 In ∆ 𝐴𝑀𝐶, 𝐴𝐶 + 𝑀𝐶 > 𝐴𝑀 ∴ AB+BM+AC+MC>AM+AM (By adding 2 eqns) AB+AC+(BM+MC)>2AM AB+AC+BC>2Am Hence it is ture. 4. In ∆𝐴𝐵𝐶, 𝐴𝐵 + 𝐵𝐶 > 𝐴𝐶 In ∆𝐴𝐷𝐶, 𝐴𝐷 + 𝐷𝐶 > 𝐴𝐶 In ∆𝐷𝐶𝐵, 𝐷𝐶 + 𝐶𝐵 > 𝐷𝐵 In ∆𝐴𝐷𝐵, 𝐴𝐷 + 𝐴𝐵 > 𝐷𝐵 By adding all we get, AB+BC+AD+DC+DC+CB+AD+AB>AC+AC+DA+DB 2AB+2BC+2AD+2DC>2(AC+DB) AB+BC+CD+DC> AC+DB AB+BC+CD+DA> AC+DB It is true. 5. In ∆𝐴𝑂𝐵, 𝐴𝐵 < 𝑂𝐴 + 𝑂𝐵 ∆𝐵𝑂𝐶, BC<OB+OC ∆𝐶𝑂𝐷, 𝐶𝐷 < 𝑂𝐶 + 𝑂𝐷 ∆𝐴𝑂𝐷, 𝐷𝐴 < 𝑂𝐷 + 𝑂𝐴 By adding all

AB+BC+CD+DA<2OA+2OB+2OC+2OD AB+BC+CD+DA<2[(AO+OC) + (DO+OB)] AB+BC+CD+DA<2(AC+BD) It is true. 6. The 3rd side should be less than 12+15=27 cm. And also 3rd side cannot be less than 15-12=3 cm ∴ The 3rd side could be more than 3 cm & less than 27 cm EX 6.5 1. R By Pythagoras Property, RQ²=PR²+PQ² 24 =10²+24² =100+576

P 10 Q RQ= 676 RA= 26 2. A 25 AB²=AC²+CB² 7 25²=7²+CB² 625-49=CB²

C ? B 576=CB²

576=24=CB 3. 15²=12²+a² 225-144=a² 81=a²

a= 81 a=9 m 4. 6.5²=6²+2.5² i) 42.25=36+6.225 42.25=42.25 Yes these will form right triangles ii) 5²=2²+2² 25=4+4 25≠ 8 It will not form right triangle iii) 2.5²=2²+1.5² 6.25=4+2.25 =6.25 It will form right triangle 5. By Pythagoras property x²=5²+12² =25+144

x= 169 x=13 m Height of tree=5 m+13 m=18 m

6. ∠𝑃 + ∠𝑄 + ∠𝑅 = 180° ∠𝑃 + 25 + 65 = 180° ∠𝑃=180-90=90° ∴ It is right angled triangle ∴ QR=Hypothenuse PR, PQ=legs ∴ By Pythagoras property, (ii) QR²=PR²+PQ² is true. 7. Length =40 cm diagonal=41 cm 41²=40²+x² 1681=1600+x² 1681-1600=x² x 41

81 = 𝑥 x= 9 cm ∴ Perimeter = 2(l+b) 40 =2(40+9) =2(49) =98 cm 8. x²=8²+15² x =64+225 15 15 =289

x= 289 =17 Perimeter=17+17+17+17=68 cm CHAPTGER 9 : RATIONAL NUMBERS EX 9.1 1. 5 Rational numbers between: i) -1 and 0 0 can be written as 0

