MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 5

The sum of the interior angles of a polygon is four times the sum of its exterior angles. Find the number of sides in the polygon.

0%
$10$
0%
$12$
0%
$8$
0%
$7$

Explanation

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to

$360^o$ .
The sum of the interior angles of polygon =

$180 (n-2)$
=>

$180 (n-2) = 4 \times 360$
=>

$n -2 = 8$
=>

$n =10$
Number of sides in the polygon =

$10$

State true or false:

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

$\displaystyle 175^{\circ}$

0%
True
0%
False

Explanation

Each interior angle of regular polygon of

$n$ sides is given by

$\dfrac{180^o(n-2)}{n}$
According to question

$\dfrac{180^o(n-2)}{n} =175^0$

$\Rightarrow 180^o(n-2)=175n$

$\Rightarrow 180n-360^o=175n$

$\Rightarrow 5n=360^o$

$\Rightarrow n=72$

Clearly there is a polygon of sides

$72$ whose each interior angle is

$175^0$

$PQRS$ is a parallelogram whose diagonals intersect at M.
If

$\angle PMS = 54^{\circ}$ ,

$\angle QSR = 25^{\circ}$ and

$\angle SQR = 30^{\circ}$ : find

$\angle PSR$

0%
$45^{\circ}$
0%
$55^{\circ}$
0%
$48^{\circ}$
0%
$56^{\circ}$

Explanation

The opposite sides of a parallelogram are

parallel. So,

$PQ \parallel SR$ and

$QS$ is transversal.

$\therefore$

$\angle$

$SQR$

$=\angle$

$PSQ$

Since, they are alternate interior angles.

$\angle PSR=\angle PSQ+\angle QSR$

$=\angle SQR+\angle QSR$

$=30^\circ+25^\circ$

$=55^\circ$

In the alongside diagram,

$ABCD$ is a parallelogram, then

$AB= 2\, BC$

0%
True
0%
False

Explanation

$Given\quad ABCD\quad is\quad a\quad parallelogram,\quad AD\parallel BC\quad \& \quad AB\parallel DC,\quad \angle CBE=\angle EBA=\angle 1,\quad \angle DAE=\angle EAB=\angle 2\\ Construction-\quad Draw\quad EF\quad from\quad E\quad such\quad that\quad EF=BC\quad \& \quad EF\parallel BC.\\ So\quad BCEF\quad is\quad a\quad parallelogram.\\ Since\quad EF=BC\quad \& \quad EF\parallel BC\quad so\quad EF=AD\quad \& \quad EF\parallel AD.\\ Now\quad \angle CBE=\angle BEF=\angle 1\quad \left\{ Alternate\quad angles\quad when\quad EF\parallel BC,\quad BE\quad is\quad the\quad transversal \right\} \\ So\quad \angle BEF=\angle EBF=\angle 1\\ \Rightarrow \triangle EFB\quad is\quad a\quad isosceles\quad triangle\quad \left\{ Since\quad base\quad angles\quad are\quad equal \right\} \\ \therefore \quad in\quad \triangle EFB,\quad EF=FB\Rightarrow FB=BC---\left( 1 \right) \\ Now\quad \angle FEA=\angle EAD=\angle 2\quad \left\{ Alternate\quad angles\quad when\quad EF\parallel AD,\quad EA\quad is\quad the\quad transversal \right\} \\ So\quad \angle FEA=\angle EAF=\angle 2\\ \Rightarrow \triangle EAB\quad is\quad a\quad isosceles\quad triangle\quad \left\{ Since\quad base\quad angles\quad are\quad equal \right\} \\ \therefore \quad in\quad \triangle EAB,\quad EF=AF\Rightarrow AF=BC---\left( 2 \right) \\ Adding\quad \left( 1 \right) \quad \& \quad \left( 2 \right) ,\quad we\quad get\\ FB+AF=BC+BC\Rightarrow AB=2BC$

The perimeter of a parallelogram

$ABCD= 40$ cm,

$AB= 3x$ cm,

$BC= 2x$ cm and

$CD= 2\left ( y\, +\, 1 \right )$ cm. Find the values of

$x$ and

$y$ .

