Match the following. (1) Rectangle (p) A quadrilateral having its opposite sides equal and parallel. (2) Square (q) A parallelogram having its opposite sides equal and each of the angle is a right angle. (3) Parallelogram (r) A parallelogram having all sides equal and each of the angle is a right angle. (4) Rhombus (s) A quadrilateral in which a pair of opposite sides are parallel. (5) Trapezium (t) A parallelogram having all sides equal.

$$1\rightarrow (q), 2\rightarrow (r), 3\rightarrow (p), 4\rightarrow (t), 5\rightarrow (s)$$

$$1\rightarrow (t), 2\rightarrow (s), 3\rightarrow (r), 4\rightarrow (p), 5\rightarrow (q)$$

$$1\rightarrow (p), 2\rightarrow (q), 3\rightarrow (r), 4\rightarrow (s), 5\rightarrow (t)$$

$$1\rightarrow (r), 2\rightarrow (q), 3\rightarrow (t), 4\rightarrow (p), 5\rightarrow (s)$$

MCQ Questions for Class 9 Maths Quadrilaterals Quiz 7

All squares are not parallelograms.

0%
True
0%
False

Explanation

$\Rightarrow$ A parallelogram is a quadrilateral with opposite sides parallel and sum of its adjacent sides are supplementary.

$\Rightarrow$ In square opposite sides are parallel and measure of each angle is

$90^o$ , so sum of its adjacent also supplementary.

$\Rightarrow$ So, square contains all the properties of a parallelogram.

$\therefore$ All squares are parallelogram.

So, given statement is false.

Which of the following group is true for quadrilateral

$ABCD$ ?

(1) $ABCD$ is a Rhombus	(a) $\overline { AC }$ and $\overline { BD }$ bisect each other.
(2) $ABCD$ is a Parallelogram	(b) $\overline { AC }$ and $\overline { BD }$ bisect each other at an right angle.
(3) $ABCD$ is Rectangle	(c) $\overline { AC }$ and $\overline { BD }$ are congruent and bisect each other at an right angle.
(4) $ABCD$ is Square	(d) $\overline { AC }$ and $\overline { BD }$ are congruent and bisect each other.

0%
$1-d,2-a,c-d,4-c$
0%
$1-c,2-d,3-a,4-b$
0%
$1-b,2-a,3-d,4-c$
0%
$1-b,2-c,3-d,4-a$

Explanation

Diagonals of rhombus

$ABCD$ intersect at right angles.

$1 \rightarrow b$

Diagonals of parallelogram

$ABCD$ bisect each other.

$2\rightarrow a$

Diagonals of rectangle

$ABCD$ are equal/congruent and bisect each other,

$3\rightarrow d$

Diagonals of square

$ABCD$ bisect at right angles and are equal/congruent.

$4\rightarrow c$

All squares are trapeziums.

0%
True
0%
False

Explanation

$\Rightarrow$ A trapezium is a quadrilateral with at least one pair of parallel sides.

$\Rightarrow$ In a square, there are always two pairs of parallel sides, so every square is trapezium.

$\therefore$ All squares are trapeziums.

1264251_703201_ans_48ca2b911d444ed3890fae6e10ad37a8.jpeg

Match the following.

(1)	Rectangle	(p)	A quadrilateral having its opposite sides equal and parallel.
(2)	Square	(q)	A parallelogram having its opposite sides equal and each of the angle is a right angle.
(3)	Parallelogram	(r)	A parallelogram having all sides equal and each of the angle is a right angle.
(4)	Rhombus	(s)	A quadrilateral in which a pair of opposite sides are parallel.
(5)	Trapezium	(t)	A parallelogram having all sides equal.

0%
$1\rightarrow (t), 2\rightarrow (s), 3\rightarrow (r), 4\rightarrow (p), 5\rightarrow (q)$
0%
$1\rightarrow (p), 2\rightarrow (q), 3\rightarrow (r), 4\rightarrow (s), 5\rightarrow (t)$
0%
$1\rightarrow (r), 2\rightarrow (q), 3\rightarrow (t), 4\rightarrow (p), 5\rightarrow (s)$
0%
$1\rightarrow (q), 2\rightarrow (r), 3\rightarrow (p), 4\rightarrow (t), 5\rightarrow (s)$

Explanation

$(1)$ Rectangle - A parallelogram having its opposite sides equal and each of the angle is a right angle.