-1 can be written as −6

6

ii) -2 and -1

Ans: -11

6 , -1

2

6 , -1

3

6 , -1

4

6 , -1

5

6

iii) −4

5 and

−2

3

−4

5=

−12

15

−2

3=

−10

15 ⇒

−12

15<

−11

15<

−10

15

Also −4

5=

−24

30

−2

3=

−20

30

−24

30 <

−23

30<

−22

30<

−21

30<

−20

30

Also

−4

5=

−36

45

−2

3=

−30

45

−36

45<

−35

45<

−34

45<

−33

45<

−32

45<

−31

45<

−30

45

Ans: −35

45 ,

−34

45 ,

−33

45,−32

45 ,

−31

45

iv) −1

2 and

2

3

−1

2=

−3

6

2

3=

4

6

−3

6<

−2

6<

−1

6< 0 <

1

6<

2

6<

3

6<

4

6

Ans: −2

6 ,

−1

6 ,

1

6,

2

6,

3

6

2. i) −3

5 ,

−6

10,−9

15,−12

20

∴ −3

5

5

5=

−15

25

−3

5

6

6=

−18

30

−3

5

7

7=

−21

35

−3

5

8

8=

−24

40

ii) −1

4 ,

−2

8 ,

−3

12

Ans: −4

16 ,

−5

20,−6

24,−7

28

iii) −1

6 ,

2

−12,

3

−18,−4

24, …………

Std form is

−1

6 ,

−2

12 ,

−3

18,−4

24 , ………

iv) −2

3,

2

−3 ,

4

−6 ,

6

−9

Ans: 8

−12 ,

10

−15 ,

12

−18 ,

14

−21

3. Equivalent rational numbers:

i) −2

7

2

2=

−4

14

−2

7

3

3=

−6

21

−2

7

4

4=

−8

28

−2

7

5

5=

−10

35

ii) 5

−3

2

2=

10

−6

5

−3

3

3=

15

−9

5

−3

4

4=

20

−12

5

−3

5

5=

25

−15

iii) 4

9

2

2=

8

18

4

9

3

3=

12

27

4

9

4

4=

16

36

4

9

5

5=

20

45

4) Number Line:

i) −2

7

ii) −5

8

iii) −7

4= −1

3

4

iv) 7

8= 1

7

8

5. P=21

3 𝑅 = −1

1

3

Q=22

3 𝑆 = −1

2

3

6.i) −7

21=

−1

3 &

3

9=

1

3

−1

3 ≠

1

3

ii) −16

20=

−4

5

20

−25=

4

−5

Represent same rational number

iii) −2

−3 and

2

3

Same rational number

iv) −3

5 and

−12

20=

−3

5

v) 8

5 and

−24

15=

−8

5

Same Rational Number

vi) 1

3 ≠

−1

9

vii) −5

−9 ≠

5

−9

7. Simplest form:

i) −8

6=

−4

3 ii)

25

45=

5

9

iii) −44

72=

−11

18 iv)

−8

10=

−4

5

8. i) −5

7<

2

3 ii)

−4

5<

−5

−7

iii) −7

8=

14

−16 iv)

−8

5>

−7

4

v) 1

−3<

−1

4 𝑣𝑖)

5

−11=

−5

11

vii) 0>−7

6

9. Greater:

i) 2

3 ,

5

2 LCM=6

2

3=

4

6

5

2=

15

6

2

3<

5

2

ii) −5

6 ,

−4

3 LCM=6

−4

3=

−8

6

−5

6>

−4

3

iii) −3

4 ,

2

−3 LCM=12

−3

4=

−9

12

−2

3=

−8

12

−3

4<

−2

3

iv) −1

4<

1

4

v) −32

7, −3

4

5 ⟹

−23

7 ,

−19

5 LCM=35

−23

7=

−115

35

−19

5=

−133

35

−23

7>

−19

5, −3

2

7> −3

4

5

10. Ascending order Ans:

i) −3

5 ,

−2

5,−1

5

ii) −1

3 ,

−2

9 ,

−4

3 LCM=9

⟹ −3

9 ,

−2

9 ,

−12

9 Ans:

−12

9,−3

9,−2

9 ⟹

−4

3,−1

3,−2

9

iii) −3

7 ,

−3

2 ,

−3

4

⟹ −12

28,−42

28 ,

−21

28

Ans: −42

28 ,

−21

28,−12

28

⟹ −3

2,−3

4 ,

−3

7

EX: 9.2 1. Sum

i) 5

4+

−11

4=

5+(−11)

