MCQ Questions for Class 9 Maths Quadrilaterals Quiz 8

The angles of a quadrilateral are in the ratio

$1 : 2 : 3 : 4$ . Then the smallest angle is:

0%
$72^{\circ}$
0%
$144^{\circ}$
0%
$36^{\circ}$
0%
$18^{\circ}$

Explanation

Given ratio of angles of quadrilateral $ABCD$ is $1:2:3:4$

Let the angles of quadrilateral $ABCD$ be $1x,2x,3x,4x$ , respectively.

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

$\Rightarrow 1x+2x+3x+4x=360^o$

$\Rightarrow 10x=360^o$

$\therefore x=36^o$ .

$\therefore \angle A =1x=1 \times 36^o =36^o$ ,

$\angle B =2x=2\times 36^o =72^o$ ,

$\angle C =3x=3 \times 36^o =108^o$

and $\angle D =4x=4\times 36^o =144^o$ .

$\therefore$ The smallest angle $=36^o$ .

Hence, option $C$ is correct.

What is the maximum number of obtuse angles that a quadrilateral can have ?

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

Explanation

We know that, the sum of all the

$4$ interior angles of a quadrilateral is

$360^{\circ}$

$\implies \theta_1+\theta_2+\theta_3+\theta_4=360^{\circ}$

An angle

$\theta$ is obtuse if

$90^{\circ}< \theta < 180^{\circ}$

If all the angles of a quadrilateral are obtuse

$90^{\circ}< \theta_1 < 180^{\circ}$ -----(1)

$90^{\circ}< \theta_2 < 180^{\circ}$ -----(2)

$90^{\circ}< \theta_3 < 180^{\circ}$ -----(3)

$90^{\circ}< \theta_4 < 180^{\circ}$ -----(4)

Adding (1), (2), (3) and (4)

$\implies 360^{\circ}<\theta_1+\theta_2+\theta_3+\theta_4<720^{\circ}$

Therefore all the

$4$ angles can not be obtuse

If three angles are obtuse say each of

$100^{\circ}$ , then the fourth angle will be

$=360^{\circ}-300^{\circ}=60^{\circ}$

Hence, utmost

$3$ angles can be obtuse

If two adjacent angles of a parallelogram are in the ratio

$2 : 3$ , then the measure of angles are:

0%
$72^{\circ} , 108^{\circ}$
0%
$36^{\circ} , 54^{\circ}$
0%
$80^{\circ} , 120^{\circ}$
0%
$96^{\circ} , 144^{\circ}$

Explanation

Let the angles be

$2x , 3x$

$2 x + 3 x = 180$

$\Rightarrow \, 5x = 180$

$\Rightarrow \, 5x = 180$
So , the angles are

$2 \times 36 = 72$

$3 \times 36 = 108$

Which of the following can be four interior angles of a quadrilateral ?

0%
$140^{\circ} , 40^{\circ} , 20^{\circ} , 160^{\circ}$
0%
$270^{\circ} , 150^{\circ} , 30^{\circ} , 20^{\circ}$
0%
$40^{\circ} , 70^{\circ} , 90^{\circ} , 60^{\circ}$
0%
$110^{\circ} , 40^{\circ} , 30^{\circ} , 180^{\circ}$

Explanation

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

$360^o$ .

Consider option

$A$ .

Then,

$140^o+40^o+20^o+160^o$

$=360^o$ .

Hence, the angles are the four interior angles of a quadrilateral.

Consider option

$B$ .

Then,

$270^o+150^o+30^o+20^o$

$=470^o \ne 360^o$ .

Hence, the angles are not the four interior angles of a quadrilateral.

Consider option

$C$ .

Then,

$40^o+70^o+90^o+60^o$

$=260^o \ne 360^o$ .

Hence, the angles are not the four interior angles of a quadrilateral.

Consider option

$D$ .

Then,

$110^o+40^o+30^o+180^o$

$=360^o$ .

Hence, the angles are the four interior angles of a quadrilateral.

Therefore, options

$A$ and

$D$ are correct.

State whether the statements are true (T) or (F) false:
Sum of all the angles of a quadrilateral is

$180^{\circ}$ .

0%
True
0%
False

Explanation

Consider a quadrilateral

$ABCD$ .

To prove this, we join

$A$ and

$C$ , i.e. we draw the diagonal

$AC$ .

$\triangle ABC$ ,

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

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

$(1)$ .

$\triangle ACD$ ,

$\angle CAD+\angle ADC+\angle DCA=180^{\circ}$ [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 the angles of a quadrilateral is

$360^o$ .

