MCQ Questions for Class 9 Maths Quadrilaterals Quiz 4

State true or false:

In triangle

$ABC$ , angle

$B$ is obtuse.

$D$ and

$E$ are mid-points of sides

$AB$ and

$BC$ respectively and

$F$ is a point in side

$AC$ such that

$EF$ is parallel to

$AB$ . Then,

$BEFD$ is a parallelogram.

0%
True
0%
False

Explanation

Given:

$D$ is mid point of

$AB$ and

$E$ is mid point of

$BC$ ,

$F$ is any point on

$AC$ and

$EF \parallel AB$

Now, in

$\triangle ABC$ ,
E is mid point of BC and

$EF \parallel AB$
By Mid point Theorem,

$F$ is mid point of

$AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem,

$DF \parallel BE$
Since,

$DF \parallel BE$ and

$EF \parallel AB or BD$
Hence, BEFD is parallelogram.

State true or false:

In triangle

$ABC$ ,

$P$ is the mid-point of side

$BC$ . A line through

$P$ and Parallel to

$CA$ meets

$AB$ at point

$Q$ ; and a line through

$Q$ and parallel to

$BC$ meets median

$AP$ at point

$R$ . Can it be concluded that,

$AP= 2AR$ ?

0%
True
0%
False

Explanation

Given:

$\triangle ABC$ , P is mid point of BC,

$QR \parallel BC$ and

$PQ \parallel AC$

Since,

$PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In

$\triangle ABP$
Since,

$QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence,

$AP = 2 AR$

State true or false:

The bisectors of angles B and C of a parallelogram ABCD meet at point O. If triangle OBC formed is an isosceles right-angled triangle, then the quadrilateral ABCD is a rectangle.

0%
True
0%
False

Explanation

$OB$ bisects

$\angle B$ and

$OC$ bisects

$\angle C$
Given:

$\angle OBC = \angle OCB$ isosceles right-angled triangle

$2 \angle OBC = 2 \angle OCB$

$\angle B = \angle C$

Given,

$ABCD$ is a parallelogram
Thus,

$\angle B + \angle C = 180$

$2 \angle B = 180$

$\angle B = 90^{\circ}$
Hence,

$\angle B = \angle C = 90^{\circ}$
Thus,

$ABCD$ is a rectangle.

Angles of a quadrilateral are

$(4x)^{\circ},\ 5(x + 2)^{\circ},\ (7x - 20)^{\circ}$ and

$6(x + 3)^{\circ}$ . Find the value of

$x$ (in degrees).

0%
$14$
0%
$15$
0%
$16$
0%
$18$

Explanation

We know, the sum of interior angles of a quadrilateral

$={ 360 }^{ o }$ .

Given the angles are:

$4x, 5(x+2), (7x-20)$ and

$6(x+3).$

Then,

$4x+5\left( x+2 \right) +\left( 7x-20 \right) +6\left( x+3 \right) =360$

$\Rightarrow 4x+5x+10+7x-20+6x+18=360\\ \Rightarrow 22x+8=360\\ \Rightarrow 22x=360-8=352\\ \Rightarrow x=\dfrac { 352 }{ 22 } =\dfrac { 176 }{ 11 } =16$

Therefore,

$x=16$ .

Hence, option

$C$ is correct.

Two angles of a quadrilateral are

$68^{\circ}$ and

$76^{\circ}$ . If the other two angles are in the ratio

$5 : 7$ , find the measure of largest of them.

0%
$126^{\circ}$
0%
$90^{\circ}$
0%
$63^{\circ}$
0%
$100^{\circ}$

Explanation

We know, by angle sum property, the sum of interior angles of a quadrilateral=

${ 360 }^{ o }$ .
Given two angles are

${ 68 }^{ o }$ &

${ 76 }^{ o }$ and the other two angles are in the ratio of

$5:7$ .
Let the other two angles be

$5x^{ o }$ &

$7x^{ o }$ .
Then,

$5x^{ o }+7{ x }^{ o }+68^o+76^o=360^o$

$\Rightarrow 12x+144^o=360^o$

$\Rightarrow 12x=360^o-144^o=216^o$

$\Rightarrow x=\dfrac { 216^o }{ 12 } =18^o$ .