4=

−6

4

=−3

2= −1

1

2

ii) 3

5+

5

3 LCM=15

3

5=

9

15

5

3=

25

15

9

15+

25

15 =

34

15= 2

4

15

iii) −9

10+

22

15 ⟹ LCM=30

−9

10=

−27

30

22

15=

44

30

−27

30+

44

30=

17

30

iv) −3

−11+

5

9 LCM=99

3

11=

27

99

5

9=

55

99

27

99+

55

99=

82

99

v) −8

−19+

−2

57 LCM=57

−8

19=

−24

57

−24

57+

−2

57=

−26

57

vi) −2

3+ 0 =

−2

3

2. Subtraction:

i) 7

24−

17

36 LCM=72

7

24=

21

72

17

36=

34

72

21

72 -

34

72=

21−34

72=

−13

72

ii) 5

63−

−6

21 =

5

63+

6

21

6

21=

18

63 LCM=63

5

63+

18

63=

23

63

iii) −6

13−

−7

15 =

−6

13+

7

15

−6

13=

−90

195 LCM=195

7

15=

91

195

−90

−195+

91

195=

1

195

iv) −3

8−

7

11 LCM=88

−3

8=

−33

88

7

11=

56

88

−33

88−

56

88=

−33−56

88=

−89

88

v) −21

9− 6 ⟹

−19

9−

6

1

6

1=

54

9 LCM=9

−19

9−

54

9=

−19−54

9=

−73

9

=−81

9

3. Product:

i) 9

2

−7

4=

9(−7)

24=

−63

8

=−77

8

ii) 3

10 −9 =

3(−9)

10=

−27

10= −2

7

10

iii) −6

5

9

11=

−54

55

iv) 3

7 (

−2

5) =

−6

35

v) 3

11

2

5=

6

55

vi) 3

−5

−5

3= 1

4. Divide:

i) (-4)÷2

3

-43

2= −6

ii) −3

5÷ 2

−3

5

1

2=

−3

10

iii) −4

5 ÷ −3

−4

5

−1

3=

4

15

iv) −1

8÷

3

4

−1

8

4

3=

−1

6

v) −2

13÷

1

7 ⟹

−2

13

7

1=

−14

13

vi) −7

12÷

−2

13

−7

12

13

−2=

−91

−24=

91

24

=319

24

vii) 3

13÷

−4

65

3

13

65

−4=

15

−4=

−15

4

=−21

4

CHAPTER 10- PRACTICAL GEOMETRY EX 10.1 1.

Construction: 1) Draw a line segment AB and mark a point C outside ‘AB’.

2) Take any point D on 𝐴𝐵 and join CD.

3) With D as centre and a convenient radius, draw an arc cutting 𝐴𝐵 at G and H. 4) Now with C as centre and the same radius as in step 3, draw an arc. It will cut CD at I. 5) Measure GH using compass. 6) Without changing measurement, Place at I cut the arc. Name it as J. 7) Join CJ & extend the line.

𝐸𝐹 is parallel to AB. 2. Construction: 1) Draw a line 𝑙. Mark any point P on the line. 2) Draw a perpendicular to 𝑙 using compass at the point p. 3) Taking P as centre, 4 as radius, cut the perpendicular line. Mark it as X.