Therefore, the statement is false.

Hence, option

$B$ is correct.

1964057_1792427_ans_764a522cf5524899ab7d15ce1f427160.png

State whether the statements are true (T) or (F) false:
A quadrilateral can have all four angles as obtuse.

0%
True
0%
False

Explanation

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

The given angles are obtuse.

Let the angles be $\angle A=90^o+a, \angle B=90^o+b, \angle C=90^o+c, \angle D=90^o+d$ .

Then, $\angle A+\angle B+\angle C+\angle D$

$=90^o+a+90^o+b+90^o+c+90^o+d$

$=360^o +a+b+c+d >360^o.$

Therefore, all the four angles of a quadrilateral cannot be obtuse.

That is, the statement is false.

Hence, option

$B$ is correct.

Area of a quadrilateral ABCD is

$20 cm^2$ and perpendiculars on BD from opposite vertices are 1 cm and 1.5 cm. The length of BD is

0%
4 cm
0%
15 cm
0%
16 cm
0%
18 cm

Explanation

Area of the quadrilateral given=

$\dfrac{1}{2} (sum\, of\, altitudes)\times corresponding\, diagonal$

$\rightarrow$

$20=\dfrac{1}{2}(1+1.5)\times BD$

$\rightarrow$

$\dfrac{1} {2}(2.5) \times BD = 20 cm ^{2}$

$\rightarrow BD = 20 \times\dfrac{2}{2.5}$ =

$\dfrac{40}{2.5} = 16 cm$

Write True or False and justify your answer in each of the following :
In Fig.

$10.10$ , if AOB is a diameter and

$\angle$ ADC =

$120^{\circ}$ , then

$\angle$ CAB =

$30^{\circ}$ .

0%
True
0%
False

Explanation

AOB is a diameter of circle with center O .

$\angle$ ADC +

$\angle$ ABC =

$180^{\circ}$

[

$\because$ ABCD is a cyclic quadrilateral]

$\Rightarrow$

$120^{\circ} + \angle$ ABC =

$180^{\circ}$

$\Rightarrow$

$\angle$ ABC =

$180^{\circ} - 120^{\circ} = 60^{\circ}$

$\Delta$ ABC , we have

$\angle$ ACB =

$90^{\circ}$

[

$\because$ Angle in a semicircle and

$\angle$ ABC =

$60^{\circ}$ (Proved above)]

$\therefore$

$\angle$ CAB =

$180^{\circ} - (90^{\circ} + 60^{\circ}) = 30^{\circ}$

Hence , the given statement is true

1660310_1795104_ans_fc395ce1e6bc42f2ae391382fde77990.png

In a quadrilateral

$PQRS, \angle P, \angle Q, \angle R$ and

$\angle S$ are interior angles. If

$\angle P: \angle Q: \angle R: \angle S = 1:2:3: 4$ , then which angle is equal to

$144^\circ$ .

0%
$\angle P$
0%
$\angle Q$
0%
$\angle R$
0%
$\angle S$

Explanation

Given ratio of angles of quadrilateral $PQRS$ is $1:2:3:4$

Let the angles of quadrilateral $PQRS$ be $1x,2x,3x,4x$ , respectively.

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

$\Rightarrow 1x+2x+3x+4x=360^o$
$\Rightarrow 10x=360^o$

$\therefore x=36^o$ .

$\therefore \angle P =1x=1 \times 36^o =36^o$ ,

$\angle Q =2x=2\times 36^o =72^o$ ,

$\angle R =3x=3 \times 36^o =108^o$

and $\angle S =4x=4\times 36^o =144^o$ .

$\therefore$ $\angle S=144^o$ .

Hence, option $D$ is correct.

Three angles of quadrilateral are

$47^{\circ}, 102^{\circ}$ and

$111^{\circ}$ , then the fourth angle is equal to:

0%
$102^{\circ}$
0%
$100^{\circ}$
0%
$360^{\circ}$
0%
$260^{\circ}$

Explanation

Let the four angles be $\angle A , \angle B , \angle C$ and $\angle D$ .

Given $\angle A=47^o , \angle B =102^o, \angle C = 111^{o}$ .

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

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

Then, $\angle D$ will be given by:

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

$\Rightarrow \angle D = 360 ^o-47^o -102^o -111^o$

$\Rightarrow \angle D = 360 - 260^o$

$\Rightarrow \angle D = 100^o$ .

The measure of fourth angle is $100 ^{o}$ .

