Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 22

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Question 1
1 / -0

The diagonals AC & BD of parallelogram ABCD intersect each other at P of $\angle DAC = 40^o$ & $\angle APB = 80^o$ . Then $\angle DBC$ is equal to ___ .

A
$40^o$

B
$80^o$

C
$120^o$

D
None

Solution

ABCD in parallelogram

$AD || BC \therefore \angle D AC = \angle BCA$

$\angle APB = \angle ACB + \angle DBC$ (Exterior angle of a triangle)

$\angle DBC = \angle APB - \angle ACB$

$\angle DBC = 80^o - \angle DAC$

$\angle DBC = 80^o - 40^o$

$\angle DBC = 40^o$
Question 2
1 / -0

In $\triangle$ ABC, BDEF and FDCE are parallelogram. Which of the following is correct?

A
AB = EF

B
DB = AB

C
ED $\neq$ BF

D
BD = CD

Solution

BDEF & FDCD are parallelogram
BF || DE, BF || DE, & FD || CE & CD || FE
$\angle BDF = \angle EFD,$ & $\angle DFE = \angle DCE$
$\angle FBD = \angle FED$ & $\angle FED = \angle FED = \dfrac{1}{2} \angle FDC$ (parallelogram purpose)
$\triangle FDB$ & $\triangle EDC$ are circular and conguent triangles.
$\because BD = BC$
Question 3
1 / -0

PQRS is a parallelogram, the bisector of $\angle$ P also bisects SQ, PQ = 5 cm. Find PS.

A
5 cm

B
7 cm

C
10 cm

D
Cant determine

Solution

If only one side of a parallelogram is given, then it's not possible to find out the other pair of opposite sides of a parallelogram. The information is incomplete.
Question 4
1 / -0

In a parallelogram PQRS, find $\angle RSQ$ if $\angle SPQ = 70^o$ & $\angle RQS = 60^o$ .

A
$50^o$

B
$60^o$

C
$70^o$

D
None

Solution

Opposite angles of a parallelogram are equal
$\therefore \angle SPQ = \angle SRQ = 70^o$
In $\triangle SQR, \angle RSQ + \angle SQR + \angle SRQ = 180^o$
$\angle RSQ = 180- 60 - 70$
$RSQ = 50^o$
Question 5
1 / -0

The value of $xy$ for parallelogram $ABCD$ is :

A
$2,695$

B
$2,940$

C
$4,704$

D
$6,468$

Solution

Opposite angles in a parallelogram are equal, implying $\angle ADC = \angle ABC$
$\therefore x + y = 112 ...(1)$
Also, $\angle DAC = \angle BCA$ since they are alternate angles.
$\therefore x - y = 42 ...(2)$
Adding equations (1) and (2), we get
$2x = 154$ or $x = 77$
$\Rightarrow y = 77 - 42 = 35$
Therefore, $xy = 77 \times 35 = 2695$
Question 6
1 / -0

If ABCD is a parallelogram, then how are the angles 'x' and 'y' related?

A
x = y

B
x > y

C
x < y

D
Can't be determined
Question 7
1 / -0

In a parallelogram $ABCD$ , $AB=4\ cm$ and $BC = 7\ cm$ . Each of its diagonals is less than which of the given options?

A
$3\ cm$

B
$6\ cm$

C
$7\ cm$

D
$11\ cm$

Solution

For a parallelogram ABCD,
$diagonal\quad AC>AB+BC\\ AC>7+4\\ AC>11cm$
So, option D is correct.
Question 8
1 / -0

The perimeter of a parallelogram,whose one side measures $12$ inches, is 72 inches. Find the length of its other three sides.

A
$12, 12, 36$

B
$12, 18, 18$

C
$12, 24, 24$

D
$12, 30, 30$
Question 9
1 / -0

In figure, if $ABCD$ is a parallelogram, $\angle ADE = 40^{\circ}$ , and $\angle C = 110^{\circ}$ , calculate the measure of $\angle BED$ .