So unknown angles are

$5x=5\times 18^o={ 90 }^{ o }$

and

$7x=7\times 18^o={ 126 }^{ o }$ .

Therefore, the largest angle is

$126^o$ .

Hence, option

$A$ is correct.

$P,Q, R$ and

$S$ are the mid-points of sides

$AB, BC, CD$ and

$DA$ respectively of rhombus

$ABCD$ . Quadrilateral

$PQRS$ is a rectangle. Under what condition will

$PQRS$ be a square ?

0%
When $ABCD$ is a square
0%
When $ABCD$ is a parallelogram
0%
When $ABCD$ is a rectangle
0%
When $ABCD$ is a square or a rectangle

Explanation

Given: ABCD is a rhombus. P, Q, R, S are mid points of AB, BC, CD, DA respectively.
Join: AC and BD

In

$\triangle ABC$
P is mid point of AB and Q is mid point of AC.
Thus, by mid point theorem,

$PQ \parallel AC$ and

$PQ = \frac{1}{2} AC$
Similarly, In

$\triangle ACD$ ,
S is mid point of AD and R is mid point of CD
Thus, by mid point theorem,

$SR \parallel AC$ and

$SR = \frac{1}{2} AC$
Hence,

$PQ \parallel SR$ and

$PQ = SR$

Similarly,

$PS = QR$ and

$PS \parallel QR$
Thus, the opposite sides of PQRS are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since

$AC \perp BD$ thus,

$PS \perp PQ$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of PQRS are equal and parallel and the sides meet each other at right angles. Hence, PQRS is a rectangle.

If PQRS had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if ABCD is a square.

From the following figure, find $$x$$.

0%
$18^o$
0%
$26^o$
0%
$24^o$
0%
$36^o$

Explanation

We know, the sum of interior angles of a quadrilateral

$=360^o$ .

Given, the angles are:

$A=48^o+x,\quad B=4x,\quad C=3x\quad \& \quad D=4x.$

Then,

$48^o+x+4x+3x+4x=360^o\\ \Rightarrow 12x+48^o=360^o\\ \Rightarrow 12x=360^o-48^o\\ \Rightarrow 12x=312^o\\ \Rightarrow x=\dfrac { 312^o }{ 12 } =26^o.$

Therefore, option

$B$ is correct.

Find

$\angle ACD$ .

0%
$56^o$
0%
$28^o$
0%
$45^o$
0%
$36^o$

Explanation

We know, the sum of interior angles of a quadrilateral

$=360^o$ .

Given, the angles are:

$A=48^o+x,\quad B=4x,\quad C=3x\quad \& \quad D=4x.$

Then,

$48^o+x+4x+3x+4x=360^o\\ \Rightarrow 12x+48^o=360^o\\ \Rightarrow 12x=360^o-48^o\\ \Rightarrow 12x=312^o\\ \Rightarrow x=\dfrac { 312^o }{ 12 } =26^o.$

Then,

$\angle D=4x=4\times 26^o={ 104 }^{ o }$ .

Now in

$\triangle ACD$ ,

$\angle ACD+\angle DAC+\angle D=180^o$ ....[Angle sum property]

$\Rightarrow \angle ACD+48^o+104^o=180^o$

$\Rightarrow \angle ACD=180^o-152^o$

$\Rightarrow \angle ACD={ 28^o }$ .

Therefore, option

$B$ is correct.

Angles of a quadrilateral are

$(4x)^{\circ}, 5(x + 2)^{\circ}, (7x - 20)^{\circ}$ and

$6(x + 3)^{\circ}$ . Find the smallest angle of the quadrilateral.

0%
$90^{\circ}$
0%
$64^{\circ}$
0%
$114^{\circ}$
0%
$78^{\circ}$

Explanation

We know, the sum of interior angles of a quadrilateral

$={ 360 }^{ o }$ .

Given the angles are:

$4x, 5(x+2), (7x-20)$ and

$6(x+3).$

Then,

$4x+5\left( x+2 \right) +\left( 7x-20 \right) +6\left( x+3 \right) =360$

$\Rightarrow 4x+5x+10+7x-20+6x+18=360\\ \Rightarrow 22x+8=360\\ \Rightarrow 22x=360-8=352\\ \Rightarrow x=\dfrac { 352 }{ 22 } =\dfrac { 176 }{ 11 } =16$

Therefore,

$x=16$ .

