MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 11

A parallelogram and a triangle stand on the same side of the base with the same base and on the same side of the base with the same height. If

${l}_{1}$ ,

${l}_{2}$ be the perimeters of the parallelogram and the rectangle respectively, then the which one of the following is correct?

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${l}_{1}< {l}_{2}$
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${l}_{1}={l}_{2}$
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${l}_{1}> {l}_{2}$ but ${l}_{1}\ne 2{l}_{2}$
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${l}_{1}=2{l}_{2}$

A school was having 4 Hexagonal buildings joined to each other. They wanted to utilize the space between the 4 buildings to make a playground. The shape is that of a parallelogram. Can you find the measure of the angles as opposite angles are equal?

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$60^o, 120^o$
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$20^o, 60^o$
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$80^o, 120^o$
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$40^o, 40^o$

Explanation

$x$ and

$y$ are exterior angles of the regular polygon.

$\displaystyle \frac{360}{6} = 60$ i.e value of

$x$ angle
Opposite angles are

$60^o+60^o$

$360^o - 120^o (60^o + 60^o)$

$360^o = 60^o + 60^o + y + y$

$2y = 360^o - 120^o$

$y =$

$\dfrac{240^o }{ 2}$

$y = 120^o$
Therefore, option A is the correct answer.

Classify the following into open and closed curves

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Open (b,p,r) Closed ( a,c,q)
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Open (b,q ,c) Closed (a,p,r)
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Open (p,q,a ) Closed (r,c,b)
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Open (r,b,a) closed (p,q,c)

Which of the following statement(s) is/ are correct?

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A parallelogram in which two adjacent angles are equal is a rectangle
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A quadrilateral in which both pairs of opposite angles are equal is parallelogram
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In a parallelogram the number of acute angles is two
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All of these

A parallelogram with all equal sides are called _________.

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Rhombus
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Rectangle
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Polygon
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None of these

Select the CORRECT match. (A) (B)

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Two pairs of parallel sides - Trapezium
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A rhombus with 4 right angles - Rectangle
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Parallelogram with 4 right angles - Rectangle
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Diagonals are perpendicular to one another -

Trapezium

In the given figure, ABCD is a parallelogram, AL

$\perp$ BC, AM

$\perp$ CD, AL

$=4$ cm and AM

$=5$ cm. If BC

$=6.5$ cm, then find CD.

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$5.2$ cm
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$8.7$ cm
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$6.5$ cm
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$3.3$ cm

Which of the following closed plane figures is/are not polygons?

Fill in the blanks.
Any drawing ( straight or non-Straight) done without lifting the pencil may be a_____P_____. A____Q____is the one that does not cross itself . A curve is said to be ___R__ if its ends are joined. A____S___is a simple closed curve made up of lines segments.

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P-Curve, Q-Open curve, R-Closed, S-Line
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P-Line, Q-Curve, R-Open, S-Line
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P-Curve, Q-Simple curve, R-Closed, S-Polygon
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P-Curved, Q-Closed curve, R-Open, S-Circle

In the given figure, if

$TS||QV, \, SV || TQ, \, \angle QRS = {115^ \circ}$ , then

$\angle TQR$ is equal to

0%
${65^ \circ}$
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${115^ \circ}$
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${45^ \circ}$
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${25^ \circ}$

Explanation

Given, A parallelogram

$QTSV,\angle QRS=115^\circ$

From parallelogram

$\displaystyle \angle S +\angle V = 180^{\circ}$

Because

$TS \parallel QV$

So,

$\displaystyle 90^{\circ}+\angle V = 180$

$\displaystyle \angle V = 90^{\circ}$

similarity

$\displaystyle \angle TQV+\angle V = 180^{\circ}$

so,

$\displaystyle \angle TQV = 90^{\circ}$

$\displaystyle \angle TRQ = 180-\angle QRS = 180-115 = 65^{\circ}$

Now we know that alternate angle will be equal as

$RQ$ is transversal to

$TS \parallel QV$

so,

$\displaystyle \angle TRQ = \angle RQV = 65^{\circ}$

But

$\displaystyle \angle TQV = 90 \Rightarrow \angle RQV+\angle TQR = 90^{\circ}$

$\displaystyle \angle TQR = 90-65$

$\displaystyle \angle TQR = 25^{\circ}$ option

$-$ D

The length of the diagonal

$BD$ of the parallelogram

$ABCD$ is

$18\ cm$ . If

$p$ and

$Q$ are the Centroid of the

$\triangle ABC$ and

$\triangle ADC$ respectively then length of the line segment

$PQ$ is:

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$4\ cm$
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$6\ cm$
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$9\ cm$
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$12\ cm$

$p$ is a point in the interior of

$\parallel gm$

$ABCD$ . If the area of

$\parallel gm$

$ABCD$ is

$60\ {cm}^{2}$ .

