Understanding Quadrilaterals(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

UNDERSTANDING QUADRILATERALS MCQs

Total Questions : 464 | Page 42 of 47 pages

Question 152 (iv)
A rectangle MORE is shown below.

Answer the following questions by giving an appropriate reason.
Is $Δ M Y O ≅ Δ R X E$ ?

Discuss Question

Answer: Option A. ->
:
Yes,

Δ M Y O ≅ Δ R X E

Δ M Y O

and

Δ R X E

MO = RE [opposite sides are equal]

∠ M O Y = ∠ R E X

[

∵

Alternate angles,

O M ∥ R E

, OE is the transversal]

∠ M Y O = ∠ R X E

[MY and RX are perpendicular to OE]

∴ Δ M Y O ≅ Δ R X E

[by AAS]

Question 412.

Question 152 (iii)
A rectangle MORE is shown below.

Answer the following questions by giving an appropriate reason.
Is ∠MOY = ∠REX?

Discuss Question

Answer: Option A. ->
:
Yes,

∠ M O Y = ∠ R E X

∵ R E ∥ O M

and EO is a transversal.

∴ ∠ M O E = ∠ O E R

[

∵

alternate interior angles]

\Rightarrow ∠ M O Y = ∠ R E X

Question 413.

Question 152 (v)
A rectangle MORE is shown below.

Answer the following questions by giving an appropriate reason.
Is MY = RX?

Discuss Question

Answer: Option A. ->
:
Yes, MY = RX
Since these are corresponding parts of congruent triangles.

Question 414.

Question 153
In parallelogram LOST, $S N ⊥ O L$ and $S M ⊥ L T$ . Find $∠ S T M, ∠ S O N$ and $∠ N S M$ .

Discuss Question

Answer: Option A. ->
:
Given,

∠ M S T = 40^{\circ}

Δ M S T

,
By the angle sum property of a triangle,

∠ T M S + ∠ M S T + ∠ S T M = 180^{\circ}

\Rightarrow ∠ S T M = 180^{\circ} - (90^{\circ} + 40^{\circ})

[∵ S M ⊥ L T, ∠ T M S = 90^{\circ}]

= 50^{\circ}

∴ ∠ S O N = ∠ S T M = 50^{\circ}

[

∵

opposite angles of a parallelogram are equal]
Now, in the

Δ O N S

∠ O N S + ∠ O S N + ∠ S O N = 180^{\circ}

[angle sum property of triangle]

∠ O S N = 180^{\circ} - (90^{\circ} + 50^{\circ})

= 180^{\circ} - 140^{\circ} = 40^{\circ}

Moreover,

∠ S O N + ∠ T S O = 180^{\circ}

[

∵

adjacent angles of a parallelogram are supplementary]

\Rightarrow ∠ S O N + ∠ T S M + ∠ N S M + ∠ O S N = 180^{\circ} \Rightarrow 50^{\circ} + 40^{\circ} + ∠ N S M + 40^{\circ} = 180^{\circ} \Rightarrow 130^{\circ} + ∠ N S M = 180^{\circ} \Rightarrow ∠ N S M = 180^{\circ} - 130^{\circ} = 50^{\circ}

Question 415.

Question 154
In trapezium HARE, EP and RP are bisectors of $∠ E$ and $∠ R$ , respectively. Find $∠ H A R$ and $∠ E H A$ .

Discuss Question

Answer: Option A. ->
:
Given,

∠ P E R = 25^{\circ}

and

∠ P R E = 30^{\circ}

Since EP and PR are angle bisectors of

∠ R E H

and

∠ A R E

respectively,

∠ P E H = 25^{\circ}

and

∠ P R A = 30^{\circ}

∠ E + ∠ H = 180^{\circ} a n d ∠ R + ∠ A = 180^{\circ}

[HARE is a trapezium]

\Rightarrow ∠ P E R + ∠ P E H + ∠ H = 180^{\circ} a n d

∠ E R P + ∠ P R A + ∠ R A H = 180^{\circ}

\Rightarrow 25^{\circ} + 25^{\circ} + ∠ H = 180^{\circ} a n d

30^{\circ} + 30^{\circ} + ∠ A = 180^{\circ}

\Rightarrow 50^{\circ} + ∠ H = 180^{\circ} a n d 60^{\circ} + ∠ A = 180^{\circ} \Rightarrow ∠ H = 130^{\circ} a n d ∠ A = 120^{\circ}

∴ ∠ E H A = 130^{\circ} a n d ∠ H A R = 120^{\circ}

Question 416.

