MCQ Questions for Class 9 Maths Quadrilaterals Quiz 3

One angle of a quadrilateral is of

$108^o$ and the remaining three angles are equal. Find each of the three equal angles.

0%
$64^o$
0%
$74^o$
0%
$84^o$
0%
$94^o$

Explanation

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

Given, one angle is $108^o$ and remaining three angles are equal.

Let the remaining angles be $x^o,y^o,z^o$ .

Since they are equal, $x^o+y^o+z^o=$ $x^o+x^o+x^o=3x^o$ .

Then, $108^o+x^o+y^o+z^o=360^o$

$\Rightarrow$ $108^o+3x^o=360^o$

$\Rightarrow$ $360^o - 108^o =3x^{\circ}$

$\Rightarrow$ $252^o =3x^{\circ}$

$\Rightarrow$ $x^o=\dfrac{252^o}{3} =84^{\circ}$ .

Therefore, each of the three remaining angle is $x^o=84^o$ .

Hence, option $C$ is correct.

If bisectors of

$\angle A$ and

$\angle B$ of a quadrilateral ABCD intersect each other at

$P$ , bisectors of

$\angle B$ and

$\angle C$ at

$Q$ , bisectors of

$\angle C$ and

$\angle D$ at

$R$ and bisectors of

$\angle D$ and

$\angle A$ at

$S$ , then PQRS is a

0%
rectangle
0%
rhombus
0%
parallelogram
0%
quadrilateral whose opposite angles are supplementary

Explanation

From question the diagram we get is above.

From figure,

$\angle QPS = \angle APB$ (Vertically opposite angles) ---- (1)

Consider,

$\triangle APB$

$\angle APB + \dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B = 180^\circ$

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

From equation (1) & (2),

$\angle QPS = 180^\circ - \dfrac{1}{2} (\angle A + \angle B)$ ----- (3)

Similarly,

$\angle QRS = 180^\circ$ -

$\dfrac{1}{2} (\angle C + \angle D)$ ---- (4)

From (3) &(4),

$\angle QPS + \angle QRS = 360^\circ$ -

$\dfrac{1}{2} (\angle A + \angle B + \angle C + \angle D)$

$= 360^\circ - \dfrac{1}{2}(360^\circ)$

$= 360^\circ - 180^\circ$

$\therefore$

$\angle$ QPS +

$\angle$ QRS =

$180^\circ$

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Three angles of a quadrilateral are

$75^{\circ}, 90^{\circ}$ and

$75^{\circ}$ . The measure of fourth angle is?

0%
$90^{\circ}$
0%
$95^{\circ}$
0%
$105^{\circ}$
0%
$120^{\circ}$

Explanation

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

$360^o$ .

The given angles are, $75^o, 90^o, 75^o$ .

Let the fourth angle be

$x^o$ .
Then,

$75^o+90^o+75^o+x^o=360^o$ .

$\Rightarrow$

$360^o - (75^o + 90^o + 75^o ) =x^{\circ}$

$\Rightarrow$

$x^o=360^o - 240^o =120^{\circ}$ .

Therefore, the fourth angle is

$x^o=120^o$ .

Hence, option

$D$ is correct.

$D$ and

$E$ are the mid-points of the sides

$AB$ and

$AC$ of

$ABC$ and

$O$ is any point on side

$BC. O$ is joined to

$A$ . If

$P$ and

$Q$ are the mid-points of

$OB$ and

$OC$ respectively, then

$DEQP$ is:

0%
a square
0%
a rectangle
0%
a rhombus
0%
a parallelogram

Explanation

Given:

$ABC$ is a triangle,

$D$ &

$E$ are the respective mid points of

$AB$ &

$AC$ .

$O$ is any point on

$BC$ ,

$P$ &

$Q$ are mid points of

$OB$ &

$OC$ respectively.

To find:

Type of quadrilateral of

$DEQP$ .

Solution:

$\Delta ABC, D \&E$ are the respective mid points of

$AB$ &

$AC.$

$\therefore DE\parallel (BC or PQ)$ ....(1)

Again in

$\Delta ABO, P \&D$ are the respective mid points of

$AB$ &

$BO$

$\therefore DP\parallel AO$ ... (2)

Similarly, in

$\Delta ACO, EQ\parallel AO$ ...(3)

$\therefore$ From (2) & (3), we get

$DP\parallel EQ$ ... (4)

$\therefore$ From (1) & (4), we have

$DE\parallel PQ$ and

$DP\parallel EQ$

So,

$DEQP$ is parallelogram.