4) Draw a line m which pass through X. 5) Now, the line m is parallel to l. 3. Construction: 1) Draw a line 𝑙. Mark a point Q on 𝑙. 2) Take any point P outside. Join PQ. 3) Taking Q as centre, with convenient radius, draw an arc cutting 𝑙

and𝑃𝑄 at H and G

3) Taking P as Centre, with same radius draw another arc cutting 𝑃𝑄 at I. 4) Measure GH using compass. 5) Taking I as centre, with same radius cut the arc. Name it as J 6) Join PJ & extend the line m. 7) Mark R on the line m. 8) Now, consider PQ as a line segment, R is a point outside of PQ and repeat the same steps above to construct a parallel line to PQ. EX 10.2 1. Construction: 1) Draw a line segment XY=4.5 cm 2) Taking X as centre, 6 as radius, draw an arc. 3) Taking Y as centre, 5 as radius, cut the previous arc. 4) Name it as Z. ∆𝑋𝑌𝑍 is the required triangle. 2. A 5.5 cm 5.5 cm B 5.5 cm C

Construction: 1) Draw a line segment BC=5.5cm 2)Taking B as centre, 5.5cm as radius, draw an arc. 3) Taking C as centre, 5.5cm as radius, (draw) cut the previous arc. 4) Name it as A. ∆𝐴𝐵𝐶 is the required equilateral triangle. 3. P 4 cm 4 cm Q 3. 5cm R Construction: 1) Draw a line segment QR=3.5 cm 2. Taking Q as centre, 4cm as radius draw an arc. 3) Taking R as centre, 4cm as radius, cut the previous arc. Name it as P. ∆𝑃𝑄𝑅 is the required isosceles triangle. 4. A 6.5 cm 2.5 cm B 6 cm C Construction: 1) Draw a line segment BC=6 cm. 2) Taking B as centre, 2.5cm as radius, draw an arc. 3) Taking C as centre, 6.5 cm as radius, cut the previous arc. Name it as A. ∆𝐴𝐵𝐶 is the required triangle. Ex 10.3 1. E 5cm D 3 cm F

90°

Construction: 1) Draw a line segment DF=3 cm. 2) Taking D as centre, measure 90° using compass. Draw the ray X. 3) Taking D as centre, 5 cm as radius, cut the ray. 4. Name it as E. Join EF. ∆𝐷𝐸𝐹 is the required right angled triangle. 5. Construction: 1)Draw a line segment AB=6.5 cm 2) Taking A as centre, measure 110° using protractor. Draw the ray X. 3) Taking A as centre, 6.5cm as radius, cut the ray name it as C 4) Join BC. ∆𝐴𝐵𝐶 is the required triangle. 3. Construction: 1) Draw a line segment BC=7.5 cm. 2. Taking C as centre, measure 60° using compass. Draw the ray X. 3) Taking C as centre, 5 cm as radius, cut the ray. Name it as A. 4. Join AB. ∆𝐴𝐵𝐶 is the required triangle.

CH -4 SIMPLE EQUATIONS EX 4.1

1) i)No ii) No iii)Yes iv)No v) Yes vi) No vii)Yes viii)No ix) No x) No xi) Yes. 2) a) n+5=19 (n=1) 1+5=19 6#19 Not a solution b) 7n+5=19 (n=2) 7(-2)+5=19 -14+5=19 -9#19 Not a soln c) 7n+5=19 (n=2) 7(2)+5=19 14+5=19 Yes it is a soln. d) 4p-3=13 (p=1) 4(1)-3=13 4-3=13 1#13 Not a soln e) 4p-3=13 (p=4) 4(-4)-3=13 -16-3=13 -1913 Not a soln. f) 4p-3=13 (p=0) 4(0)-3 =13 -313 Not a soln. 3) Trial & error method i)

Value of p

Substitution LHS 5p+2

RHS17

Result

P=1 5+2=7 17 717 P=2 10+2=12 17 121

7 P=3 15+2=17 17 17=1

7 p=3

ii) Value of m LHS 3m-

14 RHS Result

M=1 3-14=11 4 -114 M=5 15-14=1 4 14 M=6 18-14=4 4 4=4

m=6 4) i)Sum of nos x&4 is 9 x+4=9 ii)y-2=8 iii) 10a=70

iv)𝑏

5=6 v)