0%
$x= 5$ and $y= 4$
0%
$x= 4$ and $y= 5$
0%
$x= 5$ and $y= 5$
0%
$x= 4$ and $y= 4$

Explanation

In the parallelogram

$ABCD$ side

$AB$ and

$CD$ are opposite to each other. Opposite sides of parallelogram are equal.

Thus,

$AB = CD$ and

$BC = AD$

$\Rightarrow 3x = 2 (y+1)$

$\Rightarrow 2x = AD$

$\Rightarrow (y+1) = 1.5x$

Now, according to the given condition

$AB + BC + CD + AD = 40$

$\Rightarrow 3x + 2x + 2 (y+1) + 2x = 40$

$\Rightarrow 7x + 2 \times 1.5x$

$= 40$

$\Rightarrow 7x + 3x = 40$

$\Rightarrow x = 4$

Simplify further,

$(y+1) = 1.5x \\= 1.5 \times 4$

$= 6$

$\Rightarrow y = 5$

PQRS is a parallelogram whose diagonals intersect at M.
If

$\angle PMS = 54^{\circ}$ ,

$\angle QSR = 25^{\circ}$ and

$\angle SQR = 30^{\circ}$ : find

$\angle RPS$

0%
$86^o$
0%
$96^o$
0%
$92^o$
0%
$100^o$

Explanation

Given is a parallelogram PQRS whose diagonals intersect at M.

Also given

$\angle PMS={ 54 }^{ o }, \angle QSR={ 25 }^{ o }$ and

$\angle SQR={ 30 }^{ o }$

We know that in a parallelogram opposite sides are parallel.

So PQ

$\parallel$ SR & if SQ is considered as the transversal then

$\angle PSM=\angle RQM={ 30 }^{ o }$ [ Alternate Angles ]

Now in

$\triangle$ PMS,

$\angle MPS+\angle PSM+PMS={ 180 }^{ o }$

$\Rightarrow \angle MPS+{ 30 }^{ o }+{ 54 }^{ o }={ 180 }^{ o }$

$\Rightarrow \angle MPS+{ 84 }^{ o }={ 180 }^{ o }$

$\Rightarrow \angle MPS={ 180 }^{ o }-{ 84 }^{ o }$

$\Rightarrow \angle MPS={ 96 }^{ o }$

Now

$\angle RPS=\angle MPS={ 96 }^{ o }$

PQRS is a parallelogram whose diagonals intersect at M.
If

$\angle PMS = 54^{\circ}$ ,

$\angle QSR = 25^{\circ}$ and

$\angle SQR = 30^{\circ}$ : find

$\angle PRS$

0%
$39^o$
0%
$29^o$
0%
$18^o$
0%
$24^o$

Explanation

Given is a parallelogram PQRS whose diagonals intersect at M.

Also given

$\angle PMS={ 54 }^{ o }$ ,

$\angle QSR={ 25 }^{ o }$ and

$\angle SQR={ 30 }^{ o }$

We know that in a parallelogram opposite sides are parallel.

So PQ

$\parallel$ SR & if SQ is considered as the transversal then

$\angle PSM=\angle RQM={ 30 }^{ o }$ [ Alternate Angles ]

Also

$\angle PSR=\angle PSQ+\angle QSR={ 30 }^{ o }+{ 25 }^{ o }={ 55 }^{ o }$

Now in

$\triangle PMS,\\ \angle MPS+\angle PSM+PMS={ 180 }^{ o }\\ \Rightarrow \angle MPS+{ 30 }^{ o }+{ 54 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle MPS+{ 84 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle MPS={ 180 }^{ o }-{ 84 }^{ o }\\ \Rightarrow \angle MPS={ 96 }^{ o }$

Now

$\angle RPS=\angle MPS={ 96 }^{ o }$

Now in

$\triangle PSR,\\ \angle PSR+\angle PRS+RPS={ 180 }^{ o }\\ \Rightarrow \angle PRS+{ 96 }^{ o }+{ 55 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle PRS+{ 151 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle PRS={ 180 }^{ o }-{ 151 }^{ o }\\ \Rightarrow \angle PRS={ 29 }^{ o }\\$

Two alternate sides of a regular polygon, when produced, meet at a right angle. Find the number of sides of the polygon.

