MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 7

Find the angles of a parallelogram if one angle is three times another.

0%
${45}^{o}, {135}^{o}, {45}^{o}, {135}^{o}$
0%
${50}^{o}, {130}^{o}, {50}^{o}, {130}^{o}$
0%
${40}^{o}, {140}^{o}, {40}^{o}, {140}^{o}$
0%
${55}^{o}, {125}^{o}, {55}^{o}, {125}^{o}$

Explanation

Opposite angles of a parallel are equal to each other.

So, if two angles are

${x}^{o}$ each, then the other two opposite angles are

${3x}^{o}$ , since it is given that one angle is three times the other.

Sum of all four angles of a parallelogram

$= {360}^{0}$

$=> {x}^{o} + {3x}^{o}+ {x}^{o} + {3x}^{o} = {360}^{0}$

$=> {8x}^{o} = {360}^{0}$

$=> {x}^{o} = {45}^{0}$

So

${3x}^{o} = {135}^{0}$

Hence, the four angles of the parallelogram are

${45}^{0}, {135}^{0} , {45}^{0}, {135}^{0}$

$PQRS$ is a parallelogram. Find

$y$ .

0%
$27^{\circ}$
0%
$61^{\circ}$
0%
$41^{\circ}$
0%
$28^{\circ}$

Explanation

Given,

$PQRS$ is a parallelogram.
Hence,

$\angle PSR = \angle PQR = 68^{\circ}$ (Opposite angles of a parallelogram are equal)
Now,

$\angle PTS + \angle STQ = 180$ (linear pair)

$\angle PTS + 139 = 180$

$\angle PTS = 41^{\circ}$
Now,

$\angle PTS = \angle RST = 41^{\circ}$ (Alternate angles of parallel lines)
Now,

$\angle PSR = \angle PST + \angle TSR$

$68 = 41 + \angle TSP$

$\angle TSP = 27^{\circ}$

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

0%
A parallelogram in which two adjacent angles are equal is a rectangle.
0%
A quadrilateral in which both pairs of opposite angles are equal is a parallelogram.
0%
In a parallelogram, the maximum number of acute angles can be two.
0%
All of these

Explanation

$\Rightarrow$ We know that, rectangle is a parallelogram and all four angles of rectangle is right angle, so its two adjacent angles are equal.

$\Rightarrow$ A quadrilateral in which both pairs of opposite angles are equal is a parallelogram.

$\Rightarrow$ In a parallelogram, adjacent angles sum to

$180^o$ . Thus, there're

$2$ acute and

$2$ obtuse angles in a parallelogram.

$\therefore$ All the statements are correct in given options.

In the figure, ABCD is a parallelogram such that

$AF$ and

$CE$ are the bisectors of

$\angle DAB$ and

$\angle BCD$ respectively, then

$AF\parallel CE$ .

0%
True
0%
False

Explanation

Since,

$ABCD$ is a parallelogram,

$AB \parallel CD$ as opposite sides of a parallelogram are parallel.
And

$CE$ is the transversal passing through

$AB$ and

$DC$ .

$\implies \angle DCE = \angle CEB$ (Alternate Angles) -- (1)
Also since,

$ABCD$ is a parallelogram

$\angle A = \angle C$ (Opposite angles of a parallelogram as equal )
As,

$AF$ and

$CE$ are bisectors of these angles respectively,

$\angle FAB = \angle DCE$ --- (2)
From ,

$(1)$ and

$(2)$ :

$\angle FAB = \angle CEB$
This means, the corresponding angles formed by lines

$CE$ and

$FA$ cut by transversal

$AB$ are equal

$\implies AF \parallel CE$

The following figure GUNS is a parallelogram. Find

$x$ and

$y.$ (Lengths are in cm)

0%
$x= 6$ and $y= 9$
0%
$x=5$ and $y=8$
0%
$x= 9$ and $y= 6$
0%
$x= 5$ and $y= 7$

Explanation

Given:

$GUNS$ is a parallelogram

Hence, its opposite sides are equal.

i.e.

$GS = UN$ and

$GU =SN$

$\Rightarrow 3x=18$ , i.e.,

$x =6$

and

$3y-1=26$

$\Rightarrow 3y=26 + 1 = 27$

$\Rightarrow y = 9$

Hence,

$x= 6$ and

$y= 9$

The following figure RUNS is a parallelogram. Find x and y. (Lengths are in cm)

0%
$x=3$ and $y=13$
0%
$x=13$ and $y=3$
0%
$x=13$ and $y=13$
0%
$x=3$ and $y=3$

Explanation

$RUNS$ is a parallelogram and thus its diagonals bisect each other.