$(2)$ Square - A parallelogram having its opposite sides equal and each of the angle is a right angle.

$(3)$ Parallelogram - A quadrilateral having its opposite sides equal and parallel.

$(4)$ Rhombus - A parallelogram having all sides equal.

$(5)$ Trapezium - A quadrilateral in which a pair of opposite sides are parallel.

$\therefore$

$1\rightarrow (q),2\rightarrow (r),2\rightarrow (p), 4\rightarrow (t),5\rightarrow (s)$

All parallelograms having equal areas have same perimeters.

0%
True
0%
False

State whether the statements are true (T) or (F) false.
If diagonals of a quadrilateral bisect each other , it must be a parallelogram.

0%
True
0%
False

In the adjoining figure, if

$\angle A =90^o$ , then find

$x$ .

0%
$11$
0%
$19$
0%
$13$
0%
$20$

Explanation

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

The given angles are $\angle A=90^o, \angle B=6x-5^o, \angle C=7x-15^o, \angle D=2x+5^o$ .

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

$\implies$ $90^o+6x-5^o+7x-15^o+2x+5^o=360^o$

$\implies$ $15x+75^o=360^o$

$\implies$ $15x=360^o-75^o=285^o$

$\implies$ $x=19^o$ .

Therefore, option $B$ is correct.

Identify the true statements :-

0%
A rectangle is a parallelogram.
0%
A square is a rhombus
0%
A parallelogram is a rhombus
0%
A kite is a parallelogram

Explanation

$A$ is correct because the rectangle has equal parallel sides.

$B$ is correct because a square has all sides equal.

$C$ is incorrect because parallelograms do not have all sides equal as in rhombus.

$D$ is incorrect because kites do not have equal parallel side pairs.

The angles of a quadrilateral are in the ratio

$1:2:3:4$ . The largest angle is:

0%
$36^{o}$
0%
$72^{o}$
0%
$108^{o}$
0%
$144^{o}$

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 largest angle $=144^o$ .

Hence, option $D$ is correct.

Read the following statement and choose the correct alternative from those given below them:
(i) Diagonals of a rectangle are perpendicular bisectors of each other
(ii) Diagonals of a rhombus are perpendicular bisectors of each other
(iii) Diagonals of a parallelogram are perpendicular bisectors of each other

0%
Statement $(ii)$ and $(iii)$ are true
0%
Only statement $(ii)$ is true
0%
Statement $(ii)$ and $(iv)$ are true
0%
Statement $(i),(ii)$ and $(iv)$ are true

Explanation

$\Rightarrow$ Diagonals of a rectangle are equal in length but does not bisect each other perpendicularly. So, first statement is false.

$\Rightarrow$ Not all types of parallelogram diagonals bisect each other perpendicularly, so third statement is also false.

$\Rightarrow$ Diagonals of rhombus bisect each other perpendicularly.

So, only second statement is correct.

In a quadrilateral

$ABCD$ ,

$\angle{A}+\angle{C}$ is

$2$ times of

$\angle{B}+\angle{D}$ . If

$\angle {D}={40}^{o}$ , then

$\angle {B}=$ _____.

0%
${60}^{o}$
0%
${80}^{o}$
0%
${100}^{o}$
0%
${120}^{o}$

Explanation

Given,

$\angle A+\angle C=2\left(\angle B+\angle D\right)$ .

Also, by angle sum property,

$\angle A+\angle B+\angle C+\angle D={360}^{0}$

$\implies$

$\angle A+\angle C+\angle B+\angle D={360}^{0}$

$\implies$

$2\left(\angle B+\angle D \right)+ \left(\angle B+\angle D \right)= {360}^{0}$

$\implies$

$3\left(\angle B+\angle D \right)= {360}^{0}$

$\implies$

$\angle B+\angle D={120}^{0}$

$\implies$

$\angle B+{40}^{0}={120}^{0}$

$\implies$

$\angle B={80}^{0}$ .

