MCQ Questions for Class 9 Maths Quadrilaterals Quiz 5

In a quadrilateral

$ABCD$ , if

$\angle$

$A = 60$

$^o$ and

$\angle$

$B :$

$\angle$

$C :$

$\angle$

$D = 2 : 3 : 7$ , then

$\angle$

$D =$ ?

0%
$175$ $^o$
0%
$135$ $^o$
0%
$150$ $^o$
0%
None of these

Explanation

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

$360^o$ .

Given,

$\angle B:\angle C:\angle D=2:3:7$ .
Let the

$\angle B = 2x , \angle C = 3x , \angle D= 7x$ and

$\angle A = 60^o$ .

$\Rightarrow 2x + 3x + 7x + 60^o = 360^o$

$\Rightarrow 12x = 300^o$

$\Rightarrow x = 25^o$ .

$\therefore \angle D= 7x = 7 \times 25^o = 175^o$ .

Hence, option

$A$ is correct.

The angles of a quadrilateral are in the ratio

$3 : 5 : 9 : 13$ . Find all the angles of the quadrilateral in degrees.

0%

$36, 60, 108, 156$
0%
$36, 70, 108, 156$
0%
$36, 60, 118, 156$
0%
none of the above

Explanation

Let the four angles of the quadrilateral be

$3x, 5x, 9x$ and

$13x$

$3x + 5x + 9x + 13x = { 360 }^{ \circ }$

[sum of all angles of the quadrilateral is

${ 360 }^{ \circ }$ ]

$30x = { 360 }^{ \circ }$

$x = { 12 }^{ \circ }$

Hence, the angles of the quadrilateral are

$3\times{ 12 }^{ \circ }={ 36 }^{ \circ },5\times{ 12 }^{ \circ }={ 60 }^{ \circ },9\times{ 12 }^{ \circ }={ 108 }^{ \circ }$ and

$13\times{ 12 }^{ \circ }={ 156 }^{ \circ }$

If three angles of a quadrilateral are of magnitudes

$80^\circ,\ 95^\circ,\ 120^\circ,$ then what is the measure of the fourth angle?

0%
$80$ $^o$
0%
$65$ $^o$
0%
$75$ $^o$
0%
$70$ $^o$

Explanation

The given angles are, $80^\circ,\ 95^\circ,\ 120^\circ$ .

Let the fourth angle be $x^\circ$

Angle sum property states that the sum of angles of a quadrilateral is $360^\circ$ .

Then, $80^\circ+95^\circ+120^\circ+x^\circ=360^\circ$

$\Rightarrow 360^\circ - (80^\circ + 95^\circ + 120^\circ ) =x^{\circ}$

$\Rightarrow$ $x^\circ=360^\circ - 295^\circ$

$=65^{\circ}$

Therefore, the fourth angle is $x^\circ=65^\circ$

Hence, option $B$ is correct.

In a quadrilateral the angles are in the ratio

$3:4:5:6$ . Then the difference between the greatest and the smallest angle is?

0%
${108}^{o}$
0%
${60}^{o}$
0%
${80}^{o}$
0%
${90}^{o}$

Explanation

Given ratio of angles of quadrilateral $ABCD$ is $3:4:5:6$

Let the angles of quadrilateral $ABCD$ be $3x,4x,5x,6x$ respectively.

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

$\Rightarrow 3x+4x+5x+6x=360^o$

$\Rightarrow 18x=360^o$

$\therefore x=20^o$ .

$\therefore \angle A =3x=3 \times 20^o =60^o$ ,

$\angle B =4x=4\times 20^o =80^o$ ,

$\angle C =5x=5 \times 20^o =100^o$

and $\angle D =6x=6 \times 20^o =120^o$ .

$\therefore$ The greatest angle is $=120^o$

and the smallest angle is $=60^o$ .

Therefore, the difference between the greatest and the smallest angle is $=120^o-60^o=60^o$ .

Hence, option $B$ is correct.

In a square

$ABCD,$ diagonals meet at

$O. P$ is a point on

$BC$ such that

$OB = BP.$

State whether the above statement is true or false.