$\therefore$

$x+y=16$ ---- ( 1 )

and

$7+y=20$

$\therefore$

$y=20-7$

$\therefore$

$y=13\,cm$

Substituting

$y=13$ in equation ( 1 ) we get,

$\Rightarrow$

$x+13=16$

$\Rightarrow$

$x=16-13$

$\therefore$

$x=3\,cm$

$\therefore$

$x=3\,cm$ and

$y=13\,cm$

In the given figure, PQRS is a parallelogram. If perimeter of parallelogram PQRS is 40 cm and

$PQ = 12 cm$ , then PS is equal to

0%
$12 cm$
0%
$10 cm$
0%
$8 cm$
0%
$9 cm$

Explanation

Perimeter of

$\parallel gm$

$=40cm$

$2(PQ+PS)=40cm$

$2(12+PS)=40$

$PS=(40/2)-12$

$PS=8cm$

The sum of two opposite angles of a parallelogram is

$130^o$ . Find the measure of each of its angles

0%
$65^o, 115^o, 65^o, 115^o$
0%
$65^o, 105^o, 65^o, 115^o$
0%
$65^o, 115^o, 55^o, 115^o$
0%
$65^o, 115^o, 65^o, 105^o$

Explanation

$Let\quad ABCD\quad a\quad parallelogram\quad with\quad four\quad angles\quad as\quad \angle A,\angle B,\angle C\quad and\quad \angle D\\ \angle A+\angle C\quad =\quad 130(Given)\\ Also\quad \angle A\quad =\angle C\quad (Opposite\quad angles\quad of\quad \parallel gram\quad are\quad equal).......(1)\\ \Longrightarrow \angle A\quad +\quad \angle A\quad =130\\ \Longrightarrow 2\angle A\quad =\quad 130\\ \Longrightarrow \angle A\quad =\quad \frac { 130 }{ 2 } \quad =\quad 65\\ \Longrightarrow \angle A\quad =\quad \angle C\quad =\quad 65....(2)\\ Also\quad \angle A\quad +\quad \angle B\quad =180(Sum\quad of\quad cointerior\quad angles\quad is\quad 180)\\ \Longrightarrow 65\quad +\quad \angle B\quad =\quad 180\\ \Longrightarrow \angle B\quad =\quad 180\quad -\quad 65\quad =\quad 115\\ Also\quad \angle B\quad =\quad \angle D\quad =115(Opposite\quad angles\quad of\quad \parallel gram\quad are\quad equal)\\ Hence\quad all\quad the\quad angles\quad are\quad 65, 115, 65\quad and\quad 115$

In the given figure,

$ABCD$ is a parallelogram, diagonals

$BD$ and

$AC$ intersect each other at

$E$ . If

$BE + CE = 8$ cm, then

$AC + BD$ is equal to

0%
$16$ cm
0%
$14$ cm
0%
$20$ cm
0%
$24$ cm

Explanation

Given:

$ABCD$ is a parallelogram

We know that the diagonals of a parallelogram bisect each other.

$\Rightarrow$ The diagonals

$BD$ and

$AC$ bisect each other.

$\therefore AE = EC$ and

$BE = ED$

We know,

$AC+BD=AE+EC+BE+ED$

$= EC +EC + BE +BE$

$= 2 \times (EC + BE)$

$=$

$2 \times 8$ cm

$= 16$ cm

Given here are some figures:

Which one is convex polygon among

$2$ and

$3$ ?

0%
$2$
0%
$3$
0%
Both
0%
None of these

Explanation

A convex polygon is a simple polygon whose interior is a convex set. In a convex polygon, all interior angles are less than or equal to

$180$ degrees.
So, Image

$2$ is a convex polygon.

If ABCD is parallelogram. E is mid-point of AB and CE bisects

$\angle BCD$ , then

$\angle DEC$ is _____.

