MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 6

If two adjacents angles of a parallelogram are in ratio 3 : 9, then measure of smallest angle is

0%
$105$
0%
$60$
0%
$45$
0%
None

Explanation

Sum of adjacent angles of parallelogram

$=180^o$
One angle of a parallelogram

$= 3x$
other angle of a parallelogram

$= 9x$

$3x + 9x = 180$

$12 x = 180, x = \dfrac{180}{12} = 15$
one angle

$3x = 3 \times 15 = 45$

The bisectors of interior angles of a parallelogram forms a ___________.

0%
Square
0%
Rectangle
0%
Trapezium
0%
Rhombus

Explanation

$ABCD$ is a parallelogram

$AE$ bisects

$\angle BAD$

$BF$ bisects

$\angle ABC$

$CG$ bisects

$\angle BCD$

$DH$ bisects

$\angle ADC$

$\angle BAD + \angle ABC = 180{^o}$ (because adjacent angles of a parallelogram are supplementary)

$\angle BAJ = \cfrac{1}{2} \angle BAD$ because

$AE$ bisects

$\angle BAD$

$\angle ABJ = \cfrac{1}{2} \angle ABC$ because

$DH$ bisects

$\angle ABC$

$\angle BAJ$ +

$\angle ABJ$ =

$90^{o}$ [ halves of supplemetary angles are complementary]

$\triangle ABJ$ is a right triangle because its acute interior angles are complementary with

$\angle J = 90^o$

Similarly,

$\angle K =\angle L= \angle I = 90^o$

In quadrilateral

$IJLK$ , since all angles are

$90^o$ , it must be a rectangle.

Therefore, internal angle bisectors of a parallelogram forms a rectangle.

L and M are the mid-points of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC. StateTrue or False.

0%
True
0%
False

Explanation

Given: Parallelogram ABCD, L and M are mid points of

$AB$ and

$DC$ respectively.
We know,

$AB = CD$
Hence,

$DM = LB$ and

$DM \parallel LB$ (As,

$AB \parallel CD$ )
Suppose

$DL$ and

$MB$ intersect AC at P and Q respectively.
Consider,

$\triangle ABC$ ,

$PL \parallel QB$
L is mid point of

$AB$
Hence, P is mid point of AQ (Converse of mid point theorem)
Thus,

$AP = PQ$ (1)
Similarly, in

$\triangle CDP$

$CQ = QP$ (2)
From (1) and (2)

$AP = PQ = CQ$
Hence, P and Q trisect diagonal

$AC$

the area of

$\triangle ABE$ ?

0%
$60\, cm^{2}$
0%
$120\, cm^{2}$
0%
$40\, cm^{2}$
0%
$30\, cm^{2}$

Explanation

Given,

$Area \ of\ \Delta ADE = 60 cm^2$
From the figure,

$AB \parallel CF$ so,

$AB \parallel DE$

$AD\parallel BE$
So,

$ABED$ is a parallelogram.

The diagonal of a parallelogram divides into two triangles of equal areas.
Therefore,
Area of

$\Delta ABE =$ Area of

$\Delta ADE$
=> Area of

$\Delta ABE = 60 \ cm^2$

In a quadrilateral ABCD, if AB || CD,

$\angle$ D

$= 2\angle$ B, AD = b and CD = a, then the side AB is of length:

0%
$\displaystyle \frac{a}{2}\, +\, 2b$
0%
$a\, +\, 2b$
0%
$2a\, -\, b$
0%
$a\, +\, b$

Explanation

Given,

$AB\parallel CD$ ,

$\angle D=2\angle B$ ,

$AD=b$ ,

$CD=a$

Let

$\angle B=x$ and draw a line segment,

$CM$ equal and parallel to

$AD$ , as shown in the above figure.

