MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 4

All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

0%
Rectangle
0%
Trapezium
0%
Rhombus
0%
Parallelogram

Explanation

We know that the sum of the angles of the quadrilateral is

$360^{o}$

Now, the quadrilateral in given question has all angles equal.

$\therefore$ Each angle

$={ 360 }^{ o }\div 4={ 90 }^{ o }$

So, this quadrilateral would be rectangle or a square.

In the above figure, it is given that

$BDEF$ and

$FDCE$ are parallelograms. Can you say that

$BD = CD$ ? Why or why not?

0%
Yes
0%
No
0%
May be
0%
Cannot be determined

Explanation

Given:-

$ABC$ is a triangle.

$D,\;E\;F$ are points on

$BC,\;CA.\;AB$ taken in order.

$BDEF$ anf

$FDCE$ are parallelograms.

To Prove:-

$BD=CD$

Proof:-

In parallelogram

$BDEF$ ,

$BD=EF\quad\quad\quad\dots(i)$ [Opposite sides of a parallelogram are equal.]

In parallelogram

$FDCE$ ,

$CD=EF\quad\quad\quad\dots(ii)$ [Opposite sides of a parallelogram are equal.]

From

$(i)$ and

$(ii)$ ,

$BD=CD$

Hence, proved.

The adjacent figure

$\Box HOPE$ is a parallelogram. Find the angle measures

$x, y$ and

$z$ . State the properties you use to find them.

0%
$x = 110^{o}, y = 40^{o}, z = 30^{o}$
0%
$x = 110^{o}, y = 30^{o}, z = 30^{o}$
0%
$x = 110^{o}, y = 10^{o}, z = 30^{o}$
0%
$x = 110^{o}, y = 20^{o}, z = 30^{o}$

Explanation

Given-

$HOPE$ is a parallelogram with diagonal

$HP$ .

$\angle PHE={ 40 }^{ o }$

The external angle by extending

$HO$ is

${ 70 }^{ o }$

To find out-

$\angle HEP=x$

$\angle HPO=y=?$ and

$\angle PHO=z=?$

Solution-

The external

$\angle HOP={ 70 }^{ o }$

$\therefore \angle HOP={ { 180 }^{ o }-70 }^{ o }$ ....(linear pair)

$={ 110 }^{ o }$

Now

$\angle HOP=x$ ....(opposite angle of a parallelogram)

$\therefore x={ 110 }^{ o }$

Again

$HE\parallel OP$ .....(opposite sides of a parallelogram)

$\therefore y=\angle HPE$ ....(alternate angles) or

$y={ 40 }^{ o }.$

To find out

$z$ , we have

$z={ 180 }^{ o }-\angle HOP-y={ 180 }^{ o }-{ 110 }^{ o }-{ 40 }^{ o }$ ....(angle sum property of triangles)

$\Rightarrow z={ 30 }^{ o }$

Ans-

$x={ 110 }^{ o }, y={ 40 }^{ o }, z={ 30 }^{ o }$

Given a parallelogram ABCD, if

$\angle A = 60^{\circ}$ then

$\angle D$ is equal to :

0%
$110^{\circ}$
0%
$140^{\circ}$
0%
$120^{\circ}$
0%
$130^{\circ}$

Explanation

$\angle A=60^o,$ therefore the opposite angle

$C=60^o$

$\angle B=\angle D$ [Rule of parallelogram] as they are opposite angles.