0%
$3$
0%
$8$
0%
$2$
0%
$9$

Explanation

$\textbf{Step -1: Find each exterior angle of the given polygon.}$

$\text{Let the number of sides of the polygon be }n.$

$\text{Let }AB\text{ and }DC\text{ are the alternate sides of the polygon produced and meet at a right angle at }P.$

$\text{So, }\angle PBC\text{ and }\angle PCB\text{ are exterior angles of the polygon.}$

$\text{As the given polygon is regular so, all of its interior angles are equal, i.e. }\angle ABC=\angle DCB$

$\therefore\angle PBC=180^\circ-\angle ABC$ $\textbf{(Linear pair)}$

$\text{and }\angle PCB=180^\circ-\angle DCB$ $\textbf{(Linear pair)}$

$\Rightarrow \angle PCB=180^\circ-\angle ABC$

$\text{So, }\angle PBC=\angle PCB$

$\text{In }\triangle PCB,$

$\angle PBC+\angle PCB+90^\circ=180^\circ$

$\Rightarrow 2\angle PBC=90^\circ$

$\Rightarrow \angle PBC=45^\circ$

$\text{Thus, each exterior angle of the polygon}=45^\circ.$

$\textbf{Step -2: Find the number of sides of the polygon.}$

$\mathbf{\because\text{Each exterior angle of a regular polygon}=\dfrac{360^\circ}{\text{Number of sides of the polygon}}.}$

$\therefore 45^\circ=\dfrac{360^\circ}{n}$

$\Rightarrow n=8$

$\textbf{Final Answer: Number of Sides are 8. The correct option is B.}$

$\triangle$ ABC, BDEF and FDCE are parallelogram. Which of the following is correct?

0%
AB = EF
0%
DB = AB
0%
ED $\neq$ BF
0%
BD = CD

Explanation

BDEF & FDCD are parallelogram
BF || DE, BF || DE, & FD || CE & CD || FE

$\angle BDF = \angle EFD,$ &

$\angle DFE = \angle DCE$

$\angle FBD = \angle FED$ &

$\angle FED = \angle FED = \dfrac{1}{2} \angle FDC$ (parallelogram purpose)

$\triangle FDB$ &

$\triangle EDC$ are circular and conguent triangles.

$\because BD = BC$

State true or false:

In the alongside diagram,

$ABCD$ is a parallelogram in which

$AP$ bisects angle

$A$ and

$BQ$ bisects angle

$B$ , then

$AQ= BP$

0%
True
0%
False

Explanation

$Given\quad ABCD\quad is\quad a\quad parallelogram;\quad \angle PBQ=\angle QBA=\angle 1\quad \& \quad \angle QAP=\angle PAB=\angle 2\\ From\quad the\quad figure,\quad \angle PBQ=\angle AQB=\angle 1\quad \left\{ AQ\parallel BP\quad \& \quad BQ\quad is\quad the\quad transversal \right\} \\ So\quad \angle AQB=\angle QBA=\angle 1\\ \Rightarrow \triangle BAQ\quad is\quad an\quad isosceles\quad triangle\quad \left\{ since\quad base\quad angles\quad are\quad equal. \right\} \\ \therefore \quad in\quad \triangle BAQ,\quad Side\quad AQ=Side\quad AB\quad ---\left( 1 \right) \\ Again\quad from\quad the\quad figure,\quad \angle QAP=\angle APB=\angle 2\quad \left\{ AQ\parallel BP\quad \& \quad AP\quad is\quad the\quad transversal \right\} \\ So\quad \angle PAB=\angle APB=\angle 2\\ \Rightarrow \triangle ABP\quad is\quad an\quad isosceles\quad triangle\quad \left\{ since\quad base\quad angles\quad are\quad equal. \right\} \\ \therefore \quad in\quad \triangle ABP,\quad Side\quad BP=Side\quad AB\quad ---\left( 2 \right) \\ From\quad \left( 1 \right) \quad \& \quad \left( 2 \right) ,\quad we\quad get\quad \\ AQ=BP$

State true or false:

$E$ is the mid-point of side

$AB$ and

$F$ is the midpoint of side

$DC$ of parallelogram

$ABCD$ , then

$AEFD$ is a parallelogram.

