MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 9

Suppose

$ABCD$ is a parallelogram in which

$\angle A = 108^{\circ}$ . Calculate

$\angle B, \angle C$ and

$\angle D$ .

0%
$72^{\circ}, 108^{\circ}, 72^{\circ}$
0%
$22^{\circ}, 108^{\circ}, 72^{\circ}$
0%
$108^{\circ}, 72^{\circ}, 108^{\circ}$
0%
None of these

Explanation

$\Rightarrow$

$ABCD$ is a parallelogram.

$\Rightarrow$

$\angle A=108^o$ [ Given ]

$\Rightarrow$

$\angle A=\angle C$ [ Opposite angles of a parallelogram are equal ]

$\therefore$

$\angle C=108^o$

$\Rightarrow$

$\angle A+\angle D=180^o$ [ Adjacent angles of parallelogram are supplementary ]

$\Rightarrow$

$108^o+\angle D=180^o$

$\therefore$

$\angle D=72^o$

$\Rightarrow$

$\angle D=\angle B$ [ Opposite angles of a parallelogram are equal ]

$\therefore$

$\angle B=72^o$

1246091_620398_ans_ee3e7157b1024093a73c9c324143aa8c.png

Which of the following statements is definitely true, if PQRS is a parallelogram?

0%
$PQ$ and $SR$ are adjacent sides.
0%
$\angle P$ and $\angle Q$ are opposite angles
0%
$SR \parallel PQ$
0%
$PR=RQ$

Explanation

Given:

$PQRS$ is a parallelogram.

We can draw it in

$2$ possible ways.

So,

$PQ$ and

$SR$ may or may not be adjacent sides.

and

$SR$ is parallel to

$PQ$ in either case.

Also,

$\angle P$ and

$\angle Q$ are adjacent angles as they lie on the same edge.

Sides

$PR$ is not equal to

$RQ$ as one of them is side and other is the diagnol.

The angles A, B, C, D in the parallelogram ABCD are ___________.

0%
$90^o, 90^o, 90^o, 90^o$
0%
$90^o, 45^o, 30^o, 15^o$
0%
$30^o, 60^o, 20^o, 40^o$
0%
$90^o, 45^o, 15^o, 20^o$

Explanation

In parallelogram

$ABCD,\, AB ||DC$

$\therefore \angle ADB=\angle CBD$

and

$\angle ABD=\angle BDC$

$\Rightarrow \angle ABD=2a\\ \Rightarrow \angle ADB=3a$

$\therefore \angle B=2a+3a=5a$

and

$\angle D=3a+2a=5a$

$\triangle BCD:$

$3a+5a+2a=180^o\\ \Rightarrow a=18$

$\therefore \angle B=\angle C=\angle D=90^o$

For

$\angle A :$

$\angle A=360-\angle B-\angle D-\angle C\\=180-90-90-90\\=90^o$

The given polygon EFGH is a _____________.

0%
Parallelogram
0%
Concave polygon
0%
Convex polygon
0%
Trapezium

Explanation

A polygon having an angle that exceeds

$180^\circ$ is known as a concave polygon.

Since, the given diagram has

$\angle GHE > 180^\circ$

So, given polygon

$EFGH$ is a concave polygon.

Hence, option B is correct.

All squares are trapeziums.

0%
True
0%
False

Explanation

$\Rightarrow$ A trapezium is a quadrilateral with at least one pair of parallel sides.

$\Rightarrow$ In a square, there are always two pairs of parallel sides, so every square is trapezium.

$\therefore$ All squares are trapeziums.

1264251_703201_ans_48ca2b911d444ed3890fae6e10ad37a8.jpeg

ABCD is a parallelogram and L is a point on DB. The produced line AL meets BC at M and DC produced at N. Given that DL

$=3$ LB, find

$\displaystyle\frac{AB}{CN}$ .

