MCQ Questions for Class 9 Maths Quadrilaterals Quiz 6

The measures of angles of a quadrilateral in degrees are

$x, 3x - 40 , 2x$ and

$4x +20$ . Find the measure of each angle.

0%
The measures of the angles are $55^0, 74^0, 62^0$ and $100^0$
0%
The measures of the angles are $28^0, 66^0, 102^0$ and $165^0$
0%
The measures of the angles are $38^0, 74^0, 76^0$ and $172^0$
0%
The measures of the angles are $50^0, 72^0, 80^0$ and $150^0$

Explanation

Given,

$x , (3x-40) , 2x , (4x+20)$ .

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

$360^0$ .
Therefore,

$x + (3x-40) + 2x+ (4x+20) = 360^o$

$\Rightarrow 10x - 20^o = 360^o$

$\Rightarrow 10x = 360^o + 20^o =380^o$

$\Rightarrow x =\displaystyle \cfrac{380^o}{10}= 38^o$ .

Now,

$x = 38^o$ ,

$2x = 2 \times 38^o = 76^o$ ,

$3x - 40^o = 3 \times 38^o - 40^o =74^o$
and

$4x + 20^o = 4 \times 38^o + 20^o = 172^o$ .

Thus, the measures of the angles are

$38^o, 74^o, 76^o$ and

$172^o$ .

Hence, option

$C$ is correct.

In the figure

$\displaystyle \angle ADC$ is

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 70^{\circ}$
0%
$\displaystyle 80^{\circ}$

Explanation

Given,

$\angle AEB = 70^{\circ}$

$\angle AEB + \angle AEC = 180$

$70 + \angle AEC = 180$

$\angle AEC = 110^{\circ}$

Now, in quadrilateral AECD,

$\angle AEC+ \angle ADC + \angle DCE + \angle EAD = 360$

$110^\circ + \angle ADC + 90^\circ + 90^\circ = 360^\circ$

$\angle ADC = 70^{\circ}$

In a triangle a line is drawn from the mid-point of one side and parallel to another side then

0%
the line bisects the whole triangle
0%
bisects the third side
0%
bisects the opposite angle
0%
None of these

Explanation

From question,
In

$\Delta$ ABC, D is the mid-point of AB and DE is drawn parallel to BC and it meets at E on AC. Since, DE

$\parallel$ BC (By B.P. Theorem)

$\displaystyle \dfrac{AD}{DB} = \dfrac{AE}{EC}$ ......(i)
also, AD = DB (

$\because D\ is\ mid - point$ )

$\therefore \dfrac{AD}{DB} = 1$
Now

$\displaystyle \frac{AE}{EC} = 1$ ....(From ...(i))

$\Rightarrow$ AE = EC
Hence, the line biscets the third line
1074692_343535_ans_a440dae7aacd41bdb3aa53066961ab13.jpg

In the figure,

$\angle ADC$ is:

0%
$30^o$
0%
$60^o$
0%
$70^o$
0%
$80^o$

Explanation

Given ,

$\angle AEC = 60^o$ .

$\therefore \angle AEC =180 - 60 = 120^o\:$ ...[Linear pair].

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

$360^o$ .

$\therefore \angle ADC+ \angle DCE + \angle CEA+ \angle EAD = 360^o$

$\implies \angle ADC+ 90^o + 120^o+ 90^o = 360^o$

$\implies \angle ADC = 360^o - [120^o + 90^o + 90^o] \:$

$\implies \angle ADC = 360^o - 300^o = 60^o$ .

Hence, option

$B$ is correct.

The sum of all interior angles of a quadrilateral is

0%
$\displaystyle 90^{\circ}$
0%
$\displaystyle 180^{\circ}$
0%
$\displaystyle 270^{\circ}$
0%
$\displaystyle 360^{\circ}$

Explanation

The sum of all interior angles of a quadrilateral is

$360^{\circ}$ .

The proof is as below:

Consider a quadrilateral

$ABCD$ .

To prove,

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

To prove this, we join

$A$ and

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

$AC$ .

$\triangle ABC$ ,

$\angle CAB+\angle ABC+\angle BCA=180^{\circ}$ [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,

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

$\implies$

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

$\implies$

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

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

That is, the sum of all interior angles of a quadrilateral is

$360^o$ .

Hence, option

$D$ is correct.