Therefore, option

$B$ is correct.

Of the following statements, the one that is incorrect for a parallelogram is:

0%
Opposite sides are equal.
0%
Opposite angles are equal
0%
Opposite angles are bisected by the diagonals
0%
Diagonals bisect each other.

The figure formed by joining the mid-points of the sides of a quadrilateral

$ABCD$ , taken in order, is a square only if,

0%
$ABCD$ is a rhombus
0%
diagonals of $ABCD$ are equal
0%
diagonals of $ABCD$ are equal and perpendicular
0%
diagonals of $ABCD$ are perpendicular.

Explanation

In given figure,

$ABCD$ is a quadrilateral and

$P, Q, R$ &

$S$ are mid-pints of sides

$AB, BC, CD$ and

$DA$ respectively. Then,

$PQRS$ is a square.

$\therefore PQ = QR = RS = PS$ --------- (1)

and

$PR = SQ$

But

$PR = BC$ and

$SQ = AB$

$\therefore AB = BC$

Thus, all sides of quadrilateral

$ABCD$ are equal.

Hence, quadrilateral

$ABCD$ is either a square or a rhombus.

Now, in

$\triangle ADB$ ,

By using Mid-point theorem,

$\left|\right| DB; SP = \dfrac{1}{2} DB$ ------ (2)

Similarly in

$\triangle$ ABC,

$PQ \left|\right| AC; PQ = \dfrac{1}{2}AC$ ----- (3)

From equation (1),

$PS = PQ$

From (2) and (3),

$\dfrac{1}{2} DB = \dfrac{1}{2} AC$

$\therefore DB = AC$

Thus, diagonals of

$ABCD$ are equal and therefore quadrilateral

$ABCD$ is a square. So, diagonals of quadrilateral also perpendicular.

State whether the statements are true (T) or (F) false.
Every kite is a parallelogram.

0%
True
0%
False

The quadrilateral formed by joining the mid-points of the sides

$AB,BC,CD,DA$ of a quadrilateral

$ABCD$ is

0%
a trapezium but not a parallelogram
0%
a quadrilateral but not a trapezium
0%
a parallelogram only
0%
a rhombus

Explanation

Given quadrilateral formed By joining mid points of the sides

$AB, BC, CD, DA$

Let mid point of

$AB, BC, CD, DA$ are

$P, Q, R, S$ respectively.

To prove:-

$PQRS$ is a parallelogram.

Draw diagonal

$BD$ and

$PS$ is mid segment of

$\Delta ABD$ therefore

$PS || BD$

also

$QR$ is mid segment

$\Delta BCD$ therefore

$QR || BD$

$\because PS || BD$ and

$QR || BD$

$\therefore PS || QR$

Draw diagonal

$AC, SR$ is the mid segment of

$\Delta ACD$ therefore

$SR || AC$ also

$PQ$ is the mid segment of

$\Delta ABC$ therefore

$PQ || AC$

$\because SR || AC$ and

$PQ || AC$

$\therefore SR || PQ$

$\because PS || QR$ and

$SR || PQ$

$\therefore$ quadrilateral

$PQRS$ is a parallelogram as. A parallelogram is a simple quadritiral with two pairs of parallel sides.

Option C.

2106375_551125_ans_b4a52c951d64410bb65c57bccafb04c6.png

Which of the following statements is INCORRECT ?

0%
Each diagonal of a quadrilateral divides it into two triangles.
0%
Each side of a quadrilateral is less than the sum of the renraining three sides
0%
A quadriiateral can have atmost three angles
0%
A quadrilateral has two diagonals.

In the figure given,

$M$ is the midpoint of the side

$CD$ of the parallelogram

$ABCD$ . What is

$ON:OB$ ?

0%
$3:2$
0%
$2:1$
0%
$3:1$
0%
$5:2$

The angles of a quadrilateral are in the ratio 2: 4 : 5 :What is the difference between the measures of its largest and smallest angles?

0%
$20^o$
0%
$80^o$
0%
$100^o$
0%
$140^o$

$ABC$ is a triangle in which

$AB=AC$ . Let

$BC$ be produced to

$D$ . From a point

$E$ on the line

$AC$ let

$EF$ be a straight line such that

$EF$ is parallel to

$AB$ . Consider the quadrilateral

$ECDF$ thus formed. If

$\angle ABC={65}^{o}$ and

$\angle EFD={80}^{o}$ , then what is

$\angle FDC$ equal to?