0%
$3/2$
0%
$1/2$
0%
$4/5$
0%
$1/4$

Explanation

$\triangle LAB$ and

$\triangle DLN$ ,

$\angle ALB = \angle DLN$ are vertically opposite and

$\angle LAB = \angle DNL$ (alternate angles) and third angle is also equal.

So both triangles are similar.

Therefore

$\dfrac{DN}{AB} = \dfrac{DL}{LB}$

$DN=DC+CN$

So,

$1+\dfrac{CN}{AB}=3$

$\dfrac{AB}{CN}=\dfrac{1}{2}$

Hence option B is correct.

In the adjoining figure,

ABCDABCD is a parallelogram

0%
True
0%
False

Choose the correct answer from the alternatives given.
In a parallelogram

$PQRS,$ angle

$P$ is four times of angle

$Q.$ then the measure of

$\angle R$ is:

0%
$36^{\circ}$
0%
$72^{\circ}$
0%
$130^{\circ}$
0%
$144^{\circ}$

Explanation

Given:

$\angle P \, = \, 4 \angle Q$
Let

$\angle Q =x$

$\therefore \, \angle P \, = \, 4x$

$\angle P \, + \, \angle Q \, = \, 180^{\circ}$

$4x \, + \, x \, = \, 180^{\circ}$

$5x=180^o$

$\Rightarrow x \, = \, 36^{\circ}$

Thus,

$\angle P=4\times 36^o$

$\Rightarrow \angle P=144^o$

In parallelogram opposite sides are equal, thus

$\angle P=\angle R$

$\Rightarrow \angle R=144^o$

1134457_900135_ans_e39a39c982ca437586bed01156c7b229.png

$ABCD$ is a parallelogram whose diagonals intersect at

$O$ and

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

$6$ cm, then the length of diagonal

$AC$ is:

0%
$2\sqrt3$ cm
0%
$6\sqrt3$ cm
0%
$3\sqrt6$ cm
0%
$12$ cm

Explanation

$\triangle ADB$ ,

$DA = AB = DB = 6\ \text{cm}$ (Given)

Altitude of

$\triangle ADB = OA$

$= \dfrac{\sqrt3}{2} \times 6$

$= 3\sqrt 3$

$\therefore AC = 2.OA$

$=2(3\sqrt3)\ \text{cm}$

$= 6\sqrt3\ \text{cm}$

An exterior angle and an interior angle of a regular polygon are in the ratio

$2: 7$ . Find the number of sides in the polygon.

0%
$2$
0%
$7$
0%
$9$
0%
$18$

Explanation

Exterior angle

$=\dfrac {360^{\circ}}{n}$ (n is number of sides of a polygon)

Interior angle

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

$\Rightarrow \dfrac {2}{7}=\dfrac {360^{\circ}}{(n-2)180^{\circ}}$

$\Rightarrow n-2=7$

$\Rightarrow n=9$

Hence, the number of sides in the polygon is

$9$ .

Select the INCORRECT match.

0%
Quadrilateral with one pair of parallel sides is a Trapezium
0%
A rhombus with $4$ right angles is a Square
0%
Parallelogram with $4$ right angles ia a Rectangle
0%
All sides of a quadrilateral are equal then it is a Rectangle

Explanation

$\Rightarrow$ We know that, in Trapezium there is only one pair of parallel side.

$\Rightarrow$ Rhombus has its all four sides equal and if its all angle four becomes right angle then its called square.

$\Rightarrow$ In rectangle pair of opposite sides are equal, so rectangle is also a parallelogram with

$4$ right angle.

$\therefore$ Option

$A,B,C$ are correct.

$\Rightarrow$ Option

$D$ is incorrect.

In a parallelogram ABCD,

$\angle B= (2x+ 25)^0$ and

$\angle D= ( 3x-5)^0$ . Find:
(i) The value of

$x$
(ii) Measure of angle.

0%
x=30, b=91 degrees, d=89 degrees
0%
x=32, b=91 degrees, d=89 degrees
0%
x=30, b=89 degrees, d=91 degrees
0%
x=32, b=89 degrees, d=91 degrees

Explanation

$\parallel gm$ opposite angles are supplementary.

