MCQ Questions for Class 8 Maths Understanding Quadrilaterals Quiz 3

In the figure at right,

$ABCD$ is a parallelogram

$\angle BAO = 30^{\circ},\ \angle DAO=\angle 45^{\circ}$ and

$\angle COD = 105^{\circ}$ . Calculate

$\angle ABO$

0%
$45^{\circ}$
0%
$55^{\circ}$ >
0%
$25^{\circ}$
0%
$None\ of\ these$

Explanation

$\angle AOB = \angle DOC = 105^{\circ}$

In the triangle AOB

Sum of all angles of a triangle

$= 180^{\circ}$

$\therefore \angle AOB + \angle ABO + \angle BAO = 180^{\circ}$

$\therefore 105 + \angle ABO + 30 = 180$

$\angle ABO = 180 - 30 -105$

$\therefore \angle ABO = 45^{\circ}$

Identify the open curve.

0%
$1$ only
0%
$4$ only
0%
$1,4$ only
0%
$1,2,3$ only

Explanation

We know that an open curve has two end points.

Therefore the figure

$1,2,3$ are the open figures whereas

$4$ is a closed circle.

Position of point R is in the ______.

0%
interior of curve
0%
exterior of curve
0%
on the curve
0%
can not be said

Explanation

The point

$R$ is one the curve.
Hence, option C is correct.

Position of P is

0%
in the interior of curve
0%
in the exterior of curve
0%
on the curve
0%
can not be said

Explanation

$P$ is point is interior of curve.

Identify closed curve from the given figures.

0%
$1$ only
0%
$1 ,2$ only
0%
$1,2,3$ only
0%
$4$ only

Explanation

We know that an open curve has two end points and a closed curve has no end points because its starting and ending points are same.

Therefore the figure

$1,2,3$ are the open figures whereas

$4$ is a closed figure.

Hence only

$4$ is a closed figure which is a circle.

Identify open curve.

0%
$1$ only
0%
$2$ only
0%
$1,2$ only
0%
$3$ only

Explanation

We know that an open curve has two end points and a curve which begins and ends at the same point is called a closed curve.

The figure

$1,2$ only are open curve

$3$ is closed curve.

Position of Q is ______.

0%
in the interior of curve
0%
in the exterior of curve
0%
on the curve
0%
can not be said

Explanation

The point

$Q$ is on the curve.

Identify open curve.

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

Explanation

We know that an open curve has two end points and a curve which begins and ends at the same point is called a closed curve.

The curve

$2,4$ are open curve.

$1,3$ are closed curve

Identify closed curve.

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

Explanation

We know that an open curve has two endpoints and a curve that begins and ends at the same point is called a closed curve.

Thus according to the definition, the curve

$1,3$ is only a closed figure.

Identify closed curve.

0%
$1$ only
0%
$2$ only
0%
$3$ only
0%
$1\ and\ 2$ only

Explanation

A curve which begins and ends at the same point is called a closed curve.

Figure

$1,2$ are open figures and

$3$ is a closed figure.

Exterior point of curve is

0%
$P$
0%
$Q$
0%
$R$
0%
$T$

Explanation

The points

$S$ and

$T$ are in the exterior of the curve.

Point which is in the interior of curve is

0%
$P$
0%
$Q$
0%
$R$
0%
$T$

Explanation

The point

$P$ is in the interior of curve.

Point which is on the curve is

0%
$A$
0%
$B$
0%
$P$
0%
$T$

Explanation

The point which is on the curve is

$B$ .
Hence, option B is correct.

If each pair of opposite sides of a quadrilateral are equal and parallel, then it is a ____________.

0%
Kite
0%
Trapezium
0%
Parallelogram
0%
None of these

Two opposite angles of a parallelogram are

${\left( {3x - 2} \right)^ \circ }$ and

${\left( {50 - x} \right)^ \circ }$ . Find the value of

$x$

0%
$61$
0%
$13$
0%
$64$
0%
$67$

Explanation

Two opposite angles of a parallelogram are

$(3x-2)^o$ and

$(50-x)^o$

We know that, opposite angles of parallelogram are equal.

$\therefore$

$3x-2=50-x$

$\Rightarrow$

$3x+x=50+2$

$\Rightarrow$

$4x=52$

$\Rightarrow$

$x=13$

$\therefore$ The value of

$x$ is

$13^o$

The measure of an angle of a parallelogram is

$70^0$ . Its remaining angles are:

0%
$110^0,\,110^0 \text{ and } 70^0$
0%
$110^0 , 70^0\text{ and }120^0$
0%
$110^0 , 80^0\text{ and }120^0$
0%
$90^0 , 70^0\text{ and }120^0$

Explanation

Let $\angle A, \angle B, \angle C,\angle D$ be the four angles of the parallelogram ABCD, Such that $\angle B$ and $\angle D$ are opposite angles and also $\angle A$ and $\angle C$ are the opposite angles.