$\angle POC = \left ( 22\dfrac{1}{2} \right )^{\circ}$

0%
True
0%
False

Explanation

$ABCD$ ,

$OB=OP$ [Given]

$\bigtriangleup OBP , OB=OP,$ therefore

$\angle OPB=\angle BOP$ [Angles opposite to equal sides one equal]

$\angle DBP=\dfrac{90^{\circ}}{2}=45^{\circ}$ [

$\because \angle B=90^{\circ}$ and

$DB$ is angle bisector]

$\angle BOP=\dfrac{180^{\circ}-45^{\circ}}{2}=67.5^{\circ}$ [

$\because \bigtriangleup OBP$ is an isosceles

$\bigtriangleup$ ]

$\angle BOC=90^{\circ}\Rightarrow \angle BOP+\angle POC=180^{\circ}\Rightarrow \angle POC=67.5^{\circ}-45^{\circ}=22.5^{\circ}$

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

The sides of a quadrilateral are extended to make the angles as shown in the figure. Find the value of

$x.$

0%
${80}^{o}$
0%
$60^o$
0%
$40^o$
0%
$130^o$

Explanation

The sum of the exterior angles of a quadrilateral is equal to

${360}^{o}.$ So,

$\Rightarrow$

$x+{75}^{o}+{115}^{o}+{90}^{o}={360}^{o}$

$\Rightarrow$

$x+280^o={360}^{o}$

$\Rightarrow$

$x={360}^{o}-{280}^{o}$

$\Rightarrow x={80}^{o}$

If the bisectors of the angles

$A$ and

$B$ of a quadrilateral

$ABCD$ meet at

$O$ , then

$\angle AOB$ is equal to:

0%
$\angle C+\angle D$
0%
$\cfrac{1}{2}(\angle C+\angle D)$
0%
$\cfrac{1}{2}\angle C+\cfrac{1}{3}\angle D$
0%
$\cfrac{1}{3}\angle C+\cfrac{1}{2}\angle D$

Explanation

In quad

$ABCD$

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

$\Rightarrow$

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

$\Rightarrow$

$\cfrac{\angle A}{2}+\cfrac{\angle B}{2}={180}^{o}-\cfrac{1}{2}(\angle C+\angle D)$

$\Rightarrow$

$\angle OAB+\angle OBA={180}^{o}-\cfrac{1}{2}(\angle C+\angle D)$

(

$\therefore$

$OA$ bisects

$\angle A$ and

$OB$ bisects

$\angle B$ )

$\triangle AOB$ ,

$\angle AOB={180}^{o}-(\angle OAB+\angle OBA)$

$={180}^{o}-[({180}^{o}-\cfrac{1}{2}(\angle C+\angle D)]$

$=\cfrac{1}{2}(\angle C+\angle D)$

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

$2$ .

Consider the following statements and state which one is true and which one is false:
(1) The bisectors of all the four angles of a parallelogram enclose a rectangle.
(2) The figure formed by joining the midpoints of the adjacent sides of the rectangle is a rhombus.
(3) The figure formed by joining the midpoints of the adjacents sides of a rhombus is square.

0%
1 and 2
0%
2 and 3
0%
3 and 1
0%
1,2 and 3

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.

What is the value of

$x$ in the given figure?

0%
$b-a-c$
0%
$b-a+c$
0%
$b+a-c$
0%
$(a+b+c)$

Explanation

Reflex

$\angle BCD={360}^{o}-b$
In quad

$ABCD$

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

$\Rightarrow$

$a+c+{360}^{o}-b+x={360}^{o}$

$\Rightarrow$

$x={360}^{o}-{360}^{o}-a-c+b$

$=b-a-c$

The sum of the angles in a quadrilateral is equal to .......... .

0%
$2$ right angles
0%
$3$ right angles
0%
$4$ right angles
0%
$360$ right angles

Explanation

Consider a quadrilateral

$ABCD$ .

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

To prove this, we join $A$ and $C$ , i.e. we draw the diagonal $AC$ .

In $\triangle ABC$ ,

$\angle CAB+\angle ABC+\angle BCA=180^{\circ}$ [Sum of all angles of a triangle is $180^{\circ}$ ]..... $(1)$ .