0%
${42}^{o}$
0%
${41}^{o}$
0%
${37}^{o}$
0%
${35}^{o}$

$ABC$ is a triangle, right-angled at

$B$ and

$M,N$ are midpoints of

$AB$ and

$BC$ respectively, then what is

$4({AN}^{2}+{CM}^{2})$ equal to?

0%
$3{AC}^{2}$
0%
$4{AC}^{2}$
0%
$5{AC}^{2}$
0%
$6{AC}^{2}$

A quadrilateral that is not a parallelogram but has exactly two equal opposite angles is:

0%
a rhombus.
0%
a trapezium.
0%
a square.
0%
a kite.

The basic elements of a Quadrilateral are _____.

0%
4 vertices
0%
4 sides
0%
4 angles
0%
All of these

Which of the following statements is INCORRECT?

0%
Each diagonal of a quadrilateral divides it into two triangles
0%
Each side of a quadrilateral is less than the sum of the remaining three sides
0%
A quadrilateral can have atmost three angles
0%
A quadrilateral has two diagonals

If length of each side of a rhombus

$PQRS$ is

$8cm$ and

$\angle {PQR}={120}^{o}$ , then what is the length (in cm) of

$QS$ ?

0%
$4\sqrt {5}$
0%
$6$
0%
$8$
0%
$12$

Explanation

$PQS$ is an equilateral triangle
Hence,

$QS=8cm$

1158212_913553_ans_fbca83620cdf41b7a0fa52799c188291.PNG

$ABCD$ is a quadrilateral inscribed in a circle. Diagonals

$AC$ and

$BD$ are joined.
If

$\angle CAD = 60^{\circ}$ and

$\angle BDC = 25^{\circ}$ . Find

$\angle BCD$

0%
$85^{\circ}$
0%
$120^{\circ}$
0%
$115^{\circ}$
0%
$95^{\circ}$

Explanation

$\angle CAD=60^{\circ}$

$\angle BDC=25^{\circ}$

$\angle CAD=\angle DBC=60^{\circ}$ [As angles are in same segment].

$\triangle BCD$ ,

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

$\Rightarrow 60^\circ+25^\circ+\angle BCD=180^{\circ}$

$\Rightarrow \angle BCD=180^{\circ}-85^{\circ}=95^{\circ}$

1170243_882887_ans_443a1edfa1c347fda1f9b83bf0159c00.png

If the area of quadrilateral ABCD is zero, then the four points A,B,C,D are

0%
collinear
0%
not collinear
0%
nothing can be said
0%
none of these

A square is drawn by joining the mid points of the given square a third square in the same way and this process continues indefinitely. If a side of the first square is

$16$ cm , then the sum of the area of all of the squares

0%
$128$ sq. cm
0%
$256$ sq. cm
0%
$512$ sq. cm
0%
$1024$ sq. cm

In the given figure, if

${ b }^{ \circ }-{ d }^{ \circ }={ 80 }^{ \circ },{ m }^{ \circ }={ 140 }^{ \circ }and\quad { m }^{ \circ }+{ e }^{ \circ }={ 2c }^{ \circ },then\quad ({ a }^{ \circ }+{ d }^{ \circ })-({ e }^{ \circ }+{ C }^{ \circ })$

0%
${0 }^{ \circ }$
0%
${ 40 }^{ \circ }$
0%
${ 80 }^{ \circ }$
0%
${ 50 }^{ \circ }$

Let M, N, P and Q be mid points of the edges AB, CD, AC and BD respectively of the tetrahedron ABCD. Further, MN is perpendicular to both AN and CD and PQ is perpendicular to both AC and BD. Then which of the following is/are correct :

0%
AB = CD
0%
BC = DA
0%
AC = BD
0%
AN = BN

In right triangle ABC, right angle is at C, M is the mid-point Of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. point D is joined to point B show that :

0%
$\Delta AMC\cong \Delta BMD$
0%
$\angle DBC$ is a right angle
0%
$\Delta DBC\cong \Delta ACB$
0%
$CM=\dfrac { 1 }{ 2 } AB$

A,B,C,D are mid points of sides of parallelogram PQRS. If

$ar\left( {PQRS} \right) = 36\,\,c{m^2}$ then

$ar(ABCD)$

0%
$24\,\,c{m^2}$
0%
$18\,\,c{m^2}$
0%
$30\,\,c{m^2}$
0%
$36\,\,c{m^2}$

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

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Practice Class 9 Maths Quiz Questions and Answers

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

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Practice Class 9 Maths Quiz Questions and Answers

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