quadrilaterals - e-worksheet · show that each angles of a rectangle is a right angle. ... in...
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![Page 1: Quadrilaterals - E-worksheet · Show that each angles of a rectangle is a right angle. ... In figure diagonal AC of parallelogram ABCD bisects ∠Ashow that (i) ... BD = AD A C B](https://reader030.vdocuments.mx/reader030/viewer/2022021621/5b3339177f8b9ab5728dac49/html5/thumbnails/1.jpg)
Quadrilaterals
1. A quadrilateral ABCD is a parallelogram if
(a) AB = CD (b) AB �BC
(c) 0 0 060 , 60 , 120A C B∠ = ∠ = ∠ = (d) AB = AD
2. In figure, ABCD and AEFG are both parallelogram if 080 , C then DGF is∠ = ∠
(a) 0100
(b) 060
(c) 080
(d) 0120
3. In a square ABCD, the diagonals AC and BD bisects at O. Then AOB∆ is
(a) acute angled (b) obtuse angled
(c) equilateral (d) right angled
4. ABCD is a rhombus. If 030 , ACB then ADB is∠ = ∠
(a) 030 (b) 0120
(c) 060 (d) 045
5. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
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6. Show that each angles of a rectangle is a right angle.
7. A transversal cuts two parallel lines prove that the bisectors of the interior angles enclose a rectangle.
8. Prove that diagonals of a rectangle are equal in length.
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9. In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that 090APB∠ =
10. In figure diagonal AC of parallelogram ABCD bisects A∠ show that (i) if bisects C∠ and (ii) ABCD is a rhombus
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11. In figure ABCD is a parallelogram. AX and CY bisects angles A and C. prove that AYCX is a
parallelogram.
12. The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
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13. Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square.
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ANSWERS
Ans01. (c) Ans02. (c) Ans03. (d) Ans04. (c)
Ans05. Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x
0 0
0
0
0
0
360 [ 360 ]
3 5 9 13 360
30 360
12
3 3 12 36
A B C D sum of angles of a quadrilateral is
x x x x
x
x
A x
∠ + ∠ + ∠ + ∠ =+ + + =
==
∴ ∠ = = × =0
0
0
5 5 12 60
9 9 12 108
13 13 12 156
B x
C x
D x
∠ = = × =∠ = = × =∠ = = × =
Ans06. We know that rectangle is a parallelogram whose one angle is right angle.
Let ABCD be a rectangle. 090A∠ =
To prove 090B C D∠ = ∠ = ∠ = Proof: AD BC∵ � and AB is transversal
0
0 0
0 0
0
0
0
180
90 180
180 90
90
90
90
A B
B
B
C A
C
D B
D
∴ ∠ + ∠ =+ ∠ =
∠ = −=
∠ = ∠∴ ∠ =
∠ = ∠∴ ∠ =
Ans07. AB CD∵ � and EF cuts them at P and R.
[ int ]
1 1
2 2. . 1 2
[ ]
APR PRD Alternate erior angles
APR PRD
i e
PQ RS Alternate
∴ ∠ = ∠
∴ ∠ = ∠
∠ = ∠∴ �
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Ans08. ABCD is a rectangle AC and BD is diagonals.
To prove AC = BD
Proof: In DAB and CBA∆ AD = BC [In a rectangle opposite sides are equal]
0 [90 ]A B each∠ = ∠ AB = AB common [common]
[ ]
[ ]
DAB CAB By SAS
AC BD By CPCT
∴ ∆ ≅ ∆∴ =
Ans09. Given ABCD is a parallelogram is and bisectors of A and B∠ ∠ intersect each
other at P.
To prove 090APB∠ =Proof:
( )
1 11 2
2 21
( )2
A B
A B i
∠ + ∠ = ∠ + ∠
= ∠ + ∠ →
But ABCD is a parallelogram AD �BC
0
0
0
0
180
1 1 2 180
290
1 2 180 [ ]A B
In APB
APB By angle sum property∴ ∠ + ∠ =
∴ ∠ + ∠ = ×
=
∆∠ + ∠ + ∠ =
0 0
0
90 180
90
APB
APB
Hence proved
+ ∠ =∠ =
Ans10. (i) AB�DC and AC is transversal
1 2 ( )
3 4 ( )
1 3
2 4
sec
Alternate angles
and Alternate angles
But
AC bi sts C
∴ ∠ = ∠∠ = ∠∠ = ∠
∴ ∠ = ∠∴ ∠
(ii) In ABC and ADC∆ ∆ [ ]
1 3 [ ]
2 4 [ ]
[ ]
AC AC common
given
proved
ABC ADC
AB AD By CPCT
=∠ = ∠
∠ = ∠∴ ∆ ≅ ∆∴ =∴ homABCD is a r bus
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Ans11. Given in a parallelogram AX and CY bisects A and C∠ ∠ respectively and we have
to show that AYCX in a parallelogram.