$ar\left( \triangle ADP \right) +ar\left( \triangle BPC \right) =$

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$15\ { cm }^{ 2 }$
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$30\ { cm }^{ 2 }$
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$45\ { cm }^{ 2 }$
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$20\ { cm }^{ 2 }$

Explanation

In parallelogram ABCD, in quadrilateral AEFD,

$\Delta APD$ and quadrilateral AEFD lie on same base between same parallels AD & EF.

$\therefore ar(\Delta ADP)=\dfrac{1}{2}ar(\Delta EFD)$

$\Delta BPC$ and quadrilateral EFCB lie on same base between same parallel EF & BC.

$\therefore ar(\Delta BPC)=\dfrac{1}{2}ar(EFCB)$

Now,

$ar(\Delta ADP)+ar(\Delta BPC)$

$=\dfrac{1}{2}ar(AEFD)+\dfrac{1}{2}ar(EFCB)$

$=\dfrac{1}{2}ar(ABCD)$ .

$=\dfrac{1}{2}\times 60\ cm^2=30\ cm^2$ .

1367588_1237410_ans_91d7f02f6fa84c8282879c0852368a0f.png

$ABCD$ is a parallelogram

$P$ is a point of

$AD$ such that

$\dfrac {AP}{AD}=\dfrac {1}{2015}.Q$ is the point of intersection of

$AC$ and

$BP$ . Then

$\dfrac {AQ}{AC}=$

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$\dfrac {2000}{2016}$
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$\dfrac {2015}{2016}$
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$\dfrac {2018}{2016}$
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$\dfrac {1}{2016}$

The following figure shows a ABCD in which AB is quarlled to DC and AD=BC

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$\angle DAB=\angle CBA$
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$\angle ADC=\angle BCD$
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$ADC=BD$
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$OA=OB AND OC=OD$

$ABCD$ is a parallelogram.

${ AP }$ bisects

$\angle A$ and

${ CQ }$ bisects

$\angle C.P$ lies on

${ CD }$ and Q lies on

${ AB }.$ Then

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${ AB } \parallel { CQ }$
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${ AP } \parallel { CQ }$
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${ AQ } \parallel { BC }$
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None of these

A parallelogram, the lengths of whose sides are 11 cm and 13 cm, has one diagonal 20 cm, long the length of another diagonal is

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12 cm
0%
20 cm
0%
66 cm
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25 cm

The two adjacent sides of a parallelogram are 4 cm and 9 cm. The ratio of its altitude is

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21 : 22
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4 : 9
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16 : 81
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11 : 37

In the given figure, ABCD is a parallelogram,

$\angle ADE={ 50 }^{ \circ }$ and

$\angle ACE={ \angle BED=90 }^{ \circ }.$ The value of

$\angle EAC+{ \angle ABC }-2\angle DAC$ is

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${ 20 }^{ \circ }$
0%
${ 10 }^{ \circ }$
0%
${ 30 }^{ \circ }$
0%
${ 40 }^{ \circ }$

In the given figure, ABCD is a parallelogram such that AB||CD and AD||BC with opposite angles equal. If DA=DX, then find the values of

$\angle{m}$ and

$\angle{n}$ .

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$m=24^\circ.n=34^\circ$ .
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$m=24^\circ.n=24^\circ$ .
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$m=34^\circ.n=24^\circ$ .
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$m=34^\circ.n=34^\circ$ .

In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm then AD =

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3 cm
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6 cm
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8 cm
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10.5 cm

ABCD is a parallelogram, AB=14 cm, BC=18 cm and AC=16 cm, then the length of the diagonal is_

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24cm
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36cm
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28cm
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32cm

The altitude of a parallelogram is twice of the length of the base and its area is 900

$c{m^2}$ . The length of base and altitude respectively are

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$15\sqrt {2\,} \,cm,\,\,\,30\sqrt 2 cm$
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$25\sqrt 2 cm,\,\,\,\,\,35\sqrt 2 cm$
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$45\sqrt 2 cm,\,\,\,\,\,\,55\sqrt 2 cm$
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$65\sqrt 2 cm,\,\,\,\,\,75\sqrt 2 cm$

In the given figure ,

$ABCD$ and

$BDCE$ are parallelograms with common base

$DC$ . If

$BC$

$\perp$

$BD$ , then

$\angle BEC$ =

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$60^{\circ}$
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$30^{\circ}$
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$150^{\circ}$
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$120^{\circ}$

Explanation

$\angle BAD=30^{\circ}$

$\angle BCD=30^{\circ}$ [Opposite angles of parallelogram equal]

Now,, in

$\Delta CBD$ ,

$\angle DBC+\angle BCD+\angle CDB=180^{\circ}$

$\Rightarrow 90^{\circ}+30^{\circ}+\angle CDB=180^{\circ}$

$\Rightarrow CDB=60^{\circ}$

$\therefore \angle BEC=60^{\circ}$ [Opposite angles are equal]

(A)

$60^{\circ}$

2057285_1792038_ans_040daf53c8ec4d239a8daae2a5cd969f.png

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 11 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 11 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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