0%
True
0%
False

Explanation

$AB = CD$ and

$E$ and

$F$ are the mid points of

$AB$ and

$CD$ respectively.

$\therefore AE = DF$

Also,

$AB || CD \therefore AE || DF$

Hence, AEFD is a paralleogram

In parallelogram

$ABCD$ ,

$AP$ and

$AQ$ are perpendiculars from vertex of obtuse angle

$A$ as shown in the figure. If

$\angle x\, :\, \angle y= 2\, :\, 1$ ; find smallest angles of the parallelogram.(in degrees)

0%
$55^o$
0%
$60^o$
0%
$70^o$
0%
$48^o$

Explanation

Given is a parallelogram ABCD. AP & AQ are perpendiculars drawn from vertex A.

$\angle AQD=\angle AQC=\angle APC=\angle PAB={ 90 }^{ o }$

Also given

$\angle C=x$ &

$\angle QAP=y$ and

$x:y=2:1$

Now in quadrilateral QCPA,

$\angle QAP+\angle APC+\angle PCQ+\angle CQA={ 360 }^{ o }$

$\Rightarrow \angle QAP+\angle PCQ+{ 90 }^{ o }+{ 90 }^{ o }={ 360 }^{ o }$

$\Rightarrow \angle QAP+\angle PCQ={ 180 }^{ o }$

$\Rightarrow y+x={ 180 }^{ o }$

Since

$x:y=2:1$ so

$x=\dfrac { 2 }{ 3 } \times 180^o={ 120 }^{ o }$

$\Rightarrow \angle C={ 120 }^{ o }$

$\angle A=\angle C={ 120 }^{ o }$

Now

$\angle C+\angle B={ 180 }^{ o }$ [ adjacent angles in parallelogram are supplementary ]

$\Rightarrow \angle B=180^o-120^o={ 60 }^{ o }$

$\angle B=\angle D={ 60 }^{ o }$

State true or false:

The bisectors of interior angles of a parallelogram doesn't form a rectangle.

0%
True
0%
False

Explanation

Let

$P, Q, R$ &

$S$ be the points of intersection of the bisectors of

$\angle A, \angle B, \angle C \& \angle D$ respectively of parallelogram ABCD.

Since DS bisects

$\angle D$ & AS bisects

$\angle A$ so

$\angle DAB=2\angle DAS \& \angle ADC=2\angle ADS$

Here

$\angle DAB+\angle ADC={ 180 }^{ \circ }$ [ adjacent angles of the parallelogram

$={ 180 }^{ \circ }$ ]

$\Rightarrow \angle DAB+\angle ADC={ 180 }^{ \circ }$

$\Rightarrow 2\angle DAS+2\angle ADS={ 180 }^{ \circ }$

$\Rightarrow \angle DAS+\angle ADS={ 90 }^{ \circ }$ ---

$\left( 1 \right)$

$\triangle ASD$ ,

$\angle DAS+\angle ADS+\angle ASD={ 180 }^{ \circ }$

$\Rightarrow \angle ASD+{ 90 }^{ \circ }={ 180 }^{ \circ }$ [ from

$\left( 1 \right)$ ]

$\Rightarrow \angle ASD={ 90 }^{ \circ }$

$\Rightarrow \angle PSR={ 90 }^{ \circ }$ [

$\angle ASD \& \angle PSR$ are vertically opposite angles ]

Similarly it can be shown that

$\angle APB=\angle SPQ={ 90 }^{ \circ }, \angle PQR={ 90 }^{ \circ } \& \angle SRQ={ 90 }^{ \circ }$ .

So in quadrilateral PQRS all interior angles are right angles.

Hence

$PQRS$ is a rectangle.