Let

$\angle B$ and

$\angle D$ be

$x$

$\Rightarrow x+x+\angle A+\angle B=360^o$ [ sum of all sides of a polygon ]

$\Rightarrow 2x+60^o+60^o=360^o$

$\Rightarrow 2x=240^o$

$\Rightarrow x=120^o$

$\therefore \angle D=120^o$

Hence, the answer is

$120^o.$

Two consecutive angles of a parallelogram are in the ratio

$1 : 3$ , then the smaller angle is :

0%
$50^{\circ}$
0%
$90^{\circ}$
0%
$60^{\circ}$
0%
$45^{\circ}$

Explanation

$\textbf{Step -1: Formulating the sum of angles}$

$\text{We know that, sum of two consecutive angles of a parallelogram = 180}^\circ$

$\text{Given angles are in the ratio 1 : 3}$

$\therefore \text{The angles are }x\text{ and }3x.$

$\therefore x + 3x = 180^\circ$

$\Rightarrow 4x = 180^\circ$

$\Rightarrow x = \dfrac{180}{4}$

$\therefore x = 45^\circ$

$\therefore \text{The smaller angle is 45}^\circ$

$\textbf{Hence, The smaller angle is 45}^\circ.$

$\textbf{Hence, option D is correct.}$

In figure, ABCD is a parallelogram. If

$\angle DAB = 60^{\circ}\ and\ \angle DBC = 80^{\circ},\angle CDB$ is :

0%
$40^{\circ}$
0%
$80^{\circ}$
0%
$60^{\circ}$
0%
$20^{\circ}$

Explanation

Given

$ABCD$ is a parallelogram and

$BD$ is the diagonal.

$AB\parallel CD$

$\Rightarrow \angle CBD=\angle ADB=80^o$

We have

$\angle DAB=\angle BCD=60^o$

Diagonal

$BD$ forms a triangle

$CBD,$

$\Rightarrow \angle DCB+\angle CBD + \angle CDB =180^o$ [ Sum of

$\angle$ les of a triangle }

$\Rightarrow 60^o+80^o+\angle CDB=180^o$

$\therefore \angle CBD=40^o$

$,$

$\angle ADB=80^o$

Hence, the answer is

$40^o.$

In figure, PQRS is a parallelogram in which

$\angle PSR=125^{\circ} , \angle RQT$ is equal to

0%
$75^{\circ}$
0%
$65^{\circ}$
0%
$55^{\circ}$
0%
$125^{\circ}$

Explanation

Given PQRS is a parallelogram.

$\Rightarrow \angle PSR=\angle PQR=125°$ [ i.e, opposite angles ]

$\therefore \angle RQT=180°-125°$

$=55°.$

Hence, the answer is

$55°.$

$ABCD$ is a parallelogram as shown in figure. If

$AB = 2AD$ and

$P$ is mid-point of

$AB$ , then

$\angle{CPD}$ is equal to

0%
$90^{\circ}$
0%
$60^{\circ}$
0%
$45^{\circ}$
0%
$135^{\circ}$

In parallelogram ABCD, if

$\angle A = 2x + 15^{\circ}, \angle B = 3x - 25^{\circ},$ then value of

$x$ is :

0%
$91^{\circ}$
0%
$89^{\circ}$
0%
$34^{\circ}$
0%
$38^{\circ}$

Explanation

Given the parallelogram ABCD.

In case of a parallelogram the sum of adjacent angles will be

$180°$

i.e,

$\angle A +\angle B=180°$

$\Rightarrow 2x+15+3x-25^o=180°$

$\therefore 5x=190°$

i.e,

$x=38°$

Hence, the answer is

$38°.$

ABCD is a parallelogram in which

$\angle DAC = 40^{\circ} ; \angle BAC = 30^{\circ}; \angle DOC = 105^{\circ}$ ; then

$\angle CDO$ equals :

0%
$75^{\circ}$
0%
$70^{\circ}$
0%
$40^{\circ}$
0%
$45^{\circ}$

Explanation

$\angle AOB = \angle COD = 105^{\circ}$ (Vertically Opposite Angles)

In triangle

$AOB,$

$\angle BAO + \angle AOB + \angle OBA = 180^{\circ}$

$30^{\circ} + 105^{\circ} + \angle OBA = 180^{\circ}$

$\angle OBA = 45^{\circ}$

$AB$ and

$CD$ are parallel

$\therefore \angle OBA = \angle CDO = 45^{\circ}$ (Alternate Interior angles)