Therefore, option

$B$ is correct.

The line

$3x+2y=6$ will divide the quadrilateral formed by the lines

$x+y=5,y-2x=8,3xy+2x=0$ and

$4y-x=0$ in

0%
two quadrilaterals
0%
one pentagon and one triangle
0%
two traingles
0%
one triangle and one quadrilateral

Explanation

$\begin{array}{l} we\, \, have........... \\ x+y=5\, \, \, \, \, \, (draw\, \, a\, line\, \, on\, \, x-axis\, and\, y-axis\, as\, \, \, (0,5)\, \, \, (5,0) \\ y-2x=8\, \, \, \, (draw\, \, a\, line\, \, on\, \, x-axis\, and\, y-axis\, as\, \, \, (-4,0)\, \, \, (0,8) \\ 3y+2x=0\Rightarrow y=\frac { { 2x } }{ 3 } ,\, \\ 4y-x=0\Rightarrow y=\frac { x }{ 4 } ,\, \, \, \, \end{array}$

we get the quadrilateral point ABCD,

and we find out the

$3x + 2y = 6$

$\begin{array}{l} 3x+2y=6-----(i) \\ equ\, \, (i)\, \, divide\, by\, \, 6 \\ \frac { x }{ 2 } +\frac { y }{ 3 } =1\, \, \, then,\, we\, get\, two\, quadrilateral\, \, as\, (AFED,EFBC) \\ So,\, that\, the\, \, \, correct\, option\, is\, A. \end{array}$

1173808_1187710_ans_bb310b2c22b346ba8dfe889221b422b0.jpg

$A B C D$ is a quadrilateral. If

$\angle A C B = \angle A D B ,$ then

$A B C D$ is

0%
A cyclic quadrilateral
0%
A parallelogram
0%
A square
0%
Not a cyclic quadrilateral

Explanation

$\angle ACB=\angle ADB$ ,

Then the angles in the same segment are equal.

i.e.,

$A, B, C, D$ are concyclic.

Hence ABCD is a cyclic quadrilateral.

1189527_1258661_ans_56a4712e0ec04059b9a3cf7300ded9fc.png

X, Y are the mid-points of opposite sides AB and DC respectively of a parallelogram ABCD. AY and DX are intersecting at S; CX and BY are intersecting at R. Then SXRY is a_____________.

0%
Rectangle
0%
Kite
0%
Parallelogram
0%
Square

Explanation

Given:

$X$ and

$Y$ are the mid-points of opposite sides

$AB$ and

$DC$ of a parallelogram

$ABCD.$

$AX = XB$ and

$DY = YC$

$AB = DC$

$XB = DY......(i)$

Also

$AB || DC......$ (opp. sides of a ïïgm)

$XB || DY.....(ii)$

Since in quadrilateral

$XBYD, XB = DY$ and

$XB || DY$

$SYRX$ is a parallelogram.

The sides of a quadrilateral are extended in order to form exterior angles. The sum of these exterior angle is

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

State whether the statements are true (T) or (F) false.
All kites are squares.

0%
True
0%
False

State whether the statements are true (T) or (F) false.
A quadrilateral has two diagonals.

0%
True
0%
False

A diagonal of a paralleogram divides it into two congruent triangles.

0%
True
0%
False

Explanation

Let us consider a parallelogram

$ABCD$ with

$AC$ as its diagonal.

To prove:

$\triangle ABC \cong \triangle ADC$

Proof:

We know that the opposite sides of a parallelogram are parallel.

Hence

$AB \parallel DC$ and

$AD \parallel BC$

$AC$ is the transversal of the parallel lines

$AB$ and

$DC$

$\angle BAC =\angle DCA$ (Alternate angles)

$...(1)$

$AC$ is the transversal of the parallel lines

$AD$ and

$BC$

$\angle DAC =\angle BCA$

$...(2)$

$\triangle ABC$ and

$\triangle ADC,$

$\angle BAC =\angle DCA$

$AC$ is the common side

$\angle DAC =\angle BCA$

Therefore,

$\triangle ABC \cong \triangle ADC$

1465959_1303911_ans_680c786888b948c19ec04eb66a1745bb.png

If the angles of quadrilateral are

$x,x+20,2x+45,x+15$ . Find

$x$ .