( ) [ log ]
1 [ ] ( )
21
BCY= C 2
In ADX and CBY
D B i opposite angles of a paralle ram
DAX A given ii
and
∆ ∆∠ = ∠ →
∠ = ∠ →
∠ ∠ [given] (iii)
But A= C
By (2) and (3), we get
→
∠ ∠∴
( )
[ log ] ( )
( ), ( ) ( ),
[ ]
DAX BCY iv
Also AD BC opposite sides of paralle ram v
From i iv and v we get
ADX CBY By ASA
DX BY
∠ = ∠ →= →
∴∆ ≅ ∆
∴ = [ ]
[ log ]
[ ]
log
CPCT
But AB CD oppsite sides of paralle ram
AB BY CD DX
or
AY CX
But AY XC ABCD is a gm
AYCX is a paralle ram
=− = −
=
∴� ∵ �
Ans12. Given ∆ ABC in which E and F are mid points of side AB and AC respectively.
To prove: EF||BC
Construction: Produce EF to D such that EF = FD. Join CD
Proof: In AEF and CDF∆ ∆ [ int ]
1 2 [ ]
[ ]
[ ]
AF FC F is mid po of AC
vertically opposite angles
EF FD BY construction
AEF CDF By SAS
= −∠ = ∠
=∴ ∆ ≅ ∆∴
∵
[ ]
[ int]
[ ]
log
AE CD By CPCT
and AE BE E is the mid po
BE cd
and AB CD BAC ACD
BCDE is a paralle ram
EF BC Hencep
== −
∴ =∴∠ = ∠
∴
∵
�
� roved
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Ans13. Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and 090AOB∠ =To prove: ABCD is a square.
Proof: In AOB and COD∆ ∆
[ ]
[ ]
[ ]
[ ]
[
OA OC given
OB OD given
and AOB COD vertically opposite angles
AOB COD By SAS
AB CD By CPC
==
∠ = ∠∴ ∆ ≅ ∆∴ = ]
1 2 [ ]
is a parallelogram whose diagonals bisects each other at right angles
rhombus
T
By CPCT
But there are alternate angles AB CD
ABCD
ABCD is a
Again in ABD and
∠ = ∠∴
∴∆ ∆
�
[ hom ]
[ hom ]
[ ]
[
BCA
AB BC Sides of a r bus
AD AB Sides of a r bus
and BD CA Given
ABD BCA
BAD CBA By CPCT
===
∴ ∆ ≅ ∆∴ ∠ = ∠
0 0
]
180 90
There are alternate angles of these same side of transversal
BAD CBA or BAD CBA
Hence ABCD is a square
∴ ∠ + ∠ = ∠ = ∠ =
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1. In fig ABCD is a parallelogram. It 60DAB∠ = ° and 80DBC∠ = ° then CDB∠ is
( )( )( )( )
A 80°
B 60°
C 20°
D 40°
2. If the diagonals of a quadrilateral bisect each other, then the quadrilateral must
be.
(a) Square
(b) Parallelogram
(c) Rhombus
(d) Rut angle
A B
CD
600
800
Quadrilateral
3. The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular
bisector of each other then quadrilateral ABCD is a
(a) Kite
(b) Square
(c) Trapezium
(d) Rut angle
4. The quadrilateral formed by joining the mid points of the sides of a quadrilateral
ABCD taken in order, is a rectangle if
(a) ABCD is a parallelogram
(b) ABCD is a rut angle
(c) Diagonals AC and BD are perpendicular
(d) AC=BD
5. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
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6. In fig ABCD is a parallelogram and X,Y are the points on the diagonal BD such that DX<BY show
that AYCX is a parallelogram.
AB
CD
Y
X
O
7. Show that the line segments joining the mid points of opposite sides of a quadrilateral
bisect each other.
A B
CD
E
F
G
H
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8. ABCD is a rhombus show that diagonal AC bisects A∠ as well as C∠ and diagonal BD bisects B∠ as well as D∠
9. Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles.
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10. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides.
11. In fig B∠ is a right angle in .ABC D∆ is the mid-point of .AC DE AB� intersects BC at E. show that
(i) E is the mid-point of BC
(ii) DE ⊥ BC
(iii) BD = AD
A
BC
D
E
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12. ABC is a triangle and through vertices A, B and C lines are drawn parallel to BC, AC and AB respectively intersecting
at D, E and F. prove that perimeter of DEF∆ is double the perimeter of ABC∆ .