Square is not a

0%
rhombus
0%
rectangle
0%
trapezium
0%
parallelogram

Explanation

$\textbf{Step-1: Comparing with trapezium}$

$\text{The definition of a square does not comply with the}$

$\text{definition of a trapezoid. The definition of a trapezoid}$

$\text{is a quadrilateral (a closed plane figure with 4 sides) with exactly one pair of parallel sides.}$

$\textbf{Step-2: Comparing with other option}$

$\text{On the other hand, a square is a very special kind of}$

$\text{quadrilateral; A square is also a parallelogram because}$

$\text{it has two sets of parallel sides and four right angles. A}$

$\text{square is also a parallelogram because opposite sides}$

$\text{are parallel. A square is always a rhombus. A rhombus}$

$\text{is a quadrilateral with four congruent sides. If the rhombus has 4}$

$\text{right angles it may also be called a square. So every square is a rhombus.}$

$\textbf{Hence , Square is not a (C) trapezium}$

The sum of angles of a quadrilateral is:

0%
$180^\circ$
0%
$360^\circ$
0%
$270^\circ$
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None of these

Explanation

$\textbf{Step-1: Dividing quadrilateral into two triangles and applying sum of angles is 180.}$

$\text{ Now in}$

$\Delta ABC,$

$\begin{align} & \angle CAB+\angle ABC+\angle BCA={{180}^{\circ }}........(1) \\ & \text{In}\Delta ACD, \\ & \angle CAD+\angle ADC+\angle DCA={{180}^{\circ }}.......(2) \\ \end{align}$

$\text{Adding (1) and (2), we get,}$

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

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

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

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

$\textbf{Therefore , sum of all angles of quadrilateral is}$

${{360}^{\circ } (Option B)}.$

$ABCD$ is a parallelogram.

$'P'$ is a point on

$AD$ such that

$AP=\displaystyle\frac{1}{3}AD\;and\;'Q'$ is a point on

$BC$ such that

$CQ=\displaystyle\frac{1}{3}BC$ . Then

$AQCP$ is a

0%
Parallelogram
0%
Rhombus
0%
Rectangle
0%
Square

Two angles of a quadrilateral are equal and the other two angles are

${ 70 }^{ o}$ and

${ 80 }^{o }$ . Then the equal angles are:

0%
${ 210 }^{\circ }$
0%
${ 105 }^{\circ }$
0%
${ 15 }^{\circ }$
0%
${ 150 }^{\circ }$

Explanation

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

Given two angles are equal.

Then $\angle A =\angle B=x$ .

Also, $\angle C = 70^{o}$ and $\angle D = 80^{o}$ .

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

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

$x + x + 70^o + 80^o = 360^{o}$

$2x+ 150^o = 360^{o}$

$\Rightarrow 2x = 360 ^o-150^o$

$\Rightarrow 2x = 210^o$

$\Rightarrow x = 105^o$ .

Hence, the equal angles are $= 105^o$ each

Therefore, option

$B$ is correct.

In the given figure,

$\angle ADC = ?$

0%
$30^o$
0%
$60^o$
0%
$70^o$
0%
$80^o$

Explanation

$\angle ADC = ?$

Given ,

$\angle AEB = 60^o$

$\therefore \angle AEC =180^o - 60^o \\= 120^o\:$

It is known that sum of interior angles angles of a quadrilateral is

$360^o$

$\therefore \angle ADC+ \angle DCE + \angle CEA+ \angle EAD = 360^o$

$\angle ADC = 360 - 120^o - 90^o - 90^o \:$ (property of quadrilateral)

$\angle ADC = 360 - 300 \\= 60^o$

What is the other name for rectangle with four equal sides?

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Rectangle
0%
Parallelogram
0%
Square
0%
Triangle

Explanation

A rectangle is made up of

$4$ sides.

When its four sides become equal, it is known as a square.

Hence, the answer is square.

A square can be a .......................

0%
triangle
0%
heptagon
0%
quadrilateral
0%
hexagon

Explanation

$\textbf{Step -1: Find the correct option.}$

$\text{A square has four sides.}$

$\Rightarrow \text{It is a quadrilateral.}$

$\textbf{Hence, the correct option is C.}$

In a quadrilateral the measures of 3 angles are

$70^o, 60^o, 90^o$ . What would be the measure of the fourth angle?