Hence, option D.

Can all the angles of a quadrilateral be acute angles?

0%
Yes
0%
No
0%
May be
0%
Cannot be determined

Explanation

Let us take a qudrilateral

$ABCD$ with

$BD$ as diagonal.

$\Delta ABD$ we have,

$\angle ADB+\angle ABD+\angle DAB={ 180 }^{ O }$ .......

$(i)$

And in

$\Delta DBC$ we have

$\angle DCB+\angle CBD+\angle BDC={ 180 }^{ O }$ ......

$(ii)$

Adding

$(i)$ &

$(ii)$ , we get,

$\angle ADB+\angle ABD+\angle DAB+\angle DCB+\angle CBD+\angle BDC={ 360 }^{ O }$

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

So, the sum of the angles of a quadrilateral is

${ 360 }^{ O }$ .....

$(iii)$ .

Now, if the angles are acute then

$\angle A<{ 90 }^{ O }, \angle B<{ 90 }^{ O }, \angle C<{ 90 }^{ O }, \angle D<{ 90 }^{ O }$

$\Longrightarrow \angle A+\angle B+\angle C+\angle D<{ 90 }^{ O }+{ 90 }^{ O }+{ 90 }^{ O }+{ 90 }^{ O }$

$\Longrightarrow \angle A+\angle B+\angle C+\angle D<{ 360 }^{ O }.$

But this does not comply with

$(iii)$ .

Therefore, all the angles of a quadrilateral cannot be acute angles.

Hence, option

$B$ is correct.

Can the angles

$110^o, 80^o, 70^o$ and

$95^o$ be the angles of a quadrilateral?

0%
Yes
0%
No
0%
May be
0%
Cannot be determined

Explanation

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

The given angles are $110^o, 80^o, 70^o, 95^o$ .

Then,

$110^o+80^o+70^o+95^o$ .

$=190^o + 165^o$

$=355^o \ne 360^o$ .

Therefore, the given angles are not the angles of a quadrilateral.

Hence, option $B$ is correct.

All squares are rhombuses and also rectangles.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

Explanation

The diagonals of square bisect each other at

$90^{\circ}$ , same as rhombus. Hence, all squares are rhombuses.
The adjacent sides of square are perpendicular to each and the opposite sides are equal and parallel to each other, same as rectangles. Hence, all squares are rectangles too.

Hence, the given statement is

$True$ .

In a quadrilateral three angles are in the ratio

$3 : 3 : 1$ and one of the angles is

$80^{\circ}$ , then other angles are :

0%
$120^{\circ},120^{\circ},40^{\circ}$
0%
$100^{\circ},100^{\circ},80^{\circ}$
0%
$110^{\circ},110^{\circ},60^{\circ}$
0%
$90^{\circ},90^{\circ},30^{\circ}$

Explanation

In figure

$ABCD$ is a quadrilateral.

We have given ratio of three angle

$3 : 3 : 1$ .

Let

$\angle A= 3x, \angle B=3x, \angle C=x$ and

$\angle D=80^\circ$ .

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

$360^\circ$ ]

$\Rightarrow$

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

$\Rightarrow$

$3x+3x+x+80^\circ = 360^\circ$

$\Rightarrow$

$7x =280^\circ$

$\therefore$

$x=40^\circ = \angle C$ . ---- ( 1 )

$\Rightarrow$

$\angle A=3x=3\times 40^\circ = 120^\circ$ -- ( 2 )

$\Rightarrow$

$\angle B=3x=3\times 40^\circ =120^\circ$ -- ( 3 )

From ( 1 ), ( 2 ) and ( 3 ),

$\Rightarrow$ Other angles are

$120^\circ, 120^\circ, 40^\circ$ .

Hence, option

$A$ is correct.

Which of the following is not a parallelogram ?

0%
Rhombus
0%
Rectangle
0%
Trapezium
0%
Square

State true or false:

All squares are not parallelograms.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

State true or false:

All parallelograms are trapeziums.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B, and C. (The dotted lines are drawn additionally to help you).