0%
$52$
0%
$56$
0%
$58$
0%
$60$

Explanation

Given: The angles of quadrilateral are

$x,x+20,2x+45,x+15$

By angle sum property, the sum of all the angles of a quadrilateral is

$360^\circ$

$x+(x+20^\circ)+(2x+45^\circ)+(x+15^\circ)=360^\circ$

$\implies$

$5x+80^\circ=360^\circ$

$\implies$

$5x=360^\circ-80^\circ$

$\implies$

$5x=280^\circ$

$\implies$

$x=56^\circ$

Therefore, option

$B$ is correct.

In a quadrilateral

$ABCD,\angle A = 2x - {35^ \circ },\angle B = 3x - {5^ \circ },\angle C = x + {10^ \circ },\angle D = 4x + {20^ \circ },$ find the value of

$x$ ,

0%
$35$
0%
$37$
0%
$39$
0%
None of these

The angles of quadrilateral are

$30,150,60$ find the third angle

0%
$120$
0%
$200$
0%
$140$
0%
$50$

Explanation

The angles of quadiateral are

$30,150,60$

The sum of angles is

$360$

$\implies x+30+150+60=360\\x=360-240\\x=120$

If the angle of a quadrilateral are

${ x }^{o },(x-{ 10) }^{ o },(x+{ 30 })^{o }$ and

$2{ x }^{ o }$ , then the greatest angle is:

0%
${ 136 }^{ 0 }$
0%
${ 180 }^{ 0 }$
0%
${ 68 }^{ 0 }$
0%
${ 148 }^{ 0 }$

Explanation

The given angles of a quadrilateral are

${ x }^{ o },(x-{ 10) }^{ o },(x+{ 30 })^{ o }$ and

$2{ x }^{ o }.$

We know, that sum of four interior angles of a quadrilateral is

$360^o$ .

$\therefore$

$x+(x-10)+(x+30)+2x=360^o$

$\Rightarrow$

$5x+20^o=360^o$

$\Rightarrow$

$5x=340^o$

$\Rightarrow$

$x=\dfrac{340^o}{5}$

$\Rightarrow$

$x=68^o$ .

Now, measure of an angles:

$x^o=68^o$ ,

$(x-10)^o=(68-10)^o=58^o$ ,

$(x+30)^o=(68+30)^o=98^o$

and

$2x^o=2(68^o)=136^o$ .

$\therefore$ The greatest angle is

$136^o$ .

Hence, option

$A$ is correct.

The three angles of a quadrilateral are

${ 80 }^{ \circ }$ ,

${ 70 }^{ \circ }$ and

${ 120 }^{ \circ }$ . The fourth angle is:

0%
${ 110 }^{ \circ }$
0%
${ 100 }^{ \circ }$
0%
${ 90 }^{ \circ }$
0%
${ 80 }^{ \circ }$

Explanation

Three angles of quadrilateral are :

$\theta _{1} = 80^o$ ,

$\theta _{2} = 70^o$ ,

$\theta _{3} = 120^o$ .

To find

$\theta _{4} :$

We know, by angle sum property, total angle of quadrilateral

$= 360^o$

Then,

$\theta _{1} + \theta _{2} + \theta _{3} + \theta _{4} = 360^o$

$\implies$

$80^o + 70^o + 120^o + \theta _{4} = 360^o$

$\implies$

$270^o + \theta _{4} = 360^o$

$\implies$

$\theta _{4} = 360^o - 270^o$

$\therefore \theta _{4} = 90^o$ .

Therefore, the fourth angle is

$90^o$ .

Hence, option

$C$ is correct.

The sum of all the angles of a quadrilateral is :

0%
${ 180 }^{ \circ }$
0%
${ 270 }^{ \circ }$
0%
${ 360 }^{ \circ }$
0%
${ 400 }^{ \circ }$

Explanation

$\textbf{Step -1: Construction in the given quadrilateral.}$

$\text{Consider a quadrilateral }$

$ABCD$ .