0%
$180^o$
0%
$200^o$
0%
$100^o$
0%
$140^o$

Explanation

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

The given angles are, $70^o, 60^o, 90^o$ .

Let the fourth angle be $x^o$ .

Then, $70^o+60^o+90^o+x^o=360^o$ .

$\Rightarrow$ $360^o - (70^o + 60^o + 90^o ) =x^{\circ}$

$\Rightarrow$ $x^o=360^o - 220^o =140^{\circ}$ .

Therefore, the fourth angle is $x^o=140^o$ .

Hence, option $D$ is correct.

A Rhombus is also a

0%
Parallelogram
0%
Trapezium
0%
rectangle
0%
None of these

All squares are

0%
rectangles
0%
rhombus
0%
parallelograms
0%
All of these

Explanation

(

$1$ ) Opposite sides are equal and angles are right angles so sqaure is a rectangle.
(

$2$ ) Diagonals bisect each other so square is a rhombus
(

$3$ ) Opposite sides are parallel so square is a parallelogram.
Correct answer is option (D).

The four sides of a _______ are equal.

0%
square
0%
rectangle
0%
triangle
0%
none

Identify the Quadrilateral.

0%
parallelogram
0%
rectangle
0%
square
0%
trapezium

A four sided figure with all sides equal and opposite angles also equal.

0%
kite
0%
Rhombus
0%
Parallelogram
0%
Rectangle

Opposite angles of a quadrilateral

$PQRS$ are equal. Find

$RS$ , if

$PQ = 5$ cm.

0%
$5$ cm
0%
$10$ cm
0%
$4$ cm
0%
$6$ cm

Explanation

$\angle 1 = \angle 3$
&

$\angle 1 + \angle 2 + \angle 3 + \angle 4 = 360^o$

$\therefore \angle 2 + \angle 4 = 180^o$
&

$\angle 2 = \angle 4$

$\therefore$

$PQRS$ is parallelogram
So the opposite sides are equal and parallel.

$PQ = RS = 5$ cm

The diagonals of a parallelogram

$ABCD$ intersect at

$O$ . A line through O intersect AB at X and DC at Y. Identify the correct option

0%
$OX = OY$
0%
$OX > OY$
0%
$OX < OY$
0%
$OX - OY = 0$

Explanation

$\Rightarrow$ In given figure ABCD is a parallelogram.

$\therefore$ AB

$\parallel$ DC

$\Rightarrow$ Also AC is a transversal of AB

$\parallel$ DC.

$\therefore$

$\angle1=\angle2$ [Alternate interior angles]

$\Rightarrow$ Now in

$\triangle$ AXO and

$\triangle$ CYO, we have

$\Rightarrow$

$\angle1=\angle2$ [Alternate interior angles]

$\Rightarrow$

$\angle3=\angle4$ [Vertically opposite angles]

$\Rightarrow$

$CO=OA$ [Diagonals bisect each other]

$\therefore$

$\triangle AXO \cong \triangle CYO$ [ASA Criteria]

$\therefore$

$OX=OY$ [CPCT]

In parallelogram, both pairs of opposite sides are ....................

0%
equal
0%
parallel and equal
0%
perpendicular
0%
perpendicular and equal

Explanation

$ABCD$ is a parallelogram,

$\Rightarrow$ We know that

$AB\parallel CD$

Here

$AC$ is a transversal for the parallel lines

$AB$ and

$CD$

$\Rightarrow$ So,

$\angle BAC = \angle DCA$ [Alternate interior angles are equal]

$... ( 1 )$

$\Rightarrow$ Similarly, we know that

$BC\parallel AD$

$\Rightarrow \angle BCA = \angle DAC$

$... ( 2 )$

$\triangle ABC$ and

$\triangle ADC,$

We have

$\angle BAC = \angle DCA$ [From ( 1 )]

$\Rightarrow$

$AC$ is the common side

$\Rightarrow \angle BCA = \angle DAC$ [From ( 2 )]

$\Rightarrow$ So,

$\triangle ABC \cong \triangle CDA$ [As per ASA Congruence rule]

$\therefore AB = CD$ and

$BC = AD$ [Corresponding sides of congruent triangle are equal ]

Hence, both pairs of opposite sides in a parallelogram are parallel and equal.

$X, Y$ are mid-points of opposite sides

$AB$ and

$DC$ of a parallelogram

$ABCD. AY$ and

$DX$ are joined intersecting in P;

$CX$ and

$BY$ are joined intersecting in Q. What is

$PXQY$ ?