Therefore,

$\angle B+\angle D=180^{\circ}$

$2x+25+3x-5=180^{\circ}$

$x=\frac{180-20}{5}=32$

$\angle B=2x+25=(2*32)+25=89^{\circ}$

$\angle D=3x-5=(3*32)-5=91^{\circ}$

969096_1032251_ans_820cf5bee86c4ae1b57b285ea3466eed.png

$ABCD$ is a parallelogram and

$P$ is a point on the segment

$\overline {AD}$ dividing it internally in the ratio

$3 : 1$ the line

$\overline {BP}$ meets the diagonal

$\overline {AC}$ in

$Q$ . Then the ratio

$AQ:QC$ is

0%
$3:4$
0%
$4:3$
0%
$3:2$
0%
$2:3$

Explanation

According to question,

${l} p=\frac { { 3\overline { d } } }{ 4 } \\ lets, \\ AC=\lambda \, (\overline { b } +\overline { d } ) \\ PB=\frac { { 3\overline { d } } }{ 4 } +\gamma (\overline { b } -\frac { { 3\overline { d } } }{ 4 } ) \\ then,co-ordinate\, are, \\ AC=PB \\ \lambda \, (\overline { b } +\overline { d } )=\frac { { 3\overline { d } } }{ 4 } +\gamma (\overline { b } -\frac { { 3\overline { d } } }{ 4 } ) \\ \lambda \, (\overline { b } +\overline { d } )=\frac { { 3\overline { d } } }{ 4 } +\gamma \overline { b } -\frac { { 3\gamma \overline { d } } }{ 4 } \\ Now,compare\, both\, \, side, \\ \lambda =\gamma ,\, \, \, \, \, \, \, \lambda =\frac { 3 }{ 4 } -\frac { { 3\gamma } }{ 4 } \\ \, \, \, \, \, \, \, \, \lambda =\frac { 3 }{ 4 } -\frac { { 3\lambda } }{ 4 } \\ \, \, \, \, \frac { { 7\lambda } }{ 4 } =\frac { 3 }{ 4 } \\ \therefore \, \, \, \, \lambda =\frac { 3 }{ 7 } \\ it\, means,\, Q\, po{ { int } }\, is\, \frac { 3 }{ 7 } (\overline { b } +\overline { c } ) \\ So,the\, AQ:QC\, is\, 3:4 \\ and,\, the\, \, correct\, \, option\, is\, A.$

1177944_1046689_ans_3affe7eaa71947c6ae3b794bfa0183e2.png

Opposite sides of Rectangle and Parallelogram equal and Parallel?

0%
True
0%
False

$ABCD$ is a parallelogram

$X$ and

$Y$ are the mid points of

$BC$ and

$CD$ respectively. Then, ar(parallelogram

$ABCD$ ) is

0%
$4\times ar(\triangle AXY)$
0%
$2\times ar(\triangle AXY)$
0%
$\dfrac{8}{3}\times ar(\triangle AXY)$
0%
$None\ of\ these$

Explanation

$X$ and

$Y$ are the mid-points of sides

$BC$ and

$CD$ .

$\triangle BCD,$

$XY\parallel BD$ and

$XY=\dfrac{1}{2}BD$ {From the mid point theorem}

$\Rightarrow$

$ar(\triangle CYX)=\dfrac{1}{4}ar(\triangle DBC)$ {property of triangle having mid points}

$\Rightarrow$

$ar(\triangle CYX)=\dfrac{1}{8}ar(\parallel^{gm}\,ABCD)$ [ Area of parallelogram is twice the area of triangle made by diagona ] --- ( 1 )

Since, parallelogram

$ABCD$ and

$\triangle ABX$ are between the same parallel lines

$AD$ and

$BC$ and

$BX=\dfrac{1}{2}BC.$

$\Rightarrow$

$ar(\triangle ABX)=\dfrac{1}{4}ar(\parallel^{gm}\,ABCD)$ ----- ( 2 )