Hence,

$AMCD$ is a parallelogram. [Quadrilateral having one pair of opposite sides equal and parallel]

Now,

$AD=b$

$\therefore CM=AD=b$
and

$CD=a$

$\therefore AM=CD=a$

$\therefore \angle D=\angle AMC$ [Opposite angles of a parallelogram]
Hence,

$\angle AMC =2x\quad \quad [\angle D=2\angle B=2x]$

Now, in

$\triangle MCB$

$2x=\angle MCB +\angle x\quad (\because\$ Exterior angle is equal to sum of opposite interior angle

$)$

$\therefore \ \angle MCB=x$
Hence,

$MB=MC \quad [$ As sides opposite to equal angles are equal

$]$

$\therefore \ MB=b$

Now,

$AB=AM+MB$

$\therefore \ AB=a+b$

Hence, option D is correct.

If an angle of a parallelogram is 24

$^o$ less than twice the smallest angle, then the value of largest angle of the parallelogram is

0%
176 $^o$
0%
160 $^o$
0%
112 $^o$
0%
108 $^o$

Explanation

Let the smallest angle be

$x$

then,the largest angle will be

$=180-x$

but, the same equals to

$[2x-24]$

so we have

$[2x-24]= 180-x$

$3x= 204$

$x=68$

thus the largest angle

$180-68= 112$ degree

In the given figure, ABCD is a parallelogram in which

$\angle DAB$ = 75

$^o$ and

$\angle$ DBC = 60

$^o$ then the measure of

$\angle$ BDC is equal to ?

0%
75 $^o$
0%
60 $^o$
0%
45 $^o$
0%
55 $^o$

Explanation

Opposite angles of parallelogram are equal.

$\angle DAB = \angle BCD = {75}^{o}$ and

$\angle DBC = {60}^{o}$

$\angle BDC + \angle DBC + \angle BCD = {180}^{o}$

$\angle BDC = {180}^{o} - \angle DBC - \angle BCD = {180}^{o} - {75}^{o} - {60}^{o} = {45}^{o}$

In the given figure, D is mid-point of side AB of

$\Delta\,ABC$ and BDEC is a parallelogram . Is the Area of

$\Delta\,ABC$ = Area of parallelogram

$BDEC$ ?

0%
True
0%
False

State true or false:

In any quadrilateral, if a pair of opposite sides is equal, then it is a parallelogram.

0%
True
0%
False

State true or false:

If three sides of quadrilateral are equal, it is a parallelogram

0%
True
0%
False

Explanation

The statement is false.

In a parallelogram, opposite sides are equal, opposite angles are equal and diagonals bisect each other. A quadrilateral can be constructed uniquely if the length of

$3$ sides and two diagonals are given.

Hence, the answer is the statement is false.

$ABCD$ is parallelogram. If

$L$ and

$M$ are the middle points of

$BC$ and

$CD$ , then $$ \overrightarrow {AL}+ \overrightarrow {AM}= $$

0%
$$ \dfrac { 1 }{ 2 } \overrightarrow{AC}$$
0%
$$ \dfrac { 3 }{ 2 } \overrightarrow{AC}$$
0%
$$\overrightarrow{AC}$$
0%
None of the above

Explanation

$\displaystyle AL=AB+BL=AB+\frac { 1 }{ 2 } BC=AB+\frac { 1 }{ 2 } AD$

$\displaystyle AM=AD+DM=AD+\frac { 1 }{ 2 } DC=AD+\frac { 1 }{ 2 } AB$

Adding,

$\displaystyle AL+AM=\frac { 3 }{ 2 } \left( AB+AD \right) =\frac { 3 }{ 2 } \left( AB+BC \right) =\frac { 3 }{ 2 } AC$

If in quadrilateral

$ABCD$ ,

$AB \parallel CD$ , then

$ABCD$ is necessarily a

0%
Parallelogram
0%
Rectangle
0%
Rhombus
0%
None of these

Explanation

If in quadrilateral

$ABCD, AB \parallel CD,$
condition

$1) BC \parallel DA$ then it becomes Parallelogram.

$2) BC$ is not parallel to

$DA$ then it is trapezium.
Hence, None of These is the correct answer.

PQRS is a parallelogram and M, N are the mid-points of PQ and RS respectively. Which of the following is not true ?