Each angle is as follows:

$4x=4\times 16={ 64 }^{ o }$ ,

$5(x+2)=5\left( 16+2 \right) =5\times 18={ 90 }^{ o },$

$(7x-20)=7\times 16-20=112-20=92^{ o }$

and

$6(x+3)=6\left( 16+3 \right) =6\times 19=114^{ o }$ .

Therefore, the smallest angle is

$64^o$ .

Hence, option

$B$ is correct.

Two angles of a quadrilateral are

$89^{\circ}$ and

$113^{\circ}$ . If the other two angles are equal, find the equal angles.

0%
$69^{\circ}$
0%
$89^{\circ}$
0%
$158^{\circ}$
0%
$79^{\circ}$

Explanation

$\textbf{Step-1: Apply the concept of interior angles of a quadrilateral.}$

$\text{We know, by angle sum property, the sum of interior angles of a quadrilateral =}$

${ 360 }^{ o }$ .

$\text{Given two angles are}$

${ 89 }^{ o }$ &

${ 113 }^{ o }$

$\text{and the other two angles are equal.}$

$\text{Let the unknown angle be}$

${ x }^{ o }$ .

$\text{So,}$

${ x }+{ x }+89^o+113^o=360^o$

$\Rightarrow 2x+202^o=360^o\\ \Rightarrow 2x=360^o-202^o=158^o\\ \Rightarrow x=\dfrac { 158^o }{ 2 } ={ 79^o }.$

$\text{So the other angles are}$

${ 79 }^{ o }$

$\text{each.}$

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

Given, in quadrilateral

$ABCD$ ,

$\angle C = 64^{\circ}, \angle D = \angle C - 8^{\circ}, \angle A = 5(a + 2)^{\circ}$ and

$\angle B = 2(2a+7)^{\circ}$ . Calculate

$\angle A$ .

0%
$120^o$
0%
$130^o$
0%
$140^o$
0%
$150^o$

Explanation

Given

$\angle C={ 64 }^{ o }, \angle D= \angle C-{ 8 }^{ o }=64^o-8^o={ 56 }^{ o }, \angle A={ 5\left( a+2 \right) }^{ o }$ and

$\angle B={ 2\left( 2a+7 \right) }^{ o }.$

We know, by angle sum property, the sum of angles of a quadrilateral is

$360^o$ .

Then,

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

$\Rightarrow { 5\left( a+2 \right) }+{ 2\left( 2a+7 \right) }+64^o+56^o=360^o$

$\Rightarrow 5a+10^o+4a+14^o+120^o=360^o$

$\Rightarrow 9a=360^o-144^o$

$\Rightarrow 9a=216^o$

$\Rightarrow a=\dfrac { 216^o }{ 9 } =24^o$ .

Therefore,

$\angle A={ 5\left( a+2 \right) }^{ o }$

$={ 5\left( 24^o+2 \right) }^{ o }$

$={ 5\times 26^o }=130^o$ .

Find

$\angle ABC$ .

0%
$92^o$
0%
$104^o$
0%
$120^o$
0%
$106^o$

Explanation

We know, the sum of interior angles of a quadrilateral

$=360^o$ .

Given, the angles are:

$A=48^o+x,\quad B=4x,\quad C=3x\quad \& \quad D=4x.$

Then,

$48^o+x+4x+3x+4x=360^o\\ \Rightarrow 12x+48^o=360^o\\ \Rightarrow 12x=360^o-48^o\\ \Rightarrow 12x=312^o\\ \Rightarrow x=\dfrac { 312^o }{ 12 } =26^o.$

$\therefore \angle ABC=4x=4\times 26^o={ 104 }^{ o }.$

Therefore, option

$B$ is correct.

Sum of all angles of a square is

0%
$90^o$
0%
$240^o$
0%
$120^o$
0%
$360^o$

Explanation

$\Rightarrow$

$\square ABCD$ is a square.

$\Rightarrow$ All four sides and angles of square are equal.

$\Rightarrow$ All four angles are right angle in square.