Similarly,

$ar(\triangle AYD)=\dfrac{1}{4}ar(\parallel^{gm} ABCD)$ ----- ( 3 )

Now,

$ar(\triangle AXY)=ar(\parallel^{gm}\,ABCD)-[ar(\triangle ABX)+ar(\triangle AYD)+ar(\triangle CYX)]$

$=ar(\parallel^{gn} ABCD)-[\dfrac{1}{4}ar(\parallel^{gm}ABCD+\dfrac{1}{4}ar(\parallel^{gm}ABCD)+\dfrac{1}{8}ar(\parallel^{gm}ABCD)]$ [ From ( 1 ) ( 2 ) and ( 3 ) ]

$=ar(\parallel^{gm}ABCD)-\dfrac{5}{8}(\parallel^{gm}ABCD)$

$\Rightarrow$

$ar(\triangle AXY)=\dfrac{3}{8}ar(\parallel^{gm}ABCD)$

$\therefore$

$ar(\parallel^{gm}ABCD)=\dfrac{8}{3}\times ar(\triangle AXY)$

In the following figure, ABCD is a parallelogram then
(i) AP bisects angle A
(ii) BP bisects angle B
(iii)

$\angle \,DAP\, + \,\angle \,CBP\, = \,\angle \,APB$

0%
True
0%
False

Explanation

In a parallelogram,sum of adjacent angles

$={180}^{\circ}$

$\angle{A}+\angle{B}={180}^{\circ}$ .......

$(1)$

$AP$ bisects

$\angle{A}$

$\angle{BAP}=\angle{A}-\angle{DAP}$

$BP$ bisects

$\angle{B}$

$\angle{ABP}=\angle{B}-\angle{CBP}$

$\triangle{ABP},$

$\angle{BAP}+\angle{ABP}+\angle{APB}={180}^{\circ}$

$\Rightarrow\,\angle{A}-\angle{DAP}+\angle{B}-\angle{CBP}+\angle{APB}={180}^{\circ}$

$\Rightarrow\,\left(\angle{A}+\angle{B}\right)+\angle{APB}={180}^{\circ}+\angle{DAP}+\angle{CBP}$

$\Rightarrow\,{180}^{\circ}+\angle{APB}={180}^{\circ}+\angle{DAP}+\angle{CBP}$ from

$(1)$

$\therefore\,\angle{APB}=\angle{DAP}+\angle{CBP}$

Let ABC is given triangle having respective sides a, b, c. D, E, F are points of the sides BC, CA, AB respectively so that AFDE is a parallelogram. The maximum area of the parallelogram is?

0%
$\dfrac{1}{4}bc\sin A$
0%
$\dfrac{1}{2}bc\sin A$
0%
$bc\sin A$
0%
$\dfrac{1}{8}bc\sin A$

If sides AB and CD of a parallelogram ABCD are bisected at E and F then EBFD is a parallelogram.

0%
True
0%
False

Explanation

Based on the given statement, we can draw the figure shown above.

Here,

$ABCD$ is a parallelogram

$\therefore \ EB = \dfrac{1}{2}AB = \dfrac{1}{2}DC = DF$

and

$EB\parallel DF$

$(\because AB\parallel DC )$

Thus, one pair of opposite sides of the quadrilateral

$EBFD$ is equal and parallel.

Hence,

$EBFD$ is a parallelogram.

Therefore, the given statement is true.

2100595_1145485_ans_6d51ebf44dd545a7ab8babe67112f451.png

The perimeter of a parallelogram is

$38 \text{ cm}$ . If the longer side is

$11 \text{ cm}$ , find the length of shorter side.