0%
RM trisects QS
0%
PN trisects QS
0%
$\Delta$ PSN $\sim$ $\Delta$ QMR
0%
MS is not parallel to QN

Explanation

In parallelogram

$PQRS,$

We know that

$PQ= SR$

Given, M is the mid-point of PQ and N is the mid-point of RS

$\therefore PM=MQ=SN=NR$ ...(1)

and we know that sides

$PQ$ and

$RS$ are parallel to each other.

$\therefore QM$ and

$NS$ will also be parallel to each other.

$\therefore$ We can say that

$MSNQ$ is a parallelogram

Hence its opposite sides will also be parallel

$\Rightarrow MS \parallel QN$

option D is incorrect so the correct answer according to the question is Option D.

Look at the pairs of shapes Which is a pair of rectangles?

Complete the following statement .

Number of measurements required to construct a rectangle are_______.

0%
4
0%
3
0%
2
0%
1

Explanation

If we know the length and breadth of the rectangle, we can construct it since a rectangle has opposite sides equal in length and each of the four angles is equal to

${90}^{0}$
Hence, number of measurements required to construct a rectangle is

$2$ .

A quadrilateral in which only one pair of opposite sides are parallel is called a ..........

0%
square
0%
rectangle
0%
rhombus
0%
trapezium

Explanation

$\Rightarrow$ We know that in square, rectangle and rhombus two pairs of opposite sides are parallel, so answer can not be square, rectangle or rhombus.

$\Rightarrow$ We also know the properties of trapezium in which only one pair of opposite sides are equal.

$\Rightarrow$ So, correct answer is option D trapezium.

$ABCD$ is a parallelogram with diagonals intersecting at

$O$ , then the number of distinct pairs of congruent triangles formed is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

In a parallelogram ABCD,

$\angle$ D = 135

$^o$ , determine the measure of

$\angle$ A .

0%
$\angle$ A = 45 $^o$
0%
$\angle$ A = 145 $^o$
0%
$\angle$ A = 35 $^o$
0%
$\angle$ A = 75 $^o$

Explanation

$ABCD$ is a parallelogram,

$\angle D=135$
We know, in a parallelogram, opposite angles are equal.
Thus,

$\angle B =\angle D$

$=>\angle B=135$
Also, in a parallelogram opposites sides are parallel. Thus,

$AD\parallel BC$
Now,

$AD\parallel BC$ and

$AB$ is the transversal, then

$\angle A +\angle B=180$

$=>\angle A=180-135$

$=>\angle A=45$

State true or false:

If three angles of a quadrilateral are equal, then it is a parallelogram

0%
True
0%
False

Explanation

We know that, in a parallelogram two opposite angles are equal while consecutive angles are supplementary (their sum is

$180^o$ )
Therefore, a quadrilateral with three angles equal cannot be a parallelogram.

Hence, the given statement is false.

Under what conditions must

$PQRS$ be a parallelogram?

0%
$x={45}^{o}$
0%
$y={16}^{o}$
0%
$x={45}^{o},y={16}^{o}$
0%
$x={16}^{o}, y={45}^{o}$

Explanation

Opposite angles of a parallelogram are equal

$\Rightarrow$

$\angle S=\angle Q$ and

$\angle P=\angle R$

$\Rightarrow$

$3x+{42}^{o}=4x={3}^{o}$ and

$6y=5y+{16}^{o}$

$\Rightarrow$

$x={45}^{o}$ and

$y={16}^{o}$

State true or false:

The figure formed by joining the mid-points of the consecutive sides of a quadrilateral is Parallelogram.

0%
True
0%
False

$ABCD$ is a parallelogram. IF

$P$ be a point on

$CD$ such that

$AP=AD$ , then the measure of

$\angle PAB+\angle BCD$ is:

0%
${180}^{o}$
0%
${225}^{o}$
0%
${240}^{o}$
0%
${135}^{o}$

Explanation

$\triangle ADP$ ,

$AD=AP$

$\Rightarrow$

$\angle APD=\angle ADP$
Also,

$\angle ADP+\angle APD+\angle PAD={180}^{o}$

$\Rightarrow$

$2\angle ADP={180}^{o}-\angle PAD$ (

$\therefore$

$\angle APD=\angle ADP)......(i)$
In quad

$ABCD$

$\angle A+\angle B+\angle C+\angle D={360}^{o}$

$\angle PAD+\angle PAB+2\angle D+\angle C={360}^{o}$ (

$\because$

$\angle B=\angle D$ )

$\angle PAD+\angle PAB+{180}^{o}-\angle PAD+\angle C={360}^{o}$ (From

$(i)$ )

$\Rightarrow$

$\angle PAB+\angle BCD={360}^{o}-{180}^{o}={180}^{o}$

Given that

$ABCD$ is a parallelogram whose diagonals intersect at point

$O$ .