State true or false:

In the following figure,

$ABCD$ is a parallelogram, then

$AP$ bisects angle

$A$

0%
True
0%
False

Explanation

$Given\quad ABCD\quad is\quad a\quad parallelogram,\quad AD\parallel BC,\quad AB\parallel DC\quad \& \quad AD=DP=PC\\ So\quad AD=DP=PC=BC\\ Let\quad \angle DAP=\angle 2\quad \& \quad \angle CBP=\angle 1\\ In\quad isosceles\quad \triangle ADP,\quad \angle DAP=\angle DPA=\angle 2\\ In\quad isosceles\quad \triangle PCB,\quad \angle CPB=\angle CBP=\angle 1\\ Construction-\quad Draw\quad PQ\quad from\quad P\quad such\quad that\quad PQ=BC=AD\quad \& \quad PQ\parallel BC\parallel AD.\\ So\quad ADPQ\quad \& \quad BCPQ\quad are\quad both\quad parallelograms.\\ Now\quad DP\parallel AQ\quad \& \quad AP\quad is\quad the\quad transversal.\\ \angle DPA=\angle PAQ=\angle 2\quad \left\{ Alternate\quad angles\quad when\quad DP\parallel AQ\quad \& \quad AP\quad is\quad the\quad transversal \right\} \\ \Rightarrow \angle PAQ=\angle DAP\\ \Rightarrow AP\quad is\quad the\quad angle\quad bisector\quad of\quad \angle A$

In the given figure,

$E$ is mid-point of

$AB$ and

$DE$ meets diagonal

$AC$ at point

$F$ . If

$ABCD$ is a parallelogram and area of

$\bigtriangleup ADF$ is

$60 \: cm^{2}$ , then area of parallelogram

$ABCD$ is

0%
$360\ cm^{2}$
0%
$460\ cm^{2}$
0%
$450\ cm^{2}$
0%
None of These

In a parallelogram PQRS, find

$\angle RSQ$ if

$\angle SPQ = 70^o$ &

$\angle RQS = 60^o$ .

0%
$50^o$
0%
$60^o$
0%
$70^o$
0%
None

Explanation

Opposite angles of a parallelogram are equal

$\therefore \angle SPQ = \angle SRQ = 70^o$
In

$\triangle SQR, \angle RSQ + \angle SQR + \angle SRQ = 180^o$

$\angle RSQ = 180- 60 - 70$

$RSQ = 50^o$

State true or false:

In parallelogram

$ABCD$ , the bisector of angle

$A$ meets

$DC$ in

$P$ and

$AB= 2AD$ , then

$BP$ bisects angle

$B$ .

0%
True
0%
False

Explanation

Let

$AD=x$

$;$

$AB=2AD=2x$

Also

$AP$ is the bisector

$\angle A$

$\therefore \angle 1=\angle 2$

$\therefore \angle 2=\angle 5$ [alternate angles ]

$\Rightarrow \angle 1=\angle 5$

$AD=DP=2$

[ Sides opposite to equal angles are also equal ]

$\therefore AB=CD$ ( Opposite sides of parallelogram are equal )

$\therefore CD=2x$

$\Rightarrow DP+PC=2x$

$x+PC=2x$

$\therefore PC=x$

Also,

$BC=x$

$\triangle BPC,$

$\angle 6=\angle 4$ [ Angles opposite to equal sides are equal ]

$\Rightarrow \angle 6=\angle 3$ [alternate anlges ]

$\therefore \angle 6=\angle 4$ and

$\angle 6=\angle 3$

$\Rightarrow \angle 3=\angle 4$

Hence

$BP$ bisects

$\angle B.$

Hence, the answer is the statement is true.

State true or false:

In parallelogram

$ABCD$ , the bisector of angle

$A$ meets

$DC$ in

$P$ and

$AB= 2AD$ , then

$\angle APB= 90^{0}$

0%
True
0%
False

Explanation

We have to prove that

$\angle APB=90°$

Opposite angles are supplementary

$\Rightarrow \angle 1 +\angle 2+\angle 3+\angle 4=180°$

$[\angle 1=\angle 2, \angle 3=\angle 4 ]$

$\Rightarrow 2\angle 2 +2 \angle 3=180°$

$\triangle APB$

$\Rightarrow \angle 2+\angle 3+\angle APB=180°$

$\Rightarrow \angle APB=180°-90°=90°.$

Hence, the statement is true.

State true or false:

For the case of a parallelogram the bisectors of opposite angles are not parallel to each other.

0%
True
0%
False

Explanation

GIvem

$ABCD$ is a parallelogram.