0%
Rectangle
0%
Rhombus
0%
Parallelogram
0%
Square

Explanation

$ABCD$ is the given parallelogram

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

So,

$\dfrac{1}{2}AB=\dfrac{1}{2}CD$

$AX=CY$ ...... (as

$X$ is the mid point of

$AB$ and

$Y$ is the mid point of

$CD$ )

$AX||CY$ ......... (We know AB||CD, as

$ABCD$ is the parallelogram)

Quadrilateral

$AXCY$ is a prallelogram, as a pair of opposite sides is equal and parallel.

So,

$XC=YA$ and

$XC\parallel YA$

$XQ\parallel YP$ ......

$(i)$ (As

$XQ$ is part of line

$XC$ and

$YP$ is part of line

$YA$ )

We have

$ABCD$ as the given parallelogram

Now,

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

So,

$\dfrac{1}{2}AB=\dfrac{1}{2}CD$

$DY=XB$ ....... (As

$X$ is the mid point of

$AB$ and

$Y$ is the mid point of

$CD$ )

$DY\parallel XB$ ..... (We know

$AB \parallel CD$ , as

$ABCD$ is the parallelogram)

Quadrilateral

$DXBY$ is a prallelogram, as a pair of opposite sides is equal and parallel.

So,

$DX=YB$ and

$DX||YB$

$PX||YQ$ ......

$(ii)$ (As

$XP$ is part of line

$XD$ and

$YQ$ is part of line

$YB$ )

From

$(i)$ and

$(ii)$ we get,

$PXQY$ is also a parallelogram.

In quadrilateral

$ABCD$ , if

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

$\angle B : \angle C : \angle D = 2 : 3 : 7$ , then find

$\angle D$ .

0%
$175^{\circ}$
0%
$135^{\circ}$
0%
$150^{\circ}$
0%
$120^{\circ}$

Explanation

In quadrilateral

$ABCD$ ,

$\angle A$ =

$60^\circ$ and

$\angle B: \angle C: \angle D = 2 : 3 : 7$ [Given].

Let

$\angle B, \angle C, \angle D$ are

$2x^\circ$ ,

$3x^\circ$ and

$7x^\circ$ .

$\Rightarrow$

$\angle A+ \angle B+ \angle C + \angle D$ =

$360^\circ$ [Sum of all angles of quadrilateral is

$360$

$^\circ$ ]

$\Rightarrow$

$60^\circ + 2x^\circ+ 3x^\circ +7x^\circ$ =

$360^\circ$

$\Rightarrow$

$12x^\circ$ =

$360^\circ - 60^\circ$

$\Rightarrow$

$x^\circ$ =

$\dfrac {300^\circ}{12}$

$\Rightarrow$

$x^\circ$ =

$25^\circ$ .

Now,

$\angle D$ =

$7x^\circ$ .

Substitute value of

$x$ we get,

$\Rightarrow$

$\angle D$ =

$7\times 25^o$

$\therefore$

$\angle D = 175^\circ$ .

Hence, option

$A$ is correct.

Which statement(s) is/are correct?
(1)The diagonals of a parallelogram are equal.
(2)The diagonals of a square are perpendicular to each other.
(3)If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus.
(4)Every quadrilateral is either a trapezium or a parallelogram or a kite.

0%
Only (2)
0%
Only (3)
0%
Both (2) and (3)
0%
(1), (2) and (3)

Explanation

$\Rightarrow$ In four given statement we have given that, the diagonals of a parallelogram are equal which is not correct because we know that in properties of parallelogram it has opposite sides parallel but it's diagonals are not equal.

$\Rightarrow$ Second statement says, the diagonals of a square are perpendicular to each other which is true because we know that square has all four sides equal, opposite sides parallel to each other and both diagonals are equal and perpendicular to each other.

$\Rightarrow$ Third statement says that, if the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus, which is correct because the diagonals of a square also bisect each other at right angles.

$\Rightarrow$ Fourth statement says that, every quadrilateral is either a trapezium or a parallelogram or a kite which is wrong because rectangle, rhombus and square are also types of quadrilateral.

$\Rightarrow$ So correct statements are statement ( 2) and (3) then correct option is option C.

ABCD is a quadrilateral such that

$AB = BC$ ,

$AD= \cfrac{1}{2} CD$ and

$AD = \cfrac{1}{4} AB$ . If

$BC =12$ cm, what is the measures of AD?