If PQRS is a parallelogram, then

$\angle{Q}- \angle{S}$ is equal to :

0%
$90^{\circ}$
0%
$120^{\circ}$
0%
$180^{\circ}$
0%
$0^{\circ}$

Explanation

Given that PQRS is parallelogram,

Opposite angles in a parallelogram are equal.

i.e,

$\angle P=\angle R$ and

$\angle Q=\angle S$

$\therefore$ If

$\angle Q=\angle S$ then

$\angle Q-\angle S=0°$

Hence, the answer is

$0°.$

The diagonals of a parallelogram PQRS intersect at O. If

$\angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}, \ then\ \angle ORS$ is :

0%
$90^{\circ}$
0%
$40^{\circ}$
0%
$70^{\circ}$
0%
$50^{\circ}$

Explanation

Given:

$\angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}$

Now,

$\angle SOR = 180^{\circ} - \angle QOR = 90^{\circ}$ (Since angles forming a linear pair are complementary)

Consider

$\Delta SOR$ ,

$\angle SOR + \angle OSR + \angle ORS = 180^{\circ}$ (angle sum property of triangles)

$\therefore \angle ORS = 180^{\circ} - 90^{\circ} - 50^{\circ} = 40^{\circ}$

Two adjacent angles of a parallelogram are

$(2x + 30)^{\circ}$ and

$(3x + 30)^{\circ}$ . The value of

$x$ is :

0%
$30^{\circ}$
0%
$60^{\circ}$
0%
$24^{\circ}$
0%
$36^{\circ}$

Explanation

We know that the sum of the adjacent angles of parallelogram is equal to

$180^o$

$\therefore 2x+30^o+3x+30^o=180°$

$\Rightarrow 5x=120^o$

$\therefore x=\dfrac{120^o}{5}=24°$

Hence, the answer is

$24^o.$

In the given figure, ABCD is a parallelogram. If

$\angle B = 100^{\circ}$ , then (

$\angle A + \angle C$ ) is equal to :

0%
$360^{\circ}$
0%
$200^{\circ}$
0%
$180^{\circ}$
0%
$160^{\circ}$

Explanation

Given ABCD is a parallelogram.

The sum of co-interior angles is always

$180°$ .

Given

$\angle B =100°$

and,

$\angle B +\angle C=180°$

$\Rightarrow \angle C = 80°$

The opposite angles of a parallelogram are equal

i.e,

$\angle A= \angle C=80°$

$\therefore$ Sum

$\angle A+ \angle C=80°+80°=160°$

Hence, the answer is

$160°.$

Consider the following statements:
(1) The diagonals of a parallelogram are equal.
(2) The diagonals of a square are perpendicular
to each other.
(3) If the diagonals of a quadrilateral intersect at
right angles, it is not necessarily a rhombus.
(4) Every quadrilateral is either a trapezium or a
parallelogram or a kite.
Which of the above statements is/are correct?

0%
Only (2)
0%
Only (3)
0%
Both (2) and (3)
0%
(1), (2) and (3)

If the sum of all interior angles of a convex polygon is

$1440^{\circ}$ , then the number of sides of the polygon is?

0%
$8$
0%
$10$
0%
$11$
0%
$12$

Explanation

If n is the number of sides of the polygon, then

$(2n - 4)\times 90^{\circ}= 1440^{\circ}$
or

$2n = 20$
or

$n = 10$

The measurement of each angle of a polygon is 160

$^{\circ}$ . The number of its sides is?

0%
$15$
0%
$18$
0%
$20$
0%
$30$

Explanation

Each angle of polygon

$= 160^{o}$
Let there be n sides, then sum of all the angles

$= 160n$

Sum of all the angles of any polygon

$= (n-2)(180^{o})$
Therefore,

$160 n = (n-2)180^{o}$

$160^{o}n= 180^{o}n - 360^{o}$

$20n = 360$

$n= 18$
The polygon has

$18$ sides.