$DE$ bisect

$\angle D,$

$BF$ bisect

$\angle B.$

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

$\Rightarrow \angle 1=\angle 2$ [ Equal angle bisect still equal ]

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

$\Rightarrow AD=BC$ [opposite sides of parallelogram are equal ]

$\therefore \triangle ADE\cong \triangle CBF$ [ By ASA congruence condition ]

$\Rightarrow DE =BF$ [ By CPCT proved ]

$\therefore$ The statement is false.

Hence, the answer is the statement is false.

The bisectors of opposite angles of a parallelogram are parallel.

0%
True
0%
False

Explanation

$\square$

$ABCD$ is all gm and

$DE$ and

$BF$ are angle bisectors.

$\therefore$

$\angle$

$B$ =

$\angle$

$D$

$\therefore$

$\dfrac{1}{2}$

$\angle$ B

$=\dfrac{1}{2}$

$\angle$ D

$\therefore$

$\angle$

$ADE$

$=\angle$

$FBC$

$\triangle$

$ADE$

$=\triangle$

$FBC$ ,

$AD=BC$ (

$\because$ opposite sides of llgm are congruent)

$\angle$

$ADE$ =

$\angle$

$FBC$

$\angle$

$DAE$ =

$\angle$

$BCF$ (

$\because$ opposite angles of llgm are congruent)

$\therefore$ By

$ASA$ congruence criterion,

$\triangle$

$ADE$

$\cong$

$\triangle$

$CBF$

$\therefore$

$AE=CF$ =>

$BE$ =

$DF$

Also,

$DE = BF$

$\therefore$

$\square$

$DEBF$ is a parallelogram. (

$\because$ opposite sides are congruent)

$\therefore$

$DE$

$\parallel$

$BF$

$\therefore$

$\triangle$ DEBF is a parallelogram.(

$\because$ opposite sides are congruent)

$\therefore$

$DE$

$\parallel$

$FB$

Hence,the bi rectors of opposite angles of a parallelogram are parallel.

Ans : A) True.

State true or false:

In the following figure,

$ABCD$ is a parallelogram, then

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

0%
True
0%
False

Explanation

ABCD is a parallelogram.
Draw a line PQ which is parallel to AD as well as BC.

$\angle DAP = \angle APQ$ (Alternate interior angles are equal)......(1)
Similarly,

$\angle CBP = \angle BPQ$ (Alternate interior angles are equal)......(2)
Add equation (1) and (2)

$\angle DAP + \angle CBP = \angle APQ + \angle BPQ$

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

State true or false:

For the case of a parallelogram the bisectors of any two adjacent angles intersect at

$90^{0}$ .

0%
True
0%
False

Explanation

Let

$ABCD$ be the parallelogram

$\Rightarrow \angle A +\angle B=180°$

$\therefore \dfrac{1}{2} \angle A+\dfrac{1}{2} \angle B=90°$

$\triangle AOB$

$\Rightarrow \angle BOA+\angle ABO+\angle AOB=180°$

$\Rightarrow \dfrac { 1 }{ 2 } \angle A+\dfrac { 1 }{ 2 } \angle B+\angle AOB=180°$

$\therefore \angle AOB=180°-90°=90°$

$\therefore$ Angle bisects of

$2$ adjacent angles intersect at

$90°.$

Hence, the statement is true.

In parallelogram

$ABCD$ .

$P$ is a point on side

$AB$ and

$Q$ is a point on side

$BC$ , then

$\bigtriangleup CPD\:$ and

$\bigtriangleup AQD\:$ are equal in area.

0%
True
0%
False

Explanation

Given,
ABCD is a parallelogram P is a point on side AB and Q is a point on side BC
Construction:Join diagonals AC and BD.
Proof:

$CPD = BCD$ [triangle with same base and between same set of parallel lines have equal area] (i)
Area

$BCD=\dfrac{1}{2}$ area of parallelogram ABCD.[daigonal of parallelogram bisects area in two equal parts] (ii)
From (i) and (ii)
Area

$CPD=\dfrac{1}{2}$ area ABCD (III)
Similarly,
Area

$AQD=Area ABD$
=

$\dfrac{1}{2}$ area ABCD (IV)
From (iii) and (iv)

$Area CPD =Area AQD$

In the given figure,

$M$ and

$N$ are the mid-points of the sides

$DC$ and

$AB$ respectively of the parallelogram

$ABCD$ . If the area of parallelogram

$ABCD$ is

$48 \: cm^{2}$ ; then the area of the parallelogram

$BNMC$ is equal to the area of the triangle

$BEC$ .