$\Rightarrow$ Sum of all angles of a square $$ABCD=\angle A+\angle B+\angle C+\angle D$$

$=90^o+90^o+90^o+90^o$

$=360^o$

$\therefore$ Sum of angles of square is

$360^o$

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State true or false:

In quadrilateral PQRS,

$\angle P : \angle Q : \angle R : \angle S = 3 : 4 : 6 : 7$ . The smallest angle of the quadrilateral is

$45^o$ .

0%
True
0%
False

Explanation

Given ratio of angles of quadrilateral $PQRS$ is $3:4:6:7$

Let the angles of quadrilateral $PQRS$ be $3x,4x,6x,7x$ , respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $360^o$ .

$\Rightarrow \angle P+\angle Q+\angle R+\angle S=360^o$

$\Rightarrow 3x+4x+6x+7x=360^o$

$\Rightarrow 20x=360^o$

$\Rightarrow x=\dfrac{360^o}{20}$

$\therefore x=18^o$ .

$\therefore \angle P =3x=3 \times 18^o =54^o$ ,

$\angle Q =4x=4\times 18^o =72^o$ ,

$\angle R =6x=6\times 18^o =108^o$

and $\angle S =7x=7 \times 18^o =126^o$ .

$\therefore$ The smallest angle $=54^o$ .

That is, the statement is false.

Hence, option $B$ is correct.

In the given figure:

$\angle b = 2a + 15^{\circ}$ and

$\angle c = 3a + 5^{\circ}$ , find the value of

$\angle b$ .

0%
$100^o$
0%
$105^o$
0%
$95^o$
0%
$92^o$

Explanation

Given the angles are:

${ a }, { 70 }^{ o }, b=2a+15^o$ and

$c=3a+5^o$ .

We know, by angle sum property, the sum of angles of a quadrilateral is

$360^o$ .

$a+b+c+70^o=360^o$

$\Rightarrow a+2a+15^o+3a+5^o+70^o=360^o$

$\Rightarrow 6a+90^o=360^o$

$\Rightarrow 6a=270^o$

$\Rightarrow a={ 45 }^{ o }$ .

Therefore,

$\angle b=2a+15^o$

$=2(45^o)+15^o$

$=90^o+15^o$

$={ 105 }^{ o}$

Hence, option

$B$ is correct.

Three angles of a quadrilateral are equal. If the fourth angle is

$69^{\circ}$ , find the measure of equal angles.

0%
$86^o$
0%
$97^o$
0%
$68^o$
0%
None of the above

Explanation

We know, by angle sum property, the sum of angles of a quadrilateral is $360^o$ .

Given, three angles are equal. Let the measure of each of these angles be $x$ .

Also, fourth angle $=69^o$ .

Then, $x+x+x+69^o=360^o$

$\Rightarrow 3x+69^o=360^o$ .

$\Rightarrow$ $3x=360^o - 69^o$

$\Rightarrow$ $3x=291^o$

$\Rightarrow x=\dfrac{291^o}{3}\\=97^o$ .

Therefore, the equal angles are each $=97^o$ .

Hence, option $B$ is correct.

In a quadrilateral

$ABCD$ ,

$AB= AD$ and

$CB= CD$ , then check whether

$AC$ bisects angle

$BAD$ is true/false.

0%
True
0%
False

Explanation

$In\quad \triangle ADC\quad \& \quad \triangle ABC,\\ AD=AB\quad \& \quad DC=BC\quad \left( given \right) \\ AC\quad is\quad common\quad side.\\ So\quad \triangle ADC\cong \triangle ABC\quad \left( By\quad SSS\quad rule\quad of\quad congruency \right) \\ So\quad \angle DAC=\angle BAC\\ Therefore\quad we\quad can\quad say\quad that\quad AC\quad bisects\quad \angle BAD.$

State whether the given result is true or false:

In the parallelogram

$ABCD$ , the side

$AB$ is produced to the point

$X$ , so that

$BX =AB$ . The line

$DX$ cuts

$BC$ at

$E$ , then

$DBXC$ is a parallelogram.

0%
True
0%
False

Explanation

Let

$AB =a$

then

$BX=DC=a$

In the figure,

$BX$ is parallel to

$DC$ and

$BX=DC$

$\therefore$

$DBXC$ is a parallelogram.