3

4t =15 vi)7m+7=77 vii)

1

4-4=4

viii) 6y-6=60 ix) 3+1

3z=30

5) i) The sum of p and 4is 15 ii) 7 Subtracted from m is 3 iii) 2times m is 7 iv) The number m divided by 5 gives 3 v) Three fifth of m is 6 vi) If you add 4 to 3times p you get 25. vii) If you take away 2 from 4 times p, you get 18. viii) If you add 2 to half of p, you get 8. 6) i) 7+5m=37 (m=marble) ii) 4+3y=49 (y=laxmi’s age) iii) 2l+7=87 (l=lowest score) iv) b be the base angle. EX 4.2 1) a) Add 1 on both sides b) sub 1 on both sides c) Add 1 on both sides e) sub 6 on both sides f) Add 4 on both sides g) Sub 4 on both sides h) sub 4 on both sides 2) a) Divide by 3 on both sides b) multiply by 2 on both sides

3𝑙

3 =

42

3 l=14

𝑏

2 x 2 = 6x2 b=12

c) multiply by 7 on both sides d)divide by 4

7x𝑝

7 = 4x7 p=28

4

4 =

25

4 =

25

4

e) divide by 8, f) multiply by 3

8𝑦

8 =

36

8 y=

9

4

𝑧

3 x3 =

5

4 x3 z=

15

4

g) multiply by 5, h) divide by 20,

5x 𝑎

5 =

7

15 x5 a =

7

3

20𝑡

20 =

−10

20 t =

−1

2

3) a) Add 2, b) Sub 7, 3n-2+2=46+2 5m = 17-7 =10 3n=48 divide by 5

Divide by 3 m=10

5 =2

n=48

3 = 16

c) multiply by 3, d) multiply by 10, 20p = 40x3 3p=6x10 divide by 20 divide by 3,

p=40𝑥3

20 =6 p=

6𝑥10

3 = 20

4) a) 10𝑝

10 =

100

10 p=10 b) 10p+10-10=100-10

10p = 90

10𝑝

10 =

90

10 p=9

c) 𝑝

4 = 5 multiply by 4

c) multiply by 4, d) Multiply by 3

4𝑝

4 = 5x4 p=20 3

−𝑝

3=53

4x3𝑝

4 = 6x4 -p=15

divide by 3, p=-15

3𝑝

3 =

6𝑥4

3 p=8

f) divide by 3 g) Sub 12,

3𝑠

3 =

−9

3 s=-3 3s+12-12=0-12

divide by 3,

3𝑠

3 =

12

3 s=-4

h) 3s =0 i) divide by 2,

divide by 3, 2𝑞

2 =

6

2 q=3

3𝑆

3 =

0

3 S=0

j) Add 6, 2q-6+6 =0+6 k) Sub 6, 2q+6-6=0-6

divide by 2, 2𝑞

2 =

6

2 divide by 2,

2𝑞

2 =

−6

2

q =3 q=-3 l) Sub 6, 2q+6-6= 12-6

divide by 2, 2𝑞

2 =

6

2

q=3

EX 4.3

1) a) 2y+5

2 =

37

2 b) 5t+28=10

2y = 37

2 -

5

2 =

32

2 5t = 10-28 = -18

y = 32

2 𝑥2 = 8 t =

−18

5

c) 𝑎

5 +3 = 2 d)

𝑞

4 +7 =5

𝑎

5 = 2-3 = -1

𝑞

4 = 5-7 = -2

a= -5 q = -2x4 =-8

e) 5

2 = -10 f)

5

2 =

25

4

5 = -10x2=-20 5 = 25

4 x2

= −20

5 = -4 =

25

4 x

2

5 =

5

2

g) 7m +19

2 =13 h) 6z+10=-2

7m = 13-19

2 =

26−19

2 =

7

2 6z = -2-10=-12

m= 7

2 𝑥7 =

1

2 z =

−12

6 = -2

i) 3𝑙

𝑧 =

2

3 j)

2𝑏

3 -5 =3

l = 2

3 x

2

3 =

4

9

2𝑏

3 = 3+5 =8

b = 8x3

2 = 12.