$\text{Construct a line }AC \text{ i.e. the diagonal to quadrilateral }ABCD.$

$\textbf{Step -2: Sum of angles of a triangle.}$

$\text{We know, the sum of all angles of a triangle is }$

$180^{\circ}$

$\text{So, in }$

$\triangle ABC$ ,

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

$...(i)$

$\text{And, in }$

$\triangle ACD$ ,

$\angle CAD+\angle ADC+\angle DCA=180^{\circ}$

$...(ii)$

$\textbf{Step -3: Adding above equations to get the sum of all the angles of a quadrilateral.}$

$\text{Adding (i) and (ii), we get, }$

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

$\Rightarrow$

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

$\Rightarrow$

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

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

$\text{Hence, We can say that the sum of all the angles of a quadrilateral is }360^\circ$

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

$\mathbf{360^\circ}$

If two angles of a quadrilateral are

${ 60 }^{ \circ }$ and

${ 40 }^{ \circ }$ and the other two angles in the ratio

$8 : 5$ , then the measure of other two angles are ____________.

0%
${ 80 }^{ \circ }$
0%
${ 100 }^{ \circ }$
0%
${ 160 }^{ \circ }$
0%
${ 180 }^{ \circ }$

Explanation

Given the two angles are $60^o$ and $40^o$ .

Also, the other two angles are in the ratio $8:5$ .

Let the angles be $8x,5x$ , respectively.

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

$\Rightarrow 60^o+40^o+8x+5x=360^o$

$\Rightarrow 100^o+13x=360^o$

$\Rightarrow 13x=360^o-100^o$

$\Rightarrow 13x=260^o$

$\therefore x=20^o$ .

$\therefore 8x=8 \times 20^o =160^o$ ,

and $5x=5\times 20^o =100^o$ .

$\therefore$ Tmeasure of other two angles are $160^o$ and $$100^o.

Hence, options $B$ and $C$ are correct.

Say True or False.
All the sides of a rhombus are of equal length.

0%
True
0%
False

Say True or False.
All the sides of a parallelogram are of equal length.

0%
True
0%
False

Explanation

In parallelogram,

Opposite sides are equal is length.

Therefore in parallelogram ABCD

$.AB=CD$

$.BC=AD$

$.But AB\neq BC\neq CD\neq AD$

1338501_1644524_ans_a37d21fb95714c2b93a765cf1c7acbd5.png

Sate whether the following statements are true (T) or false (F):
A vertex of a quadrilateral lies in its interior.

0%
True
0%
False

Which of the following statements is false?

0%
A quadrilateral has four sides and four vertices
0%
A quadrilateral has four angles
0%
A quadrilateral has four diagonals
0%
A quadrilateral has two diagonals

Explanation

$\textbf{Step -1: Characteristics of quadrilateral,}$

$\text{We know from the characteristics of quadrilateral that it has four sides and}$

$\text{ four vertices and four angles}$

$\text{So, option A, B are TRUE.}$

$\textbf{Step -2: Checking number of diagonals.}$

$\text{Number of sides of a quadrilateral}=4$

$\text{Then, the number of diagonals}=\dfrac{4(4-3)}{2}=2$

$[\because\textbf{number of diagonals for side n}\mathbf{=\dfrac{n(n-3)}{2}}]$

$\text{So, option C is FALSE and D is TRUE.}$

$\textbf{Hence, Option C is correct.}$

The three angles of a quadrilateral are

$80^o,70^o$ and

$120^o$ . Then the fourth angle is:

0%
$110^0$
0%
$100^0$
0%
$90^0$
0%
$80^0$

Explanation

Given the three angles of a quadrilateral are

$80^o,70^o$ and

$120^o$ .
Let the fourth angle be

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

$360^0$ .

$\implies$

$80^o+70^o+120^o+x=360^o$

$\implies$

$270^o+x=360^o$

$\implies$

$x=360^o-270^o$

$\implies$

$x=90^o$ .

$\therefore$ The fourth angle is

$90^o$ .

Hence, option

$C$ is correct.

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Quadrilaterals Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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