A
$30^{\circ}$

B
$40^{\circ}$

C
$70^{\circ}$

D
$110^{\circ}$

E
$150^{\circ}$

Solution

Given, $\angle$ $C$ $=$ $110^o, \angle ADE = 40^o$
$\angle$ $D=\angle B$ {Opposite angle}
Also,
$\angle D + \angle C = 180^o$ {because these interior angles lie on the same side of the transversal}
$\angle D=180^o-110^o = 70^o$

Therefore,
$\angle$ $EDC =\angle D- \angle ADE = 70^o-40^o = 30^o$

$\angle$ $EBC = \angle B = 70^o$ , $\angle$ $DCB = 110^o$

Therefore,
$\angle$ $BED = 360^o-110^o-70^o-30^o = 150^o$ {Sum of angle of quadrilateral EBCD = $360^o$ }
Question 10
1 / -0

In the given figure, find $\angle BDC$ .

A
${ 60 }^{ o }$

B
${ 75 }^{ o }$

C
${ 45 }^{ o }$

D
${ 50 }^{ o }$

Solution

Let given quadrilateral $ABCD$ is a parallelogram.
Then, $AB$ is parallel to $CD$ and $AD$ is parallel to $BC$ (Opposite sides of parallelogram are parallel)
$\Rightarrow \angle$ DAB $=$ $\angle$ BCD (Opposite angles of parallelogram are equal)
Since, it is given that $\angle$ DAB $=$ 75 $^\circ$ .
Then, $\angle$ BCD $=$ 75 $^\circ$ ---- (1)
$\Rightarrow$ $\angle$ DBC = 60 $^\circ$ ---- (2) (Given)
In $\triangle$ DCB,
$\angle$ BDC + $\angle$ BCD + $\angle$ DBC $= 180$ $^\circ$
$\angle$ BDC + 75 $^\circ$ + 60 $^\circ$ $=$ 180 $^\circ$ [From (1) and (2) ]
$\angle$ BDC $=$ 180 $^\circ$ - 135 $^\circ$
$\therefore$ $\angle BDC = 45^\circ$

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Understanding Quadrilaterals Test - 22

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Understanding Quadrilaterals Test - 22

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Question 1

40o40^o40o

80o80^o80o

120o120^o120o

None

Solution

Question 2

AB = EF

DB = AB

ED ≠\neq= BF

BD = CD

Solution

Question 3

5 cm

7 cm

10 cm

Cant determine

Solution

Question 4

50o50^o50o

60o60^o60o

70o70^o70o

None

Solution

Question 5

2,6952,6952,695

2,9402,9402,940

4,7044,7044,704

6,4686,4686,468

Solution

Question 6

x = y

x > y

x < y

Can't be determined

Question 7

3 cm3\ cm3 cm

6 cm6\ cm6 cm

7 cm7\ cm7 cm

11 cm11\ cm11 cm

Solution

Question 8

12,12,3612, 12, 3612,12,36

12,18,1812, 18, 1812,18,18

12,24,2412, 24, 2412,24,24

12,30,3012, 30, 3012,30,30

Question 9

30∘30^{\circ}30∘

40∘40^{\circ}40∘

70∘70^{\circ}70∘

110∘110^{\circ}110∘

150∘150^{\circ}150∘

Solution

Question 10

60o{ 60 }^{ o }60o

75o{ 75 }^{ o }75o

45o{ 45 }^{ o }45o

50o{ 50 }^{ o }50o

Solution

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$40^o$

$80^o$

$120^o$

ED $\neq$ BF

$50^o$

$60^o$

$70^o$

$2,695$

$2,940$

$4,704$

$6,468$

$3\ cm$

$6\ cm$

$7\ cm$

$11\ cm$

$12, 12, 36$

$12, 18, 18$

$12, 24, 24$

$12, 30, 30$

$30^{\circ}$

$40^{\circ}$

$70^{\circ}$

$110^{\circ}$

$150^{\circ}$

${ 60 }^{ o }$

${ 75 }^{ o }$

${ 45 }^{ o }$

${ 50 }^{ o }$