0%
True
0%
False

Explanation

Between

$\Delta$ AOD &

$\Delta$ BOC we have AO=CO (given),BO=OD (by Construction).

$\angle$ AOD

$=\angle$ BOC....Vertically opposite angle

$\therefore$ By SAS test

$\Delta$ AOD &

$\Delta$ BOC are congruent.

So AD=BC....(i)

similarly, between

$\Delta$ AOB &

$\Delta$ DOC we have AO=CO (given), BO=OD (by Construction)

$\angle$ AOB

$=\angle$ DOC

$\therefore$ By SAS test

$\Delta$ AOB &

$\Delta$ DOC are congruent.

So AB=DC.....(ii)

Also

$\angle$ ABC

$={ 90 }^{ o }$ ....(iii)

$\therefore$ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.

So the diagonals AC & BD are equal and bisect each other at O.

$\therefore$ OA=OB=OC=OD.

i.e O is equidistant from A, B & C.

State true or false:

All kites are rhombuses.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

State true or false:

Every rhombus is a kite.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

ABCD is a quadrilateral whose diagonal AC divides it into two parts equal in area, then ABCD is :

0%
always a rhombus
0%
a parallelogram
0%
a kite
0%
a trapezium

Explanation

A diagonal of a parallelogram divides it into two congruent triangles and we know congruent figures have equal area.

$\therefore$ Such a quadrilateral will be a parallelogram.

Hence, the answer is a parallelogram.

If three angles of a quadrilateral are

$70^{\circ}, 95^{\circ}$ and

$105^{\circ}$ , then the fourth angle is :

0%
$90^{\circ}$
0%
$100^{\circ}$
0%
$110^{\circ}$
0%
$85^{\circ}$

Explanation

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

The given angles are, $70^o, 95^o, 105^o$ .

Let the fourth angle be $x^o$ .

Then, $70^o+95^o+105^o+x^o=360^o$ .

$\Rightarrow$ $360^o - (70^o + 95^o + 105^o ) =x^{\circ}$

$\Rightarrow$ $x^o=360^o - 270^o =90^{\circ}$ .

Therefore, the fourth angle is $x^o=90^o$ .

Hence, option $A$ is correct.

Fill in the blank:

Line joining the mid-points of any two sides of a triangle is _____ to the third side.

0%
perpendicular
0%
parallel
0%
Inclined at $60^o$
0%
None of these

State true or false:

If the diagonals of a quadrilateral intersect at right angles then the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rhombus.

0%
True
0%
False

Explanation

Given: ABCD is a quadrilateral. AC and BD intersect at right angles. P, Q, R, S are mid points of respective sides.
In

$\triangle ABC$ ,

$PQ \parallel AC$ and

$PQ = \frac{1}{2} AC$ (By mid point theorem)
In

$\triangle ADC$ ,

$SR \parallel AC$ , and

$SR = \frac{1}{2} AC$ (By mid point theorem)
Therefore,

$PQ = SR$ and

$PQ \parallel SR$
Similarly, using diagonal

$DC$ ,

$PS = RQ$ and

$PS \parallel RQ$
hence, PQRS is a parallelogram.
Since,

$PQ \parallel AC$ and

$QR \parallel BD$
But,

$AC \perp BD$
Hence,

$PQ \perp QR$ (Angles between two lines = Angles between their parallels)
Thus,

$PQRS$ is a rectangle.

In a quadrilateral PQSR, with a diagonal PS, if

$QS = SR$ and

$\angle QSP = \angle RSP$ , then consider the following statements
Assertion (A) :

$\angle QPS = \angle RPS$
Reason (R) : Triangles QPS and RPS are congruent.Of these statements

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not a correct explanation of A
0%
A is true, but R is false
0%
A is false, but R is true

Explanation

$\triangle QPS$ and

$\triangle RPS$ ,

$\Rightarrow$

$QS=SR$ [ Given ]

$\Rightarrow$

$QSP=\angle RSP$ [ Given ]

$\Rightarrow$

$SP=PS$ [ Common side ]

$\Rightarrow$

$\triangle QPS\cong\triangle RPS$ [ By SAS Congruence rule ]

$\Rightarrow$

$\angle QPS=\angle RPS$ [ CPCT ]

$\therefore$ Both assertion and reason are correct and reason is correct explanation of assertion.