In $\triangle ACD$ ,

$\angle CAD+\angle ADC+\angle DCA=180^{\circ}$ [Sum of all angle of a triangle is $180^{\circ}$ ].... $(2)$ .

Adding $(1)$ and $(2)$ , we get,

$\left(\angle CAB+\angle ABC+\angle BCA \right)+ \left(\angle CAD+\angle ADC+\angle DCA \right)=180^{\circ}+180^{\circ}$

$\implies$ $\angle ABC+\angle ADC+(CAB+CAD)+(BCA+DCA)=360^{\circ}$

$\implies$ $\angle ABC+\angle ADC+\angle BAD+\angle BCD=360^{\circ}$

$\implies$ $\therefore \angle A+\angle B+\angle C+\angle D=360^{\circ}$ .

That is, the sum of all angles of a quadrilateral is $360^o=4\times90^o$ , i.e. $4$ right angles.

Therefore, option

$C$ is correct.

In the given figure,

$AB=\,AD$ ;

$CB=\,CD$ ,

$\angle A={ 42 }^{ 0 }$ and

$\angle C={ 108 }^{ 0 }$ , Find

$\angle ABC$

0%
${ 105 }^{ 0 }$
0%
${ 150 }^{ 0 }$
0%
${ 75 }^{ 0 }$
0%
${ 110 }^{ 0 }$

Explanation

Given:

$AB=\,AD$ ;

$CB=\,CD$ ,

$\angle A={ 42 }^{ 0 }$ and

$\angle C={ 108 }^{ 0 }$
Join AC.
Now, in

$\triangle ADC$ and

$\triangle ABC$

$AD = AB$ (Given)

$BC = CD$ (Given)

$AC = AC$ (Common)
Thus,

$\triangle ADC \cong \triangle ABC$ (SSS rule)
Hence,

$\angle CAB = \angle CAD = \frac{1}{2} \angle DAB = 21$ (By CPCT)

$\angle ACB = \angle ACD = \frac{1}{2} \angle DCB = 54$ (By CPCT)

Now, In

$\triangle ABC$

$\angle ABC + \angle CAB + \angle ACB = 180$ (Angles sum property)

$\angle ABC + 21 + 54 = 180$

$\angle ABC = 105^{\circ}$

ABCD is a quadrilateral in which

$AD=\,BC$ and

$\angle DAB=\,\angle CBA$ . Then:

0%
$\triangle ABC\cong \triangle BAD$
0%
$BD=\,AC$
0%
$\angle ABD=\angle BAC$
0%
None of these

Explanation

$\triangle ABC$ and

$\triangle BAC$ ,

$AD=\,BC$ (Given)

$\angle DAB=\angle CBA$ (Given)

$AB=\,AB$ (Common side)

$\therefore$ By SAS congruence rule,

$\triangle ABC\cong \triangle BAC$

Also, by CPCT, we have

$BD=\,AC$ and

$\angle ABD=\angle BAC$

In the adjoining figure

$ABC$ is a triangle in which

$D$ and

$E$ are the mid points

$AB$ and

$AC$ respectively if

$BC = 4.6\ \text{cm}$ then

$DE =$ _______

0%
$9.2\ \text{cm}$
0%
$2.3\ \text{cm}$
0%
$5.0\ \text{cm}$
0%
$6.4\ \text{cm}$

Explanation

According to mid point theorem the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

Here

$D$ and

$E$ are the mid points of side

$AB$ aand

$AC$

$\Rightarrow DE||BC$ and

$BC=2DE$

$BC=4.6\ \text{cm}$

$2DE=4.6\ \text{cm}$

$DE=2.3\ \text{cm}$

So option

$B$ is correct.

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

0%
In a trapezium the diagonals bisect each other
0%
In a rectangle diagonals intersect at right angles
0%
The diagonals of a rhombus are equal
0%
None of these

Which of the following statements is true?

0%
The diagonals of a rectangle are perpendicular
0%
The diagonals of a rhombus are equal
0%
Every square is a rhombus
0%
None of these

$\triangle PQR, QR=10, RP=11$ and

$PQ=12.$

$D$ is the midpoint of

$PR$ .

