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$$^\circ$$
$$\therefore$$ $$\angle BDC = 45^\circ$$