Complete the following statement.

The straight lines which join the middle points of opposite sides of a quadrilateral ,

0%
Are parallel to one another
0%
Bisect one another
0%
Trisect one another
0%
None of these

Three angles of a quadrilateral are

$60^{\circ}, 110^{\circ}$ and

$86^{\circ}$ , the fourth angle of the quadrilateral is :

0%
$104^{\circ}$
0%
$124^{\circ}$
0%
$94^{\circ}$
0%
$84^{\circ}$

Explanation

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

The given angles are, $60^o, 110^o, 86^o$ .

Let the fourth angle be $x^o$ .

Then, $60^o+110^o+86^o+x^o=360^o$ .

$\Rightarrow$ $360^o - (60^o + 110^o + 86^o ) =x^{\circ}$

$\Rightarrow$ $x^o=360^o - 256^o =104^{\circ}$ .

Therefore, the fourth angle is $x^o=104^o$ .

Hence, option $A$ is correct.

Angles of a quadrilateral are in the ratio

$3 : 6 : 8 : 13$ . The largest angle is :

0%
$178^{\circ}$
0%
$90^{\circ}$
0%
$156^{\circ}$
0%
$36^{\circ}$

Explanation

Given ratio of angles of quadrilateral

$ABCD$ is

$3:6:8:13$

Let the angles of quadrilateral

$ABCD$ be

$3x,6x,8x$ ,and

$13x$ , respectively.

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

$360^o$ .

$\Rightarrow 3x+6x+8x+13x=360^o$

$\Rightarrow 30x=360^o$

$\Rightarrow x=\dfrac{360^o}{12}$

$\therefore x=12^o$ .

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

$\angle B =6x=6\times 12^o =72^o$ ,

$\angle C =8x=8 \times 12^o =96^o$

and

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

$\therefore$ The largest angle

$=156^o$ .

Hence, option

$C$ is correct.

Three angles of a quadrilateral are

$70^{\circ}, 120^{\circ}$ and

$65^{\circ}$ . Then the fourth angle of the quadrilateral is:

0%
$95^{\circ}$
0%
$75^{\circ}$
0%
$105^{\circ}$
0%
$90^{\circ}$

Explanation

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

The given angles are, $70^o, 120^o, 65^o$ .

Let the fourth angle be $x^o$ .

Then, $70^o+120^o+65^o+x^o=360^o$ ...{Angle sum property of a quadrilateral}

$\Rightarrow$ $360^o - (70^o + 120^o + 65^o ) =x^{\circ}$

$\Rightarrow$ $x^o=360^o - 255^o =105^{\circ}$ .

Therefore, the fourth angle is $x^o=105^o$ .

Hence, option $B$ is correct.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the _______ side.

0%
first
0%
second
0%
third
0%
none of the above

The diagonals of a quadrilateral ABCD are perpendicular. The quadrilateral formed by joining the mid points of its sides, is ?

0%
Trapezium
0%
Rectangle
0%
Square
0%
Rhombus

Explanation

From above figure,

AC, BD are diagonals of quadrilateral ABCD are perpendicular.

Points P, Q, R and S are mid-points of AB and BC respectively.

$\triangle$ ABC,

P and Q are mid points of AB and BC.

So, by Mid-point theorem,

$\therefore$ PQ

$\left|\right|$ AC; PQ =

$\dfrac{1}{2}$ AC -------- (1)

Similarly in

$\triangle$ ACD,

R and S are mid point of sides CD and AD.

$\therefore$ SR

$\left|\right|$ AC; SR =

$\dfrac{1}{2}$ AC ----------- (2)

From (1) and (2), we get

$\left|\right|$ SR and PQ = SR

Hence, PQRS is parallelogram -------( pair of opposite sides are equal and parallel)

Now,RS

$\left|\right|$ AC and QR

$\left|\right|$ BD.

Also, AC

$\perp$ BD ---- (Given)

$\therefore$ RS

$\perp$ QR.

Thus, PQRS is a rectangle.