$DE$ is drawn parallel to

$PQ$ meeting

$QR$ in

$E.$

$EF$ is drawn parallel to

$RP$ meeting

$PQ$ in

$F.$ What is the length of

$DF\: ?$

0%
$\displaystyle \frac{11}{2}$
0%
$6$
0%
$\displaystyle \frac{33}{4}$
0%
$5$

Explanation

Given,

$D$ is the midpoint of

$PR$ .

and

$DE \parallel PQ$ .

$\implies E$ is the midpoint of

$QR$ .

We know,

Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

Similarly,

$F$ is the midpoint of

$PQ$

Then, from Midpoint Theorem,

$DF || AR$ and

$DF=\dfrac{1}{2}QR$

Since,

$QR=10$ .

$\implies$

$DF=\dfrac{1}{2}(10)=5$ .

Hence, option

$D$ is correct.

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)

$180$

$^\circ$

The diagram shows the construction of

0%
a trapezium.
0%
a parallelogram.
0%
a rectangle.
0%
a kite.

If angles

$\angle A,\angle B,\angle C,\angle D$ of a quadrilateral

$ABCD$ are taken in order, then they are found to be in ratio

$3:7:6:4$ , then

$ABCD$ is a:

0%
rhombus
0%
parallelogram
0%
trapezium
0%
kite

Explanation

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

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

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

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

$\Rightarrow 20x=360^o$

$\therefore x=18^o$ .

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

$\angle B =7x=7\times 18^o =126^o$ ,

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

and $\angle D =4x=4 \times 18^o =72^o$ .

We know that in a trapezium we have two parallel lines which form two pairs of co-interiors angles, which are Supplementary.

Clearly $\angle A+ \angle B=180^o$ and $\angle C+\angle D=180^o$

Therefore, the given quadrilateral is a trapezium.

Hence, option $C$ is correct.

In a quadrilateral

$PQRS$ , if

$\angle P\, =\, \angle R\, =\, 100^{\circ}$ and

$\angle S\, =\, 75^{\circ}$ , then

$\angle Q =$ ......... .

0%
$50^{\circ}$
0%
$85^{\circ}$
0%
$120^{\circ}$
0%
$360^{\circ}$

Explanation

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

The given angles are $\angle P=100^o, \angle R=100^o, \angle S=75^o, \angle Q$ .

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

$\implies$ $100^o+100^o+75^o+\angle Q=360^o$ .

$\Rightarrow$ $\angle Q=360^o - (100^o + 100^o + 75^o )$

$\Rightarrow$ $\angle Q=360^o - 275^o =85^{\circ}$ .

Therefore, the unknown angle is $\angle Q=85^o$ .

Hence, option $B$ is correct.

If the bisectors of the angles

$A, B, C$ and

$D$ of a quadrilateral meet at

$O$ , then

$\angle AOB$ is equal to:

0%
$\angle C+\angle D$
0%
$\displaystyle \frac{1}{2}(\angle C+\angle D)$
0%
$\displaystyle \frac{1}{2}\angle C+\frac{1}{3}\angle D$
0%
$\displaystyle \frac{1}{3}\angle C+\frac{1}{2}\angle D$

Explanation

In quad.

$ABCD$

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

$\Rightarrow \angle A+\angle B=360^{\circ}-(\angle C+\angle D)$ ...(i)

$\Delta AOB$ , we know

$\angle AOB+\angle OAB+\angle OBA=180^{\circ}$

$\angle AOB=180^{\circ}-(\angle OAB+\angle OBA)$

$=\displaystyle 180^{\circ}-\frac{1}{2}(\angle A+\angle B)$

(

$\because$

$OA$ bisects

$\angle A$ and

$OB$ bisects

$\angle B$ )

Thus from (i),

$\angle AOB=180^{\circ}-\dfrac{1}{2}\left \{ 360^{\circ}-(\angle C+\angle D) \right \}$

$=\displaystyle 180^{\circ}-180^{\circ}+\dfrac{1}{2}(\angle C+\angle D)$

$=\displaystyle \frac{1}{2}(\angle C+\angle D)$

Find the value of

$x$ .