2) a) 2(+4)=12 b) 3 (n-5) = 21

+4 = 12

2 =6 n-5 =

21

3 =7

=6-4=2 n = 7+5=12 C) 3(n-5) = -21 d) -4 (2+) =8

n-5 = −21

3 =-7 2+ =

8

−4 = -2

n=-7+5=-2 = -2-2 =-4 e) 4 (2-) =8

2- =8

4 =2

2-2 = 0 = 3) a) 4 = 5 (9-2) b) -4 = 5 (p-2)

4

5 = p-2

−4

5 + 2 = p p =

6

5

4

5 +2 = 9 p=

14

5

c) 16 = 4+3 (t+2) d) 4+5 (p-1) = 34 16-4 = 3(t+2) 5 ((p-1) = 34-4

12

3 = t+2 4-2= t p-1 =

30

5

2 = t p = 6+1 =7 e) 0 = 16+4 (m-6)

−16

4 = m-6

-4 +6 =m 2 =m 4) a) = 2 Multiplying both sides by 10 10=20 Adding 2 on both sides 10 + 2 =22 _____________ 1 = 2 Multiplying both sides by 5 5=10 Subtracting 3 from both sides 5 - 3 =7 _____________ 2 = 2 Dividing both sides by 5

5 =

2

5 _____________ 3

b) = -2 Multiplying both sides by 3 3 = -6 ______________ 1 = -2 Multiplying both sides by 4 4 = -8 Adding 7 on both sides 4 +7 = -1 ___________ 2 = -2 Multiplying both sides by 2 2 = -4 Adding 10 on both sides 2 +10 = 6 __________ 3

EX 4.4

1) a) 4+8 = 60 B) 1

5 -4 =3

8 = 60-4 = 56 1

5 = 3+4 = 7

= 56

8 = 7 = 7 X 5 =35

C) 3

4 +3 = 21 D) 2 -11 =15

3

4 = 21-3 =18 2 = 15+11 = 26

= 18 X 3

4 =24 =

26

2 = 13

E) 50-3 =8 F) +19

5 = 8

-3N = 8-50 + 19 = 40 = 40-19

N = −42

−3 = 14 = 21

g) 5

2 - 7 = 23

5

2 = 23+7 =30

= 30x2

5 =12

2) a) Let be the lowest mark. 87 = 2 +7 87-7 = 2

40 = 80

2 =

b) Sum of 3 angles of a is 180 ( = base angle) 400+ + =180 2 = 180-40 = 40

= 40

2 = 20

c) Two short of a double century (200-2 =198) ( S = sachin) (R = Rahul) S = 2r S+r =198 2r +r =198 Rahul = 66 3r = 198 Sachin = 132

r = 198

3 = 66

3) i) 37 = 7+5p (p= permit) 37-7 = 5p

30

5 = p p=6

ii) Let f be fathers age l be Laxmi’s age f=4+3l 49 = 4+3l 49-4 = 3l

45

3 = l l =15

iii) Non-fruit trees (n) =77 fruit trees be (f) n=2+3f 77=2+3f

77-2=3f 75

3 = f f =25

4) 7 +50 = 260 (To reach a triple century you still need forty! 300-40=260) 7 =260-50 =210