In a quadrilateral

$ABCD$ , if

$\angle A = 80^{\circ},\angle B=70^{\circ}, \angle C=130^{\circ} ,$ then

$\angle D$ is:

0%
$80^{\circ}$
0%
$70^{\circ}$
0%
$130^{\circ}$
0%
$150^{\circ}$

Explanation

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

The given angles are $\angle A=80^o, \angle B=70^o, \angle C=130^o, \angle D$ .

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

$\implies$ $80^o+70^o+130^o+\angle D=360^o$ .

$\Rightarrow$ $\angle D=360^o - (80^o + 70^o + 130^o )$

$\Rightarrow$ $\angle D=360^o - 280^o =80^{\circ}$ .

Therefore, the unknown angle is $\angle D=80^o$ .

Hence, option $A$ is correct.

State true or false:

In the figure, given below,

$2 AD= AB$ .

$P$ is mid-point of

$AB$ .

$Q$ is mid-point of

$DR$ and

$PR \parallel BS$ , then

$AQ \parallel BS$ .

0%
True
0%
False

Explanation

Given:

$PR \parallel BS$ ,

$P$ is mid point of

$AB$ and

$Q$ is mid point of

$DR$ .
Also,

$2 AD = AB$

$\implies$

$2 AD = AP + PB$ .......(

$AB=AP+PB$ )

$\implies$

$2 AD = 2 AP$ .......(

$P$ is mid point of

$AB$ , i.e.

$AP=PB$ )

$\implies$

$AD = AP$
That is,

$A$ is mid point of

$PD$ .

Now, In

$\triangle DPR$ ,

$A$ is mid point of

$PD$ and

$Q$ is mid point of

$DR$ .

Thus, by converse of mid point theorem,

$AQ \parallel PR$ .

Since

$PR \parallel BS$ ,
hence,

$AQ \parallel PR \parallel BS$ .

Therefore, the given statement is true and option

$A$ is correct.

State true or false:

In the given figure,

$AD$ and

$CE$ are medians and

$DF \parallel CE$ , then

$FB= \displaystyle \dfrac{1}{4} AB$ .

0%
True
0%
False

Explanation

Given:

$D$ and

$E$ are mid points of

$BC$ and

$AB$ , respectively (since

$AD$ and

$CE$ are medians) and

$DF \parallel CE$ .

$\triangle BEC$ ,

$FD \parallel CE$ and

$D$ is mid point of

$BC$ .
Then, by converse of mid point theorem,

$F$ is mid point of

$BE$ .
That is,

$FB = \dfrac{1}{2} (BE)$

$\implies$

$FB = \dfrac{1}{2} (\dfrac{1}{2} AB)$ ......(

$E$ is mid point of

$AB$ , i.e.

$BE=\dfrac{1}{2}AB$ )

$\implies$

$FB = \dfrac{1}{4} AB$ .

Hence, the given statement is true and option

$A$ is correct.

In triangle

$ABC$ ,

$M$ is mid-point of

$AB$ and a straight line through

$M$ and parallel to

$BC$ cuts

$AC$ in

$N$ . Find the lenghts of

$AN$ and

$MN$ if

$BC= 7$ cm and

$AC= 5$ cm.

0%
$AN= 2.5$ cm and $MN= 3.5$ cm
0%
$AN= 1.5$ cm and $MN= 3.5$ cm
0%
$AN= 2.5$ cm and $MN= 4.5$ cm
0%
none of the above

Explanation

M is the mid point of AB and MN II BC. Thus, N is the mid point of AC and

$MN = \dfrac{1}{2} BC$ (Mid point theorem)

$MN = \dfrac{1}{2} (7)$

$MN = 3.5 cm$

Also,

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

$AN = \dfrac{1}{2} (5)$

$AN = 2.5 cm$

A triangle

$ABC$ in which

$AB=AC,$

$M$ is a point on

$AB$ and

$N$ is a point on

$AC$ such that if

$BM=CN$ then

$AM=AN.$

0%
True
0%
False

Explanation

Since $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC,$ we have

$AB = AM+ MB$ and $AB = AN + NC$

So,

$\begin{aligned}{}AB&=AC\quad\quad\quad\quad\quad\quad[\text{Given}]\\AM + BM& = AN + CM\\AM+BM&=AN+BM\quad\quad\quad[\text{Given }BM=CN]\\AM &= AN\end{aligned}$

Hence proved, $AM=AN.$

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 3 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 3 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.