13. Prove that in a triangle, the line segment joining the mid points of any two sides isparallel to the third side.
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ANSWERS
Ans1. (D) Ans2. (B) Ans3. (B) Ans4. (A)
Ans5. Given A quadrilateral ABCD in which AB = DC and AD = BC
To prove: ABCD is a parallelogram
Construction: Join AC
Prof: In ABC∆ and ADC∆
( )
[ ]
given
common
===
AD BC
AB DC
AC AC
[ ][ ]by sss
By CPCT
ABC ADC
BAC DAC
∴∆ ≅ ∆
∴∠ = ∠
ABCD∴ Is a parallelogram
B
CD
A
Ans6. ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each
other
OD OB∴ =
But DX BY= [given]
OD DX OB BY
or OX OY
∴ − = −=
Now in quad AYCX, the diagonals AC and XY bisect each other
AYCX∴ is a parallelogram.
Ans7. Given ABCD is quadrilateral E, F, G, H are mid points of the side AB, BC, CD and
DA respectively
To prove: EG and HF bisect each other.
In ABC∆ , E is mid-point of AB and F is mid-point of BC
EF AC∴ � And ( )1.......
2EF AC i=
Similarly HG AC� and ( )1......
2GD AC ii=
From (i) and (ii), EF HG� and EF GH=
EFGH∴ is a parallelogram and EG and HF are its diagonals
Diagonals of a parallelogram bisect each other
Thus EG and HF bisect each other.
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Ans8. ABCD is a rhombus
In ABC∆ and ADC∆
AB AD= [Sides of a rhombus]
BC DC= [Sides of a rhombus]
AC AC= [Common]
ABC ADC∴∆ ≅ ∆ [By SSS Congruency]
CAB CAD∴∠ = ∠ And ACB ACD∠ = ∠
Hence AC bisects A∠ as well as C∠
Similarly, by joining B to D, we can prove that ABD CBD∆ ≅ ∆
Hence BD bisects B∠ as well as D∠
B
CD
A
Ans9. Given ABCD is a quadrilateral diagonals AC and BD bisect each other at O at
right angles
To Prove: ABCD is a rhombus
Proof: ∵diagonals AC and BD bisect each other at O
,OA OC OB OD∴ = = And 1 2 3 90∠ = ∠ = ∠ = °
Now In BOA∆ And BOC∆
OA OC= Given
OB OB= [Common]
B
CD
A
0
1
23
And 1 2 90∠ = ∠ = ° (Given)
BOA BOC∴∆ = ∆ (SAS)
BA BC∴ = (C.P.C.T.)
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Ans10. Given a trapezium ABCD in which AB DC� and M,N are the mid Points of the
diagonals AC and BD.
We need to prove that MN AB DC� �
Join CN and let it meet AB at E
Now in CDN∆ and ∆EBN
DCN BEN∠ = ∠ [Alternate angles]
CDN BEN∠ = ∠ [Alternate angles]
And DN BN= [given]
CDN EBN∴∆ ≅ ∆ [SAA]
CN EN∴ = [By C.P.C.T]
Now in ,ACE M∆ and N are the mid points of the sides AC and CE respectively.
MN AE∴ � Or MN AB�
Also AB DC�
MN AB DC∴ � �
EB
CD
A
M N
Ans11. Proof: DE AB∵ � and D is mid points of AC
In DCE∆ and DBE∆
CE=BE
DE= DE
And 90DEC DEB∠ = ∠ = °
DCE DBE
DCE DBE
CD BD
∴∆ = ∆∴∆ ≅ ∆∴ =
Ans12. ∵BCAF Is a parallelogram
BC AF∴ =
ABCE∵ Is a parallelogram
2
BC AE
AF AE BC
∴ =+ =
Or 2EF BC=
Similarly ED = 2AB and FD = 2AC
∴Perimeter of ABC AB BC AC∆ = + +
Perimeter of DEF DE EF DF∆ = + +
= 2AB+2BC+2AC
= 2[AB+BC+AC]
= 2 Perimeter of ABC∆
Hence Proved
E
B C
D
A
F
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Ans13. Given: A ABC∆ in which D and E are mid-points of the side AB and AC
respectively
TO Prove: DE BC�
Construction: Draw CF BA�
Proof: In ADE∆ and CFE∆
1 2∠ = ∠ [Vertically opposite
angles]
AE=CE [Given]
And 3 4∠ = ∠ [Alternate interior angles]
ADE CFE∴∆ ≅ ∆ [By ASA]
∴DE=FE [By C.P.C.T]
But DA = DB
∴DB = FC
Now DB� FC
∴DBCF is a parallelogram
∴DE�BC
Also DE = EF = 1
2BC
E
BC
D
A
3
1
2
4
F