0%
$5\text{ cm}$
0%
$2\text{ cm}$
0%
$8\text{ cm}$
0%
$4\text{ cm}$

Explanation

Given:

perimeter of parallelogram

$=38 \text{ cm}$

length of longer side

$=11 \text{ cm}$

To find: shorter side

since, perimeter of parallelogram

$=2(l + h)$

$\implies 38 = 2(l + b)$

$\implies 38 = 2(11 + b)$

$\implies 38 = 22 + 2b$

$\implies b=\dfrac{{16}}{2}$

$\implies b = 8\text{ cm}$

$\therefore$ The length of the shorter side is

$8\text{ cm}$

In the adjoining figure,

$ABCD$ is a parallelogram in which

$AB = 16\ cm, BC = 10\ cm$ and

$L$ is a point on

$AC$ such that

$CL : LA = 2:3$ . If

$BL$ produced meets

$CD$ at

$M$ and

$AD$ produced at

$N$ , :then
(i)

$\triangle CLB \sim \triangle ALN$
(ii)

$\triangle CLM \sim \triangle ALB$

both statement is ?

0%
True
0%
False

ABCD is a parallelogram, G is the point on AB such that

$AG=2GB$ . E is a point on DC such that

$CE=2DE$ and F is a point of BC such that

$BF=2FC$ .Then area of

$\triangle EFG$ =

$\dfrac{5}{18}$ of area of ABCD.

0%
True
0%
False

Explanation

REF. Image.

Let

$FB = 2CE$ , here

$CF = x \& FB = 2x$

$\because \triangle CEI \cong \triangle CBH$

$\Rightarrow CI = K \& IH = 2K$

Now in

$\triangle EGF ar(\triangle EFG) = ar(\triangle ESF)+ar(\triangle SGF)$

$= \dfrac{1}{2}SF \times K+\dfrac{1}{2}SF \times 2K$

$ar (\triangle EFG) = \dfrac{3}{2}SF \times K ...(1)$

Now

$ar (\triangle EGBC) = ar(SGBF)+ar(ESFC)$

$= \dfrac{1}{2}(SF+GB)\times 2K+\dfrac{1}{2}(SF+EC)\times K$

$\dfrac{1}{2}(ar(\triangle ABCD)) = \dfrac{3}{2}K\times SE+\dfrac{2}{3}AB\times K$

$ar(\triangle ABCD) = 3K\times SF + \dfrac{4}{3}AB \times K$

$ar(\triangle ABCD) = 3K \times SF +\dfrac{4}{9} (ar (ABCD))$

$\Rightarrow K \times SF = \dfrac{5}{27} (ar(ABCD)) ...(2)$

From (1) & (2), we get

$ar (\triangle EFG) = \dfrac{3}{2} \times \dfrac{5}{27}(ar(ABCD))$

$ar(\triangle EFG) = \dfrac{5}{18} (ar(ABCD))$

Hence Proved

1171670_1144549_ans_d96d06acfbd840d2b2927626cd189580.png

State the whether given statement is true or false

$ABCD$ is a parallelogram and the bisector of

$\angle{A}$ bisects

$BC$ at

$x$ . Prove that

$AD=2AB$ .

0%
True
0%
False

Explanation

Let the bisector of $∠A$ bisects the side $BC$ at $X$ .

Given, $ABCD$ is a parallelogram.

∴ $AD||BC$ (Opposite sides of the parallelogram are parallel)

Now, $AD||BC$ and AX is the transversal ,

∴ $∠2 = ∠3$ (Alternate angles) ............(1)

and $∠1 = ∠2$ ( $AX$ is the bisector of $∠A$ ) ................(2)

From (1) and (2), we obtain

$∠1 = ∠3$

Now, in $ΔABX$ ,

$∠1 = ∠3$

⇒ $AB = BX$ ( If two angles of a triangle are equal, then sides opposite to

them are equal)

⇒ $2AB = 2BX = BX + BX = BX + XC$ ( $X$ is the mid point of $BC$ )

⇒ $2AB = BC$

⇒ $2AB = AD$ (Opposite sides of a parallelogram are equal)

∴ $AD = 2AB.$

1075129_1158411_ans_d9062f51622542018e11ba5f409bc83d.PNG

$P$ and

$Q$ are points on oppostie sides

$AD$ and

$BC$ of a parallelogram

$ABCD$ such that

$PQ$ passes through the point of intersection

$O$ of its diagonals

$AC$ and

$BD$ .