0%
True
0%
False

Explanation

Parallelogram

$ABCD$ and

$\triangle BEC$ have same base

$BC$ and they are between same parallels

$BC$ and

$AD$ .

$\therefore$ Area of

$\triangle BEC = \dfrac{1}{2}\ \text{area of parallelogram}\ ABCD$

$= \dfrac{1}{2} \times 48 cm^2 = 24 cm^2$

Since

$N$ and

$M$ are the midpoints of

$AB$ and

$DC$ respectively,

$\Rightarrow AN = NB , DM = MC$

$NB = \dfrac{1}{2} AB, MC = \dfrac{1}{2} DC$

But

$AB = DC$ ( opposite side of parallelogram

$ABCD$ )

$\therefore NB = MC$ , also

$NB \parallel MC$

$\therefore NBCM$ is a parallelogram

$\therefore A (NBCM) = \dfrac{1}{2} A(ABCD) = A ( \triangle BEC)$

$\therefore \triangle BEC$ is equal in area of parallelogram

$BNMC$

In parallelogram

$ABCD$ .

$P$ is a point on side

$AB$ and

$Q$ is a point on side

$BC$ , then

$Area \left ( \bigtriangleup \: AQD \right )= Area \left ( \bigtriangleup \: APD \right )+Area \left ( \bigtriangleup \: CPB \right )$

0%
True
0%
False

Explanation

To prove

$Area\left( \triangle AQD \right) =Area\left( \triangle APD \right) +Area\left( \triangle CPB \right)$

$Area\left( \triangle APD \right) =Area\left( \triangle APC \right)$ Area of triangle with same base AP and between same set of parallel lines are equal.

$Area\left( \triangle APD \right) =Area\left( \triangle CPB \right)$

$=Area\left( \triangle APC \right) =Area\left( \triangle CPB \right)$

$=Area\left( \triangle ABC \right)$

$=\frac { 1 }{ 2 } Area\quad of\quad parallelogram\quad ABCD$ ( AC is diagonal and diagonals of parallelogram divides it into equal areas)

$Area\left( \triangle AQD \right) =Area\left( \triangle APD \right) +Area\left( \triangle CPB \right)$ Proved

There is a regular polygon whose each interior angle is

$175^{\circ}$

State true or false.

0%
True
0%
False

Explanation

Given, a polygon whose each interior angles is

$175^o$
Sum of interior angles of a polygon is =

$180^o (n-2)$
Each interior angle of a polygon =

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

$\dfrac {180^o (n-2)}{n} = 175^o$

$180^o n - 175^o n = 360^o$

$n = \dfrac {360}{5}$

$n = 72$
Since, n (number of sides) is an integer, therefore there exist a polygon whose each interior angles is

$175^o$

State whether true or false:

In the given figure, AM bisects

$\angle A$ and DM bisects

$\angle D$ of parallelogram ABCD.

The measure of

$\angle AMD = 90^{\circ}$ .

0%
True
0%
False

Explanation

Given, AM bisects angle A and DM bisects angle D of parallelogram ABCD.
AB is parallel to DC, so,

$\angle A + \angle D = 180^o$ (Sum of interior angles on the same side of a transversal is supplementary.
AM bisects angle A and DM bisects angle D

$\angle A = 2 \angle DAM and \angle D = 2 \angle ADM$
Now, put it in above equality,

$2 \angle DAM + 2 \angle ADM = 180^o$

$\angle DAM + \angle ADM = 90^o$
In

$\triangle AMD$

$\angle DAM + \angle ADM + \angle AMD = 180^o$

$90^o + \angle AMD = 180^o$

$\angle AMD = 90^o$

In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If

$\angle x : \angle y = 2 : 1$ ; find angles of the parallelogram.