State true or false:

In the given figure,

$AP$ is bisector of

$\angle A$ and

$CQ$ is bisector of

$\angle C$ of parallelogram

$ABCD$ , then

$APCQ$ is a parallelogram

0%
True
0%
False

Explanation

We know that opposite angles of a parallelogram are equal in measures.

Here,

$\angle A$ and

$\angle C$ are opposite angles.

Hence bisector of these angles are parallel to each other.

So APQC is a parallelogram.

Hence, the given statement is true.

The angles

$P, Q, R$ and

$S$ of

$PQRS$ are in the ratio

$1:3:7:9$ , find the measure of each angle.

0%
$19^{\circ}, 57^{\circ}, 136^{\circ}, 171^{\circ}$
0%
$18^{\circ}, 54^{\circ}, 126^{\circ}, 162^{\circ}$
0%
$17^{\circ}, 66^{\circ}, 116^{\circ}, 152^{\circ}$
0%
$12^{\circ}, 36^{\circ}, 108^{\circ}, 120^{\circ}$

Explanation

Given ratio of angles of quadrilateral $PQRS$ is $1:3:7:9$

Let the angles of quadrilateral $PQRS$ be $x,3x,7x,9x$ , respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $360^o$ .

$\Rightarrow \angle P+\angle Q+\angle R+\angle S=360^o$

$\Rightarrow x+3x+7x+9x=360^o$

$\Rightarrow 20x=360^o$

$\Rightarrow x=\dfrac{360^o}{20}$

$\therefore x=18^o$ .

$\therefore \angle P =x= 18^o$ ,

$\angle Q =3x=3\times 18^o =54^o$ ,

$\angle R =7x=7\times 18^o =126^o$

and $\angle S =9x=9 \times 18^o =162^o$ .

Hence, option $B$ is correct.

State true or false:

In a quadrilateral

$ABCD$ ,

$AB= AD$ and

$CB= CD$ , then

$AC$ is perpendicular bisector of

$BD$ .

0%
True
0%
False

Explanation

In quad. ABCD

$AB=AD$ and

$BC = CD$

AC and BD intersects each other at O.

$\bigtriangleup ACD$ and

$\bigtriangleup ABC$

$AD=AB$ (Given)

$AC=AC$ (Common)

$CD=CB$ (Given)

Therefore,

$\bigtriangleup ACD \cong \bigtriangleup ACB$ [By SSS]

$\therefore \angle DAC=\angle BAC$ [By CPCT]

$\rightarrow (1)$

Now, in

$\bigtriangleup AOD$ &

$\bigtriangleup AOB$ ,

$AD=AB$ (Given)

$\angle DAO=\angle OAB$ (From (1))

$AO=AO$ (Common)

Therefore,

$\bigtriangleup DAO \cong \bigtriangleup BAO$ (By SAS)

$\therefore DO=OB$ and

$\angle AOD=\angle AOB$ [By CPCT]

Now,

$\angle AOD+\angle AOB=80^{\circ}$ [Linear pair]

$2\angle AOD=180^{\circ}$

$[\because\angle AOD=\angle AOB]$

$\angle AOD=90^{\circ}$

Similarly

$\angle BOC=90^{\circ}$

Hence,

$\angle AOD=\angle AOB=\angle DOC=\angle BOC=90^{\circ}$

Therefore,

$AC\perp BD$

Use the information given in the following figure to find the value of

$x$ .

0%
$40^o$
0%
$50^o$
0%
$45^o$
0%
$35^o$

Explanation

${\textbf{Step-1: Draw a given diagram find their missing angles.}}$

${\textbf{}}$

Use the informations given in the following figure to find the value of x. - Studyrankersonline

${\text{Take A, B, C, D as the vertices of quadrilateral and BA is produced to E(say).}}$