$PQ$ is bisected at O.

0%
True
0%
False

Explanation

$ABCD$ is a parallelogram whose diagonals bisect each other at

$O.$

$\triangle ODP$ and

$\triangle OBQ,$

$\Rightarrow$

$\angle BOQ=\angle POD$ [ Vertically opposite angles are equal ]

$\Rightarrow$

$\angle OBQ=\angle ODP$ [ Alternate interior angles ]

$\Rightarrow$

$OB=OD$ [ Given ]

$\therefore$

$\triangle ODP\cong\triangle OBQ$ [ By

$ASA$ congruence rule ]

$\therefore$

$OP=OQ$ [ By C.P.C.T ]

Hence,

$PQ$ is bisected at

$O.$

1290048_1207747_ans_1f64e5d940b54ee89425ccdd77bb27b5.png

How many unique measurements are needed to construct a parallelogram.

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

The point of intersection of the diagonals of a quadrilateral divides one diagonal in the ratio

$1: 2 .$ Can it be a parallelogram?

0%
Yes
0%
No
0%
Sometime
0%
Data not sufficient

Explanation

Let

$ABCD$ be the quadrilateral in which

$AC$ and

$BD$ are the diagonals that intersect at

$O$ .

Now,

$\dfrac{AO}{OC} =\dfrac{1}{2}$

We know that diagonals of a parallelogram bisect eah other.

So, for

$ABCD$ to be a parallelogram,

$\dfrac{AO}{OC} = \dfrac{1}{1}$

But here

$\dfrac{AO}{OC} =\dfrac{1}{2}$

So, the quadrilateral can't be a parallelogram in which the point of intersection of the diagonals divide one diagonal in the ratio of

$1 : 2$ .

$ABCD$ is a parallelogram. If

$P$ be a point on

$CD$ such that

$AP = AD$ , then the measure of

$\angle P A B + \angle B C D$ is

0%
$180 ^ { \circ }$
0%
$225 ^ { \circ }$
0%
$240 ^ { \circ }$
0%
$135 ^ { \circ }$

Explanation

$\angle PAB=\angle APD$ [alternate interior angles]

And as in

$\triangle APD$

$AP=AD$

so,

$\angle D=\angle APD=\angle PAB$

So,

$\angle PAB+\angle BCD=\angle D+\angle BCD=\angle D+\angle C=180^\circ$ since

$\angle C$ and

$\angle D+$ alternate opposite angle for

$AD||BC$

$\therefore\ \angle PAB+\angle BCD=180^\circ$

1185523_1219897_ans_1d8722dbb8ab4aefa12886849f1a1ff7.png

In a parallelogram

$ABCD$ , the diagonals intersect at

$O$ , if

$OA=3y-4$ and

$OC=y+20$ then

$y=$

0%
$60$
0%
$24$
0%
$12$
0%
$36$

$PQRS$ is a parallelogram and

$M$ is a point on

$PQ$ such that

$PM=\dfrac{3}{4}PQ$ then find the

$ar(PQR)$ if

$ar(PMRS)=28\ cm^{2}$

0%
$12\ cm^{2}$
0%
$14\ cm^{2}$
0%
$16\ cm^{2}$
0%
$20\ cm^{2}$

$x, y$ are the mid-points of opposite sides

$AB$ and

$DC$ respectively of a parallelogram

$ABCD, AY$ and

$DX$ are intersecting at

$S, CX$ and

$BY$ are intersecting at

$R$ . Then

$SXRY$ is a _

0%
Rectangle
0%
Kite
0%
Parallelogram
0%
Square

In the adjoining figure,

$ABCD$ is a parallelogram in which

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

$\angle COD=95^{o}, \angle DAC=50^{o}$

So, the measure of

$\angle ABO$ is_____________

0%
$35^{o}$
0%
$70^{o}$
0%
$50^{o}$
0%
$60^{o}$

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 9 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 9 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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