$\angle ABC={110}^{o}$ ,

$\angle ACB={35}^{o}$ and

$\angle ADB={55}^{o}$ . The term that best describes

$ABCD$ is:

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

Explanation

$AD\parallel BC$ ,

$DB$ is the transversal, therefore,

$\angle CBD=\angle ADB={55}^{o}$ (alt

$\angle s$ )
In

$\triangle BOC$ ,

$\angle BOC={180}^{o}-(\angle OCB+\angle OBC)$

$={180}^{o}-({35}^{o}+{55}^{o})={!80}^{o}-{90}^{o}={90}^{o}$

$\Rightarrow$

$BD\bot AC$
Also, in

$\triangle ABC$ ,

$\angle BAC={180}^{o}-({110}^{o}-{35}^{o})$

$={180}^{o}-{145}^{o}={35}^{o}$
Hence,

$\angle ACB=\angle BAC\Rightarrow AB=BC$

$\therefore$

$ABCD$ is a parallelogram, whose adjacent sides are equal and diagonals are perpendicular to each other

$\Rightarrow$

$ABCD$ is a rhombus.

Two adjacent angles of a parallelogram are in the ratio 1 :The angles are

0%
$20^{\circ}$ and $40^{\circ}$
0%
$30^{\circ}$ and $60^{\circ}$
0%
$60^{\circ}$ and $120^{\circ}$
0%
None of these

Explanation

The sum of 2 adjacent angles of a parallelogram is 180 degrees.

Let the angles are

$x$ and

$2x$ .

$\therefore$

$x$

$+$

$2x$

$=$

$180^{o}$

$\therefore$

$3x = 180$

$\therefore$

$x = 60^{o}$
Hence angles are

$x = 60^{o}$ and

$2x = 120^{o}$ .

In a parallelogram, the sum of adjacent angles is:

0%
$90^{\circ}$
0%
$180^{\circ}$
0%
$270^{\circ}$
0%
$360^{\circ}$

Explanation

$ABCD$ is a parallelogram.

$\therefore$

$AB$

$\parallel$

$DC$ and

$AD$

$\parallel BC$

$\therefore$

$\angle$

$d$

$+$

$\angle$

$e$ (corresponding angle)

$\angle$

$a$

$+$

$\angle e$ =

$180$

$^\circ$ (supplementary angle or linear pair)

$\angle$

$a$

$+$

$\angle$

$d$ = 180

$^\circ$ (

$\because$

$\angle$

$d$ =

$\angle$

$e$ )

$\therefore$ sum of adjacent angle are

$180$

$^\circ$

Ans : B)

$180$

$^\circ$

In the given figure,

$PQRS$ is a parallelogram and

$\angle SPQ=60^{\circ}$ . If the bisectors of

$\angle P$ and

$\angle Q$ meet at

$A$ on

$RS$ , then which of the following is not correct?

0%
$AP=SP$
0%
$AS=AR$
0%
$AR=SP$
0%
$AQ=PQ$

Explanation

We have

$\angle SPQ=60^{\circ}$

$\angle P+\angle Q=180^{\circ}$ (adj.

$\angle s$ of a || gm are supp)

$\angle Q=180^{\circ}-60^{\circ}=120^{\circ}$
Now

$PQ||SR$ and

$AP$ is the transversal.

$\therefore SAP=\angle APQ=30^{\circ}$ (

$\because AP$ bisects

$\angle SPQ$ )

$\angle SPA=\angle SAP\Rightarrow SA=SP$ (isos.