${\text{Observe given figure.}}$

$\Rightarrow \angle EAD = 70^0$

$\Rightarrow \angle EAD + \angle DAB = 180^0$

$\Rightarrow \angle DAB = 180^0 - 70^0 =110^0$

$\boldsymbol{[\because{\textbf{EAB is a straight line and AD stands on it.]}}}$

${\textbf{Step-2: Apply the sum of interior angles.}}$

$\text{We know the sum of interior angles of a quadrilateral is }$

${360}^0.$

$\Rightarrow \angle DAB + \angle ADC + \angle DCB + \angle CBA = 360^0$

$\Rightarrow 110^0 + 80^0 + 56^0 +3x-6^0 = 360^0$

$\Rightarrow 3x-6^0 = 360^0 -110^0 - 80^0 - 56^0$

$\Rightarrow 3x- 6^0 = 114$

$\Rightarrow 3x = 120^0$

$\Rightarrow x = 40^0$

${\textbf{Therefore the value of}}$

${\mathbf{x = 40^0}.}$

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

State 'true' or 'false':

The diagonals of a quadrilateral bisect each other.

0%
True
0%
False

If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral

$ABCD$ is a rectangle, show that the diagonals

$AC$ and

$BD$ intersect at right angle. State true or false.

0%
True
0%
False

Explanation

Given:

$ABCD$ is a quadrilateral.

$P, Q, R$ and

$S$ are the mid points of

$AB, BC, CD, DA. PQRS$ is a rectangle.
Construction: Join

$AC$ and

$BD$

Now in

$\triangle ACD$

$R$ and

$S$ are mid points of

$AC$ and

$AD$ respectively.
hence, by mid point theorem,

$RS \parallel AC$

Now in

$\triangle ABC$

$P$ and

$Q$ are mid points of

$AB$ and

$BC$ respectively.
Hence, by mid point theorem,

$PQ \parallel AC$

Similarly,

$QR \parallel SP \parallel BD$
We know,

$PQRS$ is a rectangle,

$PQ \perp QR$
Thus,

$AC \perp BD$ (Angle made between two lines is same as the angle between their corresponding parallel sides)

In the figure,

$LM = LN,\ \angle PLN = 110^{\circ}$ . Then,the measure of

$\angle MLN$ in degrees is____.

0%
$110^{\circ}$
0%
$40^{\circ}$
0%
$10^{\circ}$
0%
$140^{\circ}$

Explanation

In quadrilateral

$PLQN$ ,

$\angle P + \angle Q + \angle PLN + \angle QNL = 360^o$ ...(Sum of angles of quadrilateral)

$90^o + 90^o + 110^o + \angle QNL = 360^o$

$\angle QNL = 70^{\circ}$ .

In

$\triangle LMN$ ,

$\angle QNL=\angle MNL=70^o$

$\angle LMN = \angle MNL$ ...(Isosceles triangle property)

$\therefore \angle MNL= \angle LMN = 70^{\circ}$ .

We know, sum of angles of the

$\triangle LMN$ ,

$\angle MLN + \angle LNM + \angle LMN = 180^o$

$\angle MLN + 70^o + 70^o = 180^o$

$\angle MLN = 180^o - 140^o$

$\angle MLN = 40^{\circ}$ .

Therefore, option

$B$ is correct.

The angles of a quadrilateral are in the ratio 3 : 2 : 4 :Find the angles. Assign a special name to the quadrilateral.

0%
$108^{\circ}, \, 82^{\circ},\, 144^{\circ},\, 36^{\circ}$ , Trapezium
0%
$108^{\circ}, \, 92^{\circ},\, 144^{\circ}, \, 36^{\circ}$ ,Trapezium
0%
$108^{\circ}, \, 72^{\circ},\, 144^{\circ}, \, 36^{\circ}$ ,Trapezium
0%
$108^{\circ}, \, 142^{\circ},\, 144^{\circ}, \, 36^{\circ}$ , Trapezium

Explanation

$The\quad ratio\quad of\quad angles\quad of\quad quadilateral=3:2:4:1\\ Let\quad the\quad angles\quad of\quad quadilateral\quad \\ =3x+2x+4x+x=360\\ 10x=360,x=36\\ \therefore \quad angles\quad of\quad quadilateral\quad are\\ 3x=3\times 36=108\\ 2x=2\times 36=72\\ 4x=4\times 36=144\\ In\quad the\quad adjoining\quad figure\\ \angle A+\angle B=108+72=180\\ ie\quad the\quad sum\quad of\quad interior\quad angles\quad on\quad the\quad same\quad side\quad of\quad transversal\\ AB=180\\ \therefore AD\parallel BC,\\ Hence\quad quadilateral\quad ABCD\quad is\quad a\quad trapezium$