=210

7 = 30

EXTRA QUESTIONS CHAPTER 5 LINES AND ANGLES

1. i) ∠3 = 35° (Vertically opp. angles are equal) ∠2 + 35° = 180° (Supplementary angles) ∠2= 180 − 35 = 145° ∠2 = ∠4 = 145° (Vertically opp. angles) ∠5 = 35° (Corresponding angles) ∠5 = ∠7 = 35° (Vertically opp. angles) ∠2 = ∠6 = 145° (Corresponding angles) ∠6 = ∠8 = 145° (Vertically opp. angles) ii) 35+𝑥 = 125° [By exterior angle property] 𝑥 = 125 − 35 𝑥 = 90° 2. 𝑦 = 70° [ y and 70° are alternate angles] 𝑥 = 70° [𝑥 and 70° are corresponding angles] 3. 𝑥 + 2𝑥 + 24 = 180° [Supplementary angles] 3𝑥 + 24 = 180 3𝑥 = 180 − 24 3𝑥 = 156

𝑥 =156

3

𝑥 = 52° 4. Find a: 3a+5a+4a=180° (Straing angle) 12a=180°

a=180

12=

60

4 a=15°

5. b=45° [b & 45° are alternate angle] a=35° [a & 35° are alternate angles]

CHAPTER 9 RATIONAL NUMBERS 1. Smallest and greatest rational numbers are Not defined. 2. Yes, zero is a rational number.

3.i) (7

3

11

3) ÷

1

3

77

9÷

1

3=

77

9

3

1

=77

3= 26

1

3

ii) 2[5 − {4 ÷3

2(2 +

−2

3)

=2 5 − 4 ÷3

2

6+(−2)

3

= 2[5 − 4 ÷3

2

4

3 }]

= 2[5 − 4 ÷12

6 ]

= 2 5 − 46

12

=2[5-2] =2[3] =6 4. Total apples=25

No. of rotten apples=2

5 of total

ie) 2

3 of 25=

2

525

Rotten=10 Good apples=25-10 =15 5. Total container=7

Total amount of milk=63

4𝑙

Amount of milk in each container=63

4÷ 7

=63

7

1

7=

9

4𝑙

= 21

4𝑙

6. Find (𝑥 + 𝑦) ÷ (𝑥 − 𝑦) where 𝑥 =3

4 and 𝑦 =

−7

4

3

4+

−7

4 ÷

3

4−

−7

4

3−7

4 ÷

3+7

4

−4

4 ÷

10

4 = (-1)÷

10

4

(-1)4

10=

−4

10

7. Express (−3

7) as a rational number with

a) denominator (-35) b) numerator (63)

a) −3

7

−5

−5=

15

−35

b) −3

7

9

9=

−27

63

CHAPTER 4 SIMPLE EQUATIONS

1. Let 𝑥 & 𝑥 + 1 be the two consecutive numbers. Given 𝑥 + 𝑥 + 1 = 13 ⇒ 2𝑥 + 1 = 13 2𝑥 = 13 − 1 2𝑥 = 12 𝑥 =12/2=6 No. s are 6 and 7 2. Now, Seema’s age=20 yrs Reema’s age=4 yrs Let s be the Seema age and r be the Reema age. Let 𝑥 be the no. of years. Given, 𝑠 + 𝑥 = 2 𝑟 + 𝑥 20 + 𝑥 = 2 4 + 𝑥 20 + 𝑥=8 + 2𝑥 20 − 8 = 2𝑥 − 𝑥 12=𝑥 3. The 3 consecutive even numbers are 𝑥, 𝑥 + 2, 𝑥 + 4 𝑥 + 𝑥 + 2 + 𝑥 + 4 = 96 3𝑥 + 6 = 96 3𝑥 = 96 − 6 3𝑥 = 90

𝑥 =90

3

No.s are 30, 32, 34.

4. Total Students=48 Let b be the no. of boys and g be the no. of girls. 𝑏 + 𝑔 = 48

𝑏 =3

5𝑔 =

39

5

Eq becomes,

3𝑔

5+ 𝑔 = 48

3𝑔 + 5𝑔 = 485 8𝑔 = 485

𝑔 =485

8

∴g=30 ∴b=18

5. If 1

2 is subtracted from a number and the difference is multiplied by 4, the

result is 5. What is the number?