0%
$60^o$
0%
$90^o$
0%
$100^o$
0%
$120^o$

Explanation

ABCD is parallelogram. E is mid-point of AB and CE bisects BCD
AB||CD and CE Traversal

EC Bisects

$\angle BCD$
IN parallelogram. ABCD

$\angle D$ +

$\angle C=180$

$\therefore \frac{1}{2}\angle D+\frac{1}{2}\angle C=\frac{1}{2}\times 180$

$\angle EDC+\angle ECD= 90^{0}$

$\bigtriangleup DEC$

$\angle DEC+\angle EDC+\angle ECD= 180^{0}$

$\therefore \angle DEC+90^{0}= 180^{0}$

$\therefore \angle DEC= 180^{0}- 90^{0}= 90^{0}$

In a parallelogram PQRS, X is the mid point of PS and Y is the mid point of QR, then XY divides QS in the ratio of

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

The adjacent angles of a parallelogram are as 2 : 3. Find the measures of all the angles

0%
$108^o$
0%
$60^o$
0%
$116^o$
0%
$90^o$

Explanation

Let ABCD be a parallelogram. Let

$\angle A$ and

$\angle B$ be its two adjacent angles such that

$\angle A:\angle B=2:3$
Let

$\angle A = 2x^o$ and

$\angle B = 3x^o$
Since adjacent interior angles are supplementary.

$\therefore \angle A + \angle B =180^o$

$\Rightarrow 2x^o + 3x^o =180^o$

$\Rightarrow 5x^o=180^o\Rightarrow x^o=\frac {180}{5}=36^o$

$\therefore \angle A=2x^o=72^o, \angle B=3x^o=108^o$
Since opposite angles of parallelogram are equal.

$\therefore \angle C=\angle A=72^o$ and

$\angle D=\angle B=108^o$

Square is not a

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

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

Two adjacent angles of a parallelogram are in the ratio

$4:5$ , Find all the angles of the parallelogram

0%
$\angle C = \angle A = 90^o, \angle D =\angle B = 100^o$
0%
$\angle C = \angle A = 80^o, \angle D =\angle B = 100^o$
0%
$\angle C = \angle A = \angle D =\angle B = 90^o$
0%
None of above

Explanation

Let

$ABCD$ be a parallelogram.
Given that,

$\angle A:\angle B=4:5$

Sum of the ratios

$=4+5=9$
But,

$\angle A+\angle B=180^o$
(Adjacent angles of the parallelogram,

$ABCD$ )

$\therefore \angle A=\dfrac {4}{9}\times 180^o=80^o$

$\angle B=\dfrac {5}{9}\times 180^o=100^o$
So

$\angle C = \angle A = 80^o, \angle D =\angle B = 100^o$
(Opposite angles of a parallelogram are equal).

In the given figure

$AE = BC$ and

$AE || BC$ and the three sides

$AB, CD$ and

$ED$ are equal in length If

$\displaystyle \angle A=102^{\circ}$ find the measure of

$\displaystyle \angle BCD$

0%
$\displaystyle 138^{\circ}$
0%
$\displaystyle 162^{\circ}$
0%
$\displaystyle 88^{\circ}$
0%
None of these

Explanation

As $$AE = BC

$and$ AE || BC

$So if a line will be drawn parallel to$ AB

$from$ E

$, the line$ EC

$will be pass through$ C

$and$ AB=EC$$
And parallelogram will be formed.
and in parallelogram opposite angels are congruent.
So,

$\displaystyle \angle BAE=\angle BCE=102^{\circ}$
And AB=EC=CD=ED
So, ECD is equilateral triangle and all the three angle will be equal.

$\displaystyle \angle ECD=60^{\circ}$
So,

$\displaystyle \angle BCD=\angle BCE+\angle ECD=102^{\circ}+60^{\circ}=162^{\circ}$
Answer - option

$B$ is correct

$ABCD$ is a quadrilateral,

$E, F, G$ snd

$H$ are the midpoints of

$AB, BC, CD$ and

$DA$ respectively, then

$EFGH$ is a ____ .

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

$ABCD$ is a parallelogram whose diagonals intersect at

$O$ and

$\Delta$ BCD is an equilateral triangle having each side of length

$6$ cm, then the length of diagonal

$AC$ is ____ .

0%
$3 \sqrt 3\ cm$
0%
$6 \sqrt 3\ cm$
0%
$3 \sqrt 6\ cm$
0%
$12\ cm$

Explanation

$ABCD$ is a parallelogram.

∴

$BC=AD=6\ cm$

and

$AB=DC=6\ cm$ ,

Triangle

$BCD$ is equilateral triangle

then

$BD=6\ cm$

Hence

$ABCD$ becomes a rhombus.

∴ Diagonals

$AC$ and

$BD$ bisect each other at right angle.