If one angle of the parallelogram is

$16^{\circ}$ less than three times the smallest angle, then the largest angle of the parallelogram is

0%
$131^{\circ}$
0%
$136^{\circ}$
0%
$112^{\circ}$
0%
$108^{\circ}$

Explanation

Let

$x$ and

$y$ be the largest and smallest angles of parallelogram.
By hypothesis

$x = 3y - 16$
Also in a parallelogram, we have

$x + y = 180^{\circ}$

$\Rightarrow (3y - 16) + y = 180^{\circ}$

$\Rightarrow 4y = 196$

$\Rightarrow y = 49$
Thus

$x = 49\times 3 - 16$

$= 147 - 16 \\= 131^{\circ}$

Each interior angle of a regular polygon is

$144^0$ . Find the interior angle of a regular polygon which has double the number of sides as the first polygon.

0%
$100^0$
0%
$160^0$
0%
$36^0$
0%
$162^0$

Explanation

Since each interior angle of the first polygon

$=144^o$
Each exterior angle of the first polygon

$=180-144$

$=36^o$

$\therefore$ The number of sides of the first polygon

$=\frac {360}{36}=10$

$\therefore$ The number of sides of the second polygon

$=2\times 10=20$

$\therefore$ Each exterior angle of the second polygon

$=\frac{360^o}{20}=18^o$

$\therefore$ Each interior angle of the second polygon

$=180^o-18^o=162^o$

If the sum of all the angles of a polygon except one angle is

$2220^\circ$ , then the number of sides of the polygon are:

0%
12
0%
13
0%
14
0%
15

Explanation

Sum of angles of polygon with

$n$ sides

$= (n - 2) \times 180^{\circ}\dots(1)$

The sum of the angles of a polygon should be a multiple of

$180^\circ$ .

But,

$2220^{\circ} = 180^{\circ}\times 12 + 60^{\circ}$

$2220^{\circ}$ +

$120^{\circ}$ =

$2340^{\circ}$

$( x - 2 ) = \dfrac{2340}{180}\dots$ (From

$(1)$ )

$x - 2 = 13$

$\Rightarrow x = 15$

Therefore, the number of sides of the polygon is

$15$ .

$ABCD$ is a parallelogram whose diagonals intersect at

$O$ and

$BCD$ is an equilateral triangle having each side of length

$6$ cm, then the length of diagonal

$AC$ is :

0%
$3\sqrt{3}$ cm
0%
$6\sqrt{3}$ cm
0%
$3\sqrt{6}$ cm
0%
$12$ cm

Explanation

$ABCD$ is a parallelogram

$\therefore BC = AD = 6$ cm
And

$AB = DC = 6$ cm
Hence,

$ABCD$ becomes a rhombus.

$\therefore$ Diagonals

$AC$ and

$BD$ bisect each other at right angle.

$\therefore OD = \dfrac{1}{2} BD = 3$ cm
From

$\Delta OCD$ ,

$OC^2 = CD^2 - OD^2$ .... Pythagoras theorem

$= 36 - 9$

$= 27$

$\therefore OC = 3 \sqrt{3}$
and

$AC = 2\cdot OC = 6 \sqrt{3}$

Match List I with List II and select the correct answer using the codes given below the lists:

List I (Regular plane figure)	List II (Measure of interior angles)
I . Triangle	(A) $30^{\circ}$
II. Square	(B) $60^{\circ}$
III. Pentagon	(C) $108^{\circ}$
IV. Hexagon	(D) $90^{\circ}$
	(E) $120^{\circ}$

0%
I-D, II-A, III-B, IV-E
0%
I-B, II-D, III-C, IV-E
0%
I-A, II-D, III-C, IV-B
0%
I-B, II-C, III-A, IV-D

Explanation

Each Angle (of a Regular Polygon)

$= \dfrac {(n-2) × 180°} { n}$

Triangle =

$(3-2)×180°/3=60°$

Square=

$(4-2)×180°/4=90°$

Pentagon =

$(5-2)×180°/5=108°$

Hexagon =

$(6-2)×180°/6=120°$

$\therefore$ I-B, II-D, III-C, IV-E

A parallelogram with any pair of its consecutive sides equal is a

0%
Square
0%
Rectangle
0%
Rhombus
0%
Kite

State true or false:

It is possible to have a polygon whose sum of interior angles is

$\displaystyle 320^{\circ}$ .