4 𝑥 −1

2 = 5

4 2𝑥−1

2 = 5

2 2𝑥 − 1 = 5 ⇒ 4𝑥 − 2 = 5 4𝑥 = 7

𝑥 =7

4

6. Let l be the length and b be the breadth, Given, 𝑙 = 3𝑏 − 6 2 𝑙 + 𝑏 = 148 2 3𝑏 − 6 + 𝑏 = 148

2(4𝑏 − 6) =148

2

4b=74+6=80

b=80

4

b=20 𝑙 = 3𝑏 − 6 = 3 20 − 6 = 60-6 =54 CHAPTER-12- ALGEBRAIC EXPRESSIONS

1. Let 𝑙 be the length and w be the width. Given, 𝑙 = 4 + 2𝑤 ⇒ 2𝑤 = 𝑙 − 4

𝑤 =𝑙−4

2

Perimeter=28 units i.e) 2 𝑙 + 𝑏 = 28

2 𝑙 +𝑙−4

2 = 28

2 2𝑙+𝑙−4

2 = 28

3𝑙 − 4 = 28 2. Expression: 𝑝 − 𝑞 + 𝑟 𝑐𝑚 3. Let n be the Nishant age, p be the Prashant age, h be the Hemant age.

Given, n=3p h=2+p n+p+h=62 3p+p+2+p=62 5p=62-2 5p=60

p=60

5= 12

∴Hemant age=2+p =2+12 =14 4. 𝑃 = 6𝑥3 − 8𝑥2 + 4𝑥 − 2 𝑄 = 5 − 4𝑥 + 6𝑥2 − 8𝑥3 𝑃 + 2𝑄 ⇒ 2𝑄 = 2 5 − 4𝑥 + 6𝑥2 − 8𝑥3 = 10−8𝑥 + 12𝑥2 − 16𝑥3 𝑃 + 2𝑄 = 6𝑥3 − 8𝑥2 + 4𝑥 − 2 + 10 − 8𝑥 + 12𝑥2 − 6𝑥3 = −10𝑥3 + 4𝑥2 − 4𝑥 + 8 5. Ones place is represented by 𝑥 Tens place is represented by y ∴The 2-digit number is 10𝑦 + 𝑥 6. Side of an equilateral triangle= 3𝑎 + 4𝑏 𝑚 Perimeter=𝑠 + 𝑠 + 𝑠 (3a+4b)+(3a+4b)+(3a+4b) 3a+3a+3a+4b+4b+4b 9a+12b CHAPTER 6 THE TRIANGLE AND ITS PROPERTIES

1. The two aspects are 1) sides 2) angles 2. ∠𝑃𝑄𝑅 + 120° = 180° ∠𝑃𝑄𝑅 = 180 − 120 = 60° ∴ 60 + 𝑥 = 110° [By Exterior angle property] 𝑥 = 110 − 60 𝑥 = 50° 3. By Pythagora’s theorem, 17²=15²+𝑥2 289=225+𝑥² 64=𝑥2 8=𝑥 4. 4km N end 3km ? (x) W N Start S

By Pythagora’s theorem, x²=4²+3² =16+9 x²=25 x=5 5. In ∆𝐴𝐵𝐶, line segment DE is parallel to BC. Find 𝑥, 𝑦 and z. ∠𝑧 = 70° (corresponding angles) ∠𝑦 = 50° (Corresponding angles) 𝑥 + 70 + 𝑦 = 180° (Angle sum property of triangle) x+70+50=180 x=180-120 𝑥 = 60° 6. Three angles of a triangle are in the ratio 4:3:2. Find the measure of the angles. Angles of a triangle ⇒ 4: 3: 2 By angle sum property of triangle, 4𝑥 + 3𝑥 + 2𝑥 = 180° 9𝑥 = 180°

x=180

9

𝑥 = 20° ∴The 3 angle are 4𝑥 = 420 = 80° 3𝑥 = 320 = 60° 2𝑥 = 220 = 40°