$\therefore OD=\dfrac{1}{2}BD=\dfrac{6}{2}=3\ cm$

$\Delta OCD$

$(OC)^{2}=CD^{2}-(OD)^{2}=(6)^{2}-(3)^{2}=36-9=27$

$\therefore OC=3\sqrt{3}$

$\therefore AC=2\times OC=2\times 3\sqrt{3}=6\sqrt{3}\ cm$

If the adjacent angles of the parallelogram are

$x +30^\circ$ and

$2x- { 90 }^{ \circ}$ , then the smaller angle is

0%
${ 110 }^{\circ }$
0%
${ 100 }^{\circ }$
0%
${ 70 }^{\circ }$
0%
${ 80 }^{\circ }$

Explanation

In a parallelogram the adjacent angles are supplementary:

$x+30^\circ+2x-90^\circ=180^\circ$

$\Rightarrow 3x=240^\circ$

$\Rightarrow x=80^\circ$
Therefore, angles are:

$x+30^\circ= 80^\circ+30^\circ=110^\circ$ and

$2x-90^\circ=2\times80^\circ-90^\circ=70^\circ$
Therefore, smaller angle is

$70^\circ$ .
Option C is correct.

$x$ &

$y$ are opposite angles of a parallelogram. If the adjacent angle to these angles is

$75^o$ . What is the measure of these angles?

0%
$110^o$
0%
$150^o$
0%
$85^o$
0%
$105^o$

Explanation

One opposite angle is adjacent to

$75^o$ angle. So,

$180^o-75^o=105^o$
This is the measure of opposite angle. So, opposite angles being equal have same measure

$105^o$ .
Therefore, D is the correct answer.

If the perimeter of a parallelogram is

$50\ cm$ and one side is

$12\ cm$ . what is the measure of the other side ?

0%
$15\ cm$
0%
$10\ cm$
0%
$13\ cm$
0%
$18\ cm$

Explanation

Perimeter

$= S_1 + S_2 + S_3 + S_4$

$50 = 12 + S_2 + 12 + S_4$

$\Rightarrow 50 - 24 = 2S$

$\Rightarrow \dfrac{26}{2} = S$

$\therefore 13 = S$ (other side of the parallelogram)
Therefore,

$C$ is the correct answer.

In the measure of an angle in a parallelogram is

$110^o$ , what will be the measure of its adjacent angle?

0%
$110^o$
0%
$80^o$
0%
$70^o$
0%
$250^o$

Explanation

Sum of angles

$=360^o$

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

$\Longrightarrow110^o+x+110^o+x= 360^o$

$\Longrightarrow2x = 360^o - 220^o$

$\Longrightarrow x = 70^o$
Therefore, C is the correct answer.

Which of the following is/are an example(s) of simple closed curve?

0%
Circle
0%
Rectangle
0%
Square
0%
All of the above

Explanation

A curve that does not cross itself and ends at the same point where it begins, is called a simple closed curve.
Therefore,

$D$ is the correct answer.

A Rhombus is also a

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

In parallelogram ABCD , CD = 6 cm and AF = 8 cm. If AB = 12 cm, than CE equals to

0%
28 cm
0%
2 cm
0%
4 cm
0%
14 cm

Which of the following figures satisfy the following property ?
-Only one pair of sides is parallel.

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

0%
True
0%
False

In the figure above ABCD is a parallelogram

$\displaystyle \overline{DE}, \overline{EF}$ and

$\displaystyle \overline{BG}$ are the bisectors of

$\displaystyle \angle ADC, \angle DEB$ and

$\displaystyle \angle ABC$ respectively. BG and EF intersect at H. If

$\displaystyle \angle DAB=70^{\circ}$ then find

$\displaystyle \angle BHE$

0%
$\displaystyle 50^{\circ}$
0%
$\displaystyle \left ( 62\frac{1}{2} \right )^{\circ}$
0%
$\displaystyle 75^{\circ}$
0%
$\displaystyle \left ( 87\frac{1}{2} \right )^{\circ}$

Explanation

Given

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

$\angle ADC = \angle ABC =180-70 = 110^{\circ}$ (Sum of angles formed at adjacent vertices of parallelogram)

$$\angle ADE = half of angle ADC =55 degree

For triangle ADE , exterior angle DEB = angle DAE + angle ADE

= 70 + 55 =

$125^{\circ}$

Now angle HEB is half of angle DEB

Thus

$\angle HEB = \frac{125}{2} = 62.5^{\circ}$

Angle HBC = half of angle ABC =

$55 ^{\circ}$

In triangle HEB

$\angle EHB = 180^{\circ}- \angle HEB - \angle HBE$

=180 - 62.5- 55

$62.5^{\circ}$

So the correct answer is option B

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

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

If one angle of a parallelogram is

$65^o$ . What are the other angles?

0%
$50^o, 120^o, 50^o$
0%
$115^o, 65^o, 115^o$
0%
$65^o, 115^o, 65^o$
0%
None of the above

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 7 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 7 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.