0%
True
0%
False

Explanation

A polygon whose sum of interior angles is

$320^o$
Sum of interior angle of Polygon =

$180 (n -2)$
=>

$180 (n -2) = 320$
=>

$n - 2 = \dfrac {16}{9}$
=>

$n = \dfrac {25}{9}$
Their is no possibility of a polygon whose sum of interior angles is

$320^o$ because n must be an integer.

In parallelogram ABCD,

$\angle A = 3 \angle B$ . In the same parallelogram, if AB

$= 5x-7$ and

$CD = 3x + 1$ ; find the length of CD.

0%
$CD = 16$ units
0%
$CD = 11$ units
0%
$CD = 13$ units
0%
$CD = 15$ units

Explanation

$=$ CD ( Opposite sides of parallelogram)

$\implies 5x -7 = 3x + 1$

$\implies x = 4$

$\therefore$ AB

$=$ CD

$= 13$ units.

State true or false:

Is it possible to have a regular polygon whose exterior angle is

$40\%$ of a right angle?

0%
True
0%
False

Explanation

$40\%$ of a right angle will be,

$90^\circ \times \dfrac{{40}}{{100}} = 36^\circ$

For a

$n$ -sided regular polygon all exterior angles are equal to

$\dfrac{360^\circ}{n}$ . So,

$\begin{aligned}{}36^\circ& = \frac{{360^\circ}}{n}\\n &= \frac{{360^\circ}}{{36^\circ}}\\ &= 10\end{aligned}$

We obtain an integral positive value of

$n$ so, the given statement is true for a regular

$10$ -sided polygon.

In parallelogram PQRS

$\angle Q = (4x-5)^{\circ}$ and

$\angle S = (3x+10)^{\circ}$ . Calculate

$\angle Q$

0%
$45^o$
0%
$50^o$
0%
$55^o$
0%
$60^o$

Explanation

Opposite angles of a parallelogram are equal.

Therefore,

$\angle Q=\angle S$

$(4x-5)^{\circ}=(3x+10)^{\circ}$

$x=15^{\circ}$

$\angle Q=(4*15-5)^{\circ}=55^{\circ}$

Is it possible to have a regular polygon whose each interior angle is

$170^{\circ}$
State true or false:

0%
True
0%
False

Explanation

Let number of polygon = n (which must be an integer)
Sum of all interior angles is

$n \times 170^o$
Sum of all interior angles of a polygon is

$180^o (n-2)$

$=> n \times 170^o = 180^o (n-2)$

$=> 10^o n = 360^o$

$=> n = 36^o$
Yes, Their is a regular polygon whose each interior angle is

$170^o$

Is it possible to have a polygon, whose sum of interior angles is

$4500^{\circ}$ ?
State true or false:

0%
True
0%
False

Explanation

Sum of Interior angles of a polygon, having

$n$ sides

$= 180^{\circ} (n-2)$
Given sum of interior angles is

$4500^o$

$180^{\circ} (n-2) = 4500^{\circ}$

$n - 2 = 25$

$n = 27$
Since

$n$ is an integer so their exist a polygon whose sum of interior angles is

$4500^{\circ}$

State true or false:

Is it possible to have a polygon whose sum of interior angles is

$\displaystyle 540^{\circ}$ ?

0%
True
0%
False

Explanation

A polygon whose sum of interior angles is

$540^{\circ}$
Sum of interior angle of Polygon

$=180 (n -2)$

$\implies 180 (n -2) = 540$

$\implies n - 2 = 3$

$\implies n = 5$
Their exist a polygon whose sum of interior angles is

$540^{\circ}$ and the polygon is pentagon.

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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