0%
$60^o$
0%
$120^o$
0%
$90^o$
0%
$140^o$

Explanation

Clearly, ext

$\angle A=90^o$ (given)

$\angle A=180^o-90^o=90^o$ .....[Linear pair].
We know, by angle sum property, the sum of the measures of interior angles of a quadrilateral is

$360^o$ ,

$\implies$

$90^o+60^o+70^o+x=360^o$

$\implies$

$220^o+ x= 360^o$

$\implies$

$x=360^o-220^o$

$= 140^o$ .

Therefore, option

$D$ is correct.

In the diagram,

$ABD$ and

$BCD$ are isosceles triangles triangles, where

$AB=BC=BD$ . The special name that is given to quadrilateral

$ABCD$ is:

0%
rectangle
0%
kite
0%
parallelogram
0%
trapezium

Explanation

$\Delta BAD$ ,

$\angle BDA=\angle BAD=57^{\circ}$ (isosceles triangle property)
In

$\Delta BDC$ ,

$\angle BCD=\angle BDC=66^{\circ}$ (isosceles triangle property)

$\therefore \displaystyle \angle D=\angle ADB + \angle BDC = 57^{\circ}+66^{\circ}=123^{\circ}$ and

$\angle A+\angle D=57^{\circ}+123^{\circ}=180^{\circ}$
Also,

$\angle D+ \angle C=123^{\circ}+66^{\circ}=189^{\circ}$
Hence, by the property that corresponding interior angles are supplementry, lines are parallel, we have

$AB \parallel DC$ and

$AD \not \parallel BC$
Hence,

$ABCD$ is a trapezium.

Find the value of

$x$ .

0%
$60^o$
0%
$40^o$
0%
$50^o$
0%
$90^o$

Explanation

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

The given angles are $\angle A=130^o, \angle B=120^o, \angle C=x, \angle D=50^o$ .

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

$\implies$ $130^o+120^o+x+50^o=360^o$ .

$x=360^o - (130^o + 120^o + 50^o )$

$x=360^o - 300^o =60^{\circ}$ .

Therefore, the unknown angle is $x=60^o$ .

Hence, option $A$ is correct.

$P$ is the mid-point of side

$AB$ to a parallelogram

$ABCD$ . A line through

$B$ parallel to

$PD$ meets

$DC$ at

$Q$ and

$AD$ produced at

$R$ . Then

$BR$ is equal to ______.

0%
$BQ$
0%
$\dfrac {1}{2}BQ$
0%
$2BQ$
0%
None of these

Explanation

$\triangle ABR$ ,

$P$ is the mid-point of

$AB$ and

$DP || BR$

$\therefore D$ is the mid-point of

$AR$ .....(Converse of mid-point theorem).

Then,

$AR=2AD$

$\implies$

$AR=2BC$ (Opposite sides of parallelogram are equal, i.e.

$AD=BC$ )

$\implies$

$AD = BC$ .

Also,

$D$ is the mid-point of

$AR$ and

$DQ||AB$ ....(Since,

$ABCD$ is parallelogram)

Then,

$Q$ is the mid-point of

$DC$ ....(Converse of mid-point theorem)

$\therefore BR = 2 BQ$ .

Hence, option

$C$ is correct.

The diagonals of a quadrilateral PQRS bisect each other at right angles. If

$PQ=5.5\:cm$ , then the perimeter of PQRS is

0%
$11$ cm
0%
$22$ cm
0%
$16.5$ cm
0%
$55(1+\sqrt{2})$ cm

Explanation

Given that the diagonals of a quadrilateral bisect each other at right angles.

Since, the diagonal bisect each other, the quadrilateral is a parallelogram.
The diagonals of the quadrilateral bisect at right angles, hence it must be a rhombous in which all sides are equal.
Then, perimeter

$=4S\\ = 4 \times 5.5\\ = 22\ cm$

Suppose the triangle ABC has an obtuse angle at C and let D be the midpoint of side AC. Suppose E is on BC such that the segment DE is parallel to AB. Consider the following three statements
i) E is the midpoint of BC
ii) The length of DE is half the length of AB
iii) DE bisects the altitude from C to AB

0%
only (i) is true
0%
only (i) and (ii) are true
0%
only (i) and (iii) are true
0%
all three are true

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 5 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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