0%
$\angle DAB = \angle C = 160^{\circ}$ and $\angle B = \angle D = 60^{\circ}$
0%
$\angle DAB = \angle C = 120^{\circ}$ and $\angle B = \angle D = 60^{\circ}$
0%
$\angle DAB = \angle C = 230^{\circ}$ and $\angle B = \angle D = 60^{\circ}$
0%
$\angle DAB = \angle C = 100^{\circ}$ and $\angle B = \angle D = 60^{\circ}$

Explanation

Given, ABCD is a parallelogram and

$AQ\bot CD\quad and\quad AP\bot BC$

$\therefore \angle AQC=\angle APC=90$

AQCP is forming a quadrilateral.
As we know that sum of angles of quadrilateral is 360,

$\therefore \angle QAP+\angle AQC+\angle QCP+\angle APC=360$

$\\ or,\angle QAP+90+\angle QCP+90=360\quad \quad \quad \quad \quad \quad \quad \quad (\because \angle AQC=\angle APC=90)\\ or,\angle QAP+\angle QCP=360-180\\ or,\angle QAP+\angle QCP=180\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$
Given

$\angle QCP=x,\quad \angle QAP=y\\ x\quad :\quad y\quad =\quad 2\quad :\quad 1$
Let angle be m.

$\therefore x=2m\quad and\quad y=m$
from (1),

$\quad \angle QAP+\angle QCP=180\\ or,m+2m=180\\ or,\quad 3m=180\\ or,\quad m=60$

$\angle QCP=2m\\ or,\angle C=2\times 60=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)$

In a parallelogram ABCD, Sum of any two consecutive angles is supplementary,

$\quad \therefore \angle C+\angle D=180\\ or,\angle 120+\angle D=180\\ or,\angle D=180-120\\ or,\angle D=60\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (3)$

In a paralleogram ABCD, opposite angles are equal,

$\therefore \angle C=\angle A=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (from\quad 2)\\ \quad \angle D=\angle B=60\qquad \qquad \qquad \qquad \qquad \quad \quad (from\quad 3)$

Find the value of

$x , y$

0%
$x = 12^{\circ}$ and $y = 16^{\circ}$
0%
$x = 10^{\circ}$ and $y = 16^{\circ}$
0%
$x = 15^{\circ}$ and $y = 16^{\circ}$
0%
$x = 14^{\circ}$ and $y = 16^{\circ}$

Explanation

Given, ABCD is a parallelogram with

$\angle A=4x+20,\quad \angle B=7y,\quad \angle D=6x+3y-8$

Opposite angles of a parallelogram are equal

$\therefore \angle B=\angle D$ ,

$\quad 6x+3y-8=7y\\ or,6x+3y-8-7y=0\\ or,6x-4y-8=0\quad \\ or,6x-4y=8\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$

Any two consecutive angles of a parallelogram are supplementary.

$\therefore \angle A+\angle B=180\\ \quad \quad 4x+20+7y=180\\ or,4x+7y=180-20\\ or,4x+7y-160\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)$

From (1) and (2),

$6x-4y=8...(i)$ and

$4x+7y=160.....(ii)$

Multiply eq.(i) by

$4$ and multiply eq (ii) by

$6$

$24x-16y=32....(iii)$ and

$24x+42y=960.....(iv)$

Now Eq.

$(iv)$

$-$ Eq.

$(iii)$ we get,

$x=12^{\circ} and\quad y=16^{\circ}$

The perimeter of a parallelogram ABCD = 40 cm, AB = 3x cm, BC = 2x cm and CD = 2(y +1) cm. Find the values of x and y.

0%
x $=$ 4 and y $=$ 5
0%
x $=$ 2 and y $=$ 7
0%
x $=$ 3 and y $=$ 8
0%
x $=$ 1 and y $=$ 12

Explanation

Let the parallelogram be ABCD Perimeter=

$40$ cm

$= 3x$ cm BC

$= 2x$ cm(given)

Perimeter of parallelogram

$2(3x+2x)=40$

solving equation we get

$x=4$

Also,AB=CD(opposite sides of parallelogram)

$3x=2(y+1)$ (given)

Putting

$x=4$

$3(4)=2y+2$

$12=2y+2$

$10=2y$

Hence

$y=5$ and

$x=4$

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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