$\Delta$ property) ...(i)
Also

$\angle RAQ=\angle AQP=60^{\circ}$
(

$PQ||SR,\:AQ$ is transversal, alt

$\angle s$ )
and

$\angle AQP=\dfrac{1}{2}\angle PQR=60^{\circ}$

$\Rightarrow \angle RQA=\angle RAQ$ (AQ bisect

$\angle PQR$ )

$\Rightarrow RA=RQ$ (isos.

$\Delta$ property) ....(ii)

$\therefore$ from eqn. (i) and (ii)

$AS=AR$ (

$\because SP=RQ$ , opp. side of a ||gm)
Also, in

$\Delta ARQ,\:\angle ARQ=60^{\circ}$

$\Rightarrow \Delta ARQ$ is equilateral

$\Rightarrow AR=RQ\Rightarrow AR=SP$
Therefore, the incorrect statement is

$AQ=PQ$ .

ABCD is a parallelogram. The angle bisectors of

$\angle$ A and

$\angle$ D meet at O. The measure of

$\angle$ AOD is:

0%
$45^{\circ}$
0%
$90^{\circ}$
0%
dependent on the angles A and D
0%
cannot be determined from given data
0%
none of these

Explanation

Let

$\angle A$ be

$2x$
Let

$\angle D$ be

$2y$

Since

$AB \parallel DC$ , sum of angles on same side of the transversal

$AD$ will be

${180}^{o}$

$=> 2x + 2y = {180}^{o}$

$x + y = {90}^{o}$

After bisection of angles A and D, we get

$\angle 1 = x$ and

$\angle 2 = y$ as per the given figure.

Now in

$\triangle AOD$ , we have

$x + y + \angle AOD = {180}^{o}$

${90}^{o} + \angle AOD = {180}^{o}$

$=> \angle AOD = {90}^{o}$

Which of the following properties need not be true for a parallelogram?

0%
It's diagonals are equal
0%
It's diagonals are perpendicular to each other
0%
The diagonals divide the figure into four congruent triangles.
0%
All the above

Explanation

A) The diagonals of a parallelogram need not be equal.
Option

$A$ need not be true.

B)In a parallelogram, diagonals need not be perpendicular to each other.
Option

$B$ need not be true.

C) In a parallelogram, diagonals need not divide it into four congruent triangles.
Option

$C$ need not be true.

So the correct answer is option

$D$ .

In a parallelogram ABCD, if

$\angle A\, =\, 65^{\circ}$ ,then

$\angle B,\, \angle C\, and\, \angle D$ are respectively

0%
$115^{\circ},\, 65^{\circ},\, 115^{\circ}$
0%
$112^{\circ},\, 65^{\circ},\, 112^{\circ}$
0%
$110^{\circ},\, 60^{\circ},\, 110^{\circ}$
0%
None of these

Explanation

$\because$ In a parallelogram, opposite angles are equal.

So,

$\angle A\, =\, \angle C\, =\, 65^{\circ}$

Also

$\angle B\, =\, \angle D$ we have

$\angle A\, =\, \angle B\, +\, \angle C\, +\, \angle D\, =\, 360^{\circ}$

$\Rightarrow\, 65^{\circ}\, +\, \angle B\, +\, 65^{\circ}\, +\, \angle B\, =\, 360^{\circ}$

$\Rightarrow\, 2\, \angle B\, =\, 360^{\circ}\, -\, 130^{\circ}$

$\Rightarrow\, 2\, \angle B\, =\, 230^{\circ}$

$\Rightarrow\,\angle B\, =\, \displaystyle \frac {230^{\circ}}{2}\, =\, 115^{\circ}$

$\therefore\, \angle B\, =\, \angle D\, =\, 115^{\circ}$

In a parallelogram ABCD, diagonals AC and BD intersect at O. If AO =

$6$ cm then find the length of AC.

0%
$5$ cm
0%
$10$ cm
0%
$12$ cm
0%
$14$ cm

Explanation

$\parallel gm$ ABCD, AC and BD are diagonals

We know that diagonals of a

$\parallel gm$ bisect each other

Therefore,

$AC=2AO$

$=2\times 6$

$=12cm$

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 6 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 6 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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