Question 155
In parallelogram MODE, the bisectors of $∠ M$ and $∠ O$ meet at Q. Find the measure of $∠ M Q O$

Discuss Question

Answer: Option A. ->
:
Let MODE be a parallelogram and Q be the point of intersection of the bisector of

∠ M

and

∠ O

Question 155In Parallelogram MODE, The Bisectors Of ∠M And...

Since, MODE is a parallelogram,

∴ ∠ E M O + ∠ D O M = 180^{\circ}

[

∵

adjacent angles are supplementary]

\frac{1}{2} ∠ E M O + \frac{1}{2} ∠ D O M = 90^{\circ}

[dividing both sides by 2]

\Rightarrow ∠ Q M O + ∠ Q O M = 90^{\circ}

........ (i)
Now, in

Δ M O Q

∠ Q O M + ∠ Q M O + ∠ M Q O = 180^{\circ}

[angle sum property of triangle]

\Rightarrow 90^{\circ} + ∠ M Q O = 180^{\circ}

[from Eq. (i)]

∴ ∠ M Q O = 180^{\circ} - 90^{\circ} = 90^{\circ}

Question 417.

Question 156
A playground is in the form of a rectangle ATEF. Two players are standing at the point F and B, where EF = EB. Find the values of x and y.

Discuss Question

Answer: Option A. ->
:
Given, a rectangle ATEF in which EF = EB. Then,

Δ F E B

is an isosceles triangle. Therefore, by the angle sum property of a triangle, we have

∠ E F B + ∠ E B F + ∠ F E B = 180^{\circ}

[angle sum property of triangle]

\Rightarrow ∠ E F B + ∠ E B F + 90^{\circ} = 180^{\circ}

[

∵

in a rectangle, each angle is of

90^{\circ}

]

\Rightarrow 2 ∠ E F B = 90^{\circ}

[

∵ E B = E F \Rightarrow ∠ E F B = ∠ E B F

]

∠ E F B = 45^{\circ}

and

∠ E B F = 45^{\circ}

Now,

∠ x = 180^{\circ} - 45^{\circ} = 135^{\circ}

[linear pair]
and

∠ E F B + ∠ y = 90^{\circ}

[

∵

in a rectangle, each angle is of

90^{\circ}

]

\Rightarrow ∠ y = 90^{\circ} - 45^{\circ} = 45^{\circ}

Question 418.

Question 157
In the following figure of a ship, ABDH and CEFG are two parallelograms. Find the value of x.

Discuss Question

Answer: Option A. ->
:
We have, two parallelograms ABDH and CEFG.
In parallelogram ABDH,

∠ A B D + ∠ B D H = 180^{\circ}

[

∵

adjacent angles of a parallelogram are supplementary]

\Rightarrow ∠ B D H = 180^{\circ} - 130^{\circ} \Rightarrow ∠ B D H = 50^{\circ} \dots \dots (i)

In parallelogram CEFG,

∠ G C E = ∠ G F E

[

∵

Opposite angles are equal]

\Rightarrow ∠ G C E = 30^{\circ} \dots \dots (i i)

Δ O C D

, by using angle sum property,

∠ O C D + ∠ O D C + ∠ C O D = 180^{\circ}

\Rightarrow 50^{\circ} + 30^{\circ} + x = 180^{\circ}

[From (i) and (ii)]

\Rightarrow x = 180^{\circ} - 80^{\circ} = 100^{\circ}

Question 419.

Question 158
A rangoli has been drawn on the floor of a house. ABCD and PQRS both are in the shape of a rhombus. Find the radius of semi-circle drawn on each side of rhombus ABCD.

Discuss Question

Answer: Option A. ->
:
In rhombus ABCD,
AO = OP + PA = 2 + 2 = 4 units and OB = OQ+QB = 2 + 1 = 3 units.
We know that, diagonals of the rhombus bisect each other at