In the above figure,

$ABCD$ is a parallelogram,

$AL \bot CD$ and

$AM \bot BC$ . If

$AB = 12\space cm, AD = 8\space cm$ and

$AL = 6\space cm$ , then

$AM =$

0%
$15\space cm$
0%
$9\space cm$
0%
$10\space cm$
0%
None of these

Explanation

$ABCD$ is a parallelogram, where

$AL\perp CD$ and

$AM\perp BC$

$AB=12\,cm,\,AD=8\,cm$ and

$AL=6\,cm$

We know, opposite sides of a parallelogram are equal.

$AB=CD$ and

$AD=BC$

$\therefore$

$CD=12\,cm$ and

$BC=8\,cm$

$\Rightarrow$ Area of a parallelogram

$ABCD=AL\times CD$

$=6\times 12$

$=72\,cm^2$ ------ ( 1 )

$\Rightarrow$ Area of a parallelogram

$ABCD=AM\times BC$

$=AM\times 8$ ------- ( 2 )

Comparing ( 1 ) and ( 2 ), we get

$\Rightarrow$

$AM\times 8=72$

$\Rightarrow$

$AM=9\,cm$

1309549_242404_ans_430eb9a15cc04a719c65340f906a8a54.png

Use the information given in figure to choose the correct value of

$x$ .

0%
$x = 86^{o}$
0%
$x = 60^{o}$
0%
$x = 68^{o}$
0%
$x = 107^{o}$

Explanation

Since,

$EAB$ is a straight line

$\therefore \angle DAE + \angle DAB = 180^{\circ}$ ....[Linear pair]

$\Rightarrow 73^{\circ} + \angle DAB = 180^{\circ}$ .

Since, by angle sum property, the sum of the angles of quadrilateral

$ABCD$ is

$360^o$ .

$\angle ADC + \angle DCB + \angle CBA + \angle BAD = 360^o$

$\therefore 107^{\circ} + 105^{\circ} +x+80^{\circ} = 360^{\circ}$

$\Rightarrow 292^{\circ} + x = 360^{\circ}$

$x = 360^{\circ} - 292^{\circ} = 68^{\circ}$ .

Therefore, option

$C$ is correct.

The angles of a quadrilaterals are in the ratio

$3 : 5 : 9 : 13$ . All the angles of the quadrilaterals are:

0%
$36^{\circ}, 60^{\circ}, 108^{\circ}, 156^{\circ}$
0%
$30^{\circ}, 50^{\circ}, 90^{\circ}, 130^{\circ}$
0%
$39^{\circ}, 65^{\circ}, 117^{\circ}, 169^{\circ}$
0%
$30^{\circ}, 65^{\circ}, 100^{\circ}, 150^{\circ}$

Explanation

Given ratio of angles of quadrilateral $PQRS$ is $3:5:9:13$

Let the angles of quadrilateral $PQRS$ be $3x,5x,9x,13x$ , respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $360^o$ .

$\Rightarrow \angle P+\angle Q+\angle R+\angle S=360^o$

$\Rightarrow 3x+5x+9x+13x=360^o$

$\Rightarrow 30x=360^o$

$\Rightarrow x=\dfrac{360^o}{30}$

$\therefore x=12^o$ .

$\therefore \angle P =3x= 3\times 12^o=36^o$ ,

$\angle Q =5x=5\times 12^o =60^o$ ,

$\angle R =9x=9\times 12^o =108^o$

and $\angle S =13x=13 \times 12^o =156^o$ .

Hence, option $A$ is correct.

Three angles of a quadrilateral are respectively equal to

$110^o$ ,

$50^o$ and

$40^o$ . Find the fourth angle.

0%
$160^o$
0%
$140^o$
0%
$120^o$
0%
$100^o$

Explanation

Sum of all angles of a quadrilateral is

$360^0$
Then fourth angle will be =

$360^0 - ( 110^0 + 50^0 + 40^0 )$
=

$160^0$

Hence, the fourth angle is

$160^0$

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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