0%
$6$ cm
0%
$4$ cm
0%
$12$ cm
0%
$3$ cm

Explanation

$\Rightarrow$ In quadrlateral ABCD,

$\Rightarrow$ AB = BC [Given] --- ( 1 )

$\Rightarrow$ BC = 12cm. [Given]

$\Rightarrow$ AB = 12cm. [From ( 1 )]

$\Rightarrow$ It is given that,

$AD=\dfrac{1}{4} AB$

$\Rightarrow$

$AD=\dfrac{1}{4} \times 12$ [Substituting value of AB]

$\therefore$

$AD=3cm.$

A quadrilateral has three acute angles, each measuring

$75^o$ . Find the measure of the fourth angle.

0%
$65^o$
0%
$135^o$
0%
$140^o$
0%
$225^o$

Explanation

Let the four angles be

$\angle A , \angle B , \angle C$ and

$\angle D$ .

Given

$\angle A , \angle B , \angle C = 75^{o}$ .

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:

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

$\Rightarrow \angle D = 360 ^o-75^o -75^o -75^o$

$\Rightarrow \angle D = 360 - 225^o$

$\Rightarrow \angle D = 135^o$ .

The measure of fourth angle is

$135 ^{o}$ .

Therefore, option

$B$ is correct.

The angles of a quadrilateral are in the ratio

$3:5:9:13$ . Then angles of the quadrilateral respectively are ___________.

0%
$36^o, 60^o, 108^o, 156^o$
0%
$30^o, 60^o, 45^o, 15^o$
0%
$45^o, 20^o, 30^o, 60^o$
0%
$30^o, 90^o, 60^o, 20^o$

Explanation

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

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

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

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

$\Rightarrow 30x=360^o$

$\therefore x=12^o$ .

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

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

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

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

Hence, option $A$ is correct.

In a quadrilateral, the angles are

$x^{\circ}, (x + 10)^{\circ}, (x + 20)^{\circ}, (x + 30)^{\circ}$ . Find the value of

$x$ .

0%
$175^{\circ}$
0%
$275^{\circ}$
0%
$375^{\circ}$
0%
$75^{\circ}$

Explanation

The given quadrilateral angles are

$x^o,(x+10)^o,(x+20)^o$ and

$(x+30)^o$ .

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

$360^o$ .

$\therefore$

$x^o+(x+10)^o+(x+20)^o+(x+30)^o=360^o$

$\Rightarrow$

$4x^o+60^o=360^o$

$\Rightarrow$

$4x^o=300^o$

$\Rightarrow$

$x^o=75^o$ .

$\therefore$ The value of

$x$ is

$75^o$ .

Hence, option

$D$ is correct.

In a quadrilateral

$ABCD$ , the angles

$\angle A, \angle B, \angle C$ and

$\angle D$ are in the ratio

$2 : 3 : 4 : 6$ . Find the measure of each angle of the quadrilateral.

0%
$48^{\circ}, 72^{\circ}, 96^{\circ}, 144^{\circ}$
0%
$38^{\circ}, 72^{\circ}, 96^{\circ}, 144^{\circ}$
0%
$48^{\circ}, 22^{\circ}, 96^{\circ}, 144^{\circ}$
0%
None of these

Explanation

Let the angles of quadrilateral

$ABCD$ are

$2x,3x,4x$ and

$6x$ .

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

$360^o$ .

$\therefore$

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

$\Rightarrow$

$2x+3x+4x+6x=360^o$

$\Rightarrow$

$15x=360^o$

$\therefore$

$x=24^o$ .

$\Rightarrow$

$\angle A=2x=2\times 24^o=48^o$

$\Rightarrow$

$\angle B=3x=3\times 24^o=72^o$

$\Rightarrow$

$\angle C=4x=4\times 24^o=96^o$

$\Rightarrow$

$\angle D=6x=6\times 24^o=144^o$ .

Therefore, option

$A$ is correct.

The sum of pairs of opposite angles of a cyclic quadrilateral is:

0%
${ 90 }^{ o }$
0%
${ 180 }^{ o }$
0%
${ 270 }^{ o }$
0%
${ 360 }^{ o }$

Explanation

Sum of pair of opposite angles of a cyclinc quadrilateral is

${ 180 }^{ o }$

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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