Let $\angle A = 70^o$ , then $\angle C=70^o$ {opposite angles are equal}

Angle adjacent to $\angle A = \angle D=180^o-70^o$

$=110^o$

Also, $\angle D = \angle B = 110^o$ {Opposite angles}

$So\>angles\>are\>70^\circ,110^\circ,70^\circ,110^\circ\>(complete\>set)$

What values of

$a$ would make the given quadrilateral a parallelogram.

0%
$5 \dfrac{1}{4}$
0%
$\dfrac{11}{2}$
0%
$5 \dfrac{3}{4}$
0%
$5$

Explanation

As the diagonal of quadrilateral bisects each other; 5b-4=3b+6
2b=10
$b=5$
So by the same rule;3b-4=2a
15-4=2a
11=2a
$a=\dfrac{11}{2}$

1102147_1175437_ans_eaf3c9896c8f4f35a15b2beafbc41898.png

Which of the following quadrilateral have both pairs of opposite sides parallel?

0%
Rectangle
0%
Square
0%
Parallelogram
0%
Trapezium

Which of the following is not always true for a parallelogram?

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

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD,

$CQ\parallel AP.$

0%
True
0%
False

Explanation

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

$\Rightarrow$

$OA=OC$ and

$OB=OD$

Since

$P$ and

$Q$ are point of trisection of

$BD$

$\therefore$

$BP=PQ=QD$

Now,

$OB=OD$ and

$BP=QD$

$\Rightarrow$

$OB-BP=OD-QD$

$\Rightarrow$

$OP=OQ$

In quadrilateral

$APCQ$ , we have

$OA=OC$ and

$OP=OQ$

Diagonals of quadrilateral

$APCQ$ bisect each

$\therefore$

$APCQ$ is a parallelogram.

$\Rightarrow$

$AP\parallel CQ$ [ Opposite sides are parallel in parallelogram ]

1278630_1216876_ans_afd7a738d22e41e4a575406939534d16.png

State whether the following statements are true or false
A rhombus is a regular quadrilateral.

0%
True
0%
False

State whether the statements are true (T) or (F) false.
A kite is not a convex quadrilateral.

0%
True
0%
False

State whether the statements are true (T) or (F) false.
Is a concave pentagon.

0%
True
0%
False

Explanation

Pentagon has

$5$ sides but this figures has

$6$ sides.

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)

What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

0%
$700^{0}$
0%
$800^{0}$
0%
$900^{0}$
0%
$1000^{0}$

Explanation

$After\quad analysing\quad the\quad given\quad table\quad it\quad is\quad observed\quad that\quad when\\ (i)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=3\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 180 }^{ o }=\left( 3-2 \right) \times { 180 }^{ o }\\ (ii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=4\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 360 }^{ o }=\left( 4-2 \right) \times { 180 }^{ o }\\ (iii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 540 }^{ o }=\left( 5-2 \right) \times { 180 }^{ o }\\ (iv)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 720 }^{ o }=\left( 6-2 \right) \times { 180 }^{ o }\\ \therefore \quad By\quad induction,\quad it\quad can\quad be\quad concluded\quad that\\ When\quad the\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=n\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles=\left( n-2 \right) \times { 180 }^{ o }\\ \therefore \quad For\quad a\quad 7\quad sided\quad polygon\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \rightarrow sum\quad of\quad the\quad internal\quad angles=\left( 7-2 \right) \times { 180 }^{ o }={ 900 }^{ o }\\ \\ \\ \\ \\ \\ \\$

How many simple closed curves are there?

0%
2
0%
3
0%
4
0%
5

Explanation

$A\quad simple\quad closed\quad curve\quad encloses\quad a\quad single\quad area \quad and \quad do \quad not\quad have \quad intersecting \quad lines.\\ Here\quad in\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad a\quad single\quad area\\ is\quad enclosed.\\ But\quad in\quad each\quad of\quad the\quad figures\quad 3,\quad 4,\quad 8\quad more\quad than\quad one\quad area\\ are\quad enclosed.\\ So\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad is\quad a\quad simple\quad closed \quad curve.\\$

The number of simple closed curves is thus

$5$

ABCD is a parallelogram and X is the mid-point of AB. If

$Area(\Box AXCD) =$

$24$

${cm}^2$ , Find

$Area(\Delta ABC)$ .

0%
$16$ $cm^2$
0%
$26$ $cm^2$
0%
$12$ $cm^2$
0%
$24$ $cm^2$

Explanation

We know that the area of a triangle is half of that of a parallelogram if they are on the same base and between the same parallels.

In case

$\triangle XCB$ , base is half.

$\therefore$ Area of

$\triangle XCB=\dfrac{1}{2}\times \dfrac{1}{2}\times ar(ABC)$

$=\dfrac{1}{4}\times ar(ABC)$

Now,

$ar(AXCD)=ar(ABCD)-ar(XCB)$

$=ar(ABCD)-\dfrac{1}{4}\times ar(ABCD)$

$=\dfrac{3}{4}\times ar(ABCD)$

$\Rightarrow$

$ar(AXCD)=\dfrac{3}{4}\times ar(ABCD)$

$\Rightarrow$

$24=\dfrac{3}{4}\times ar(ABCD)$

$\Rightarrow$

$ar(ABCD)=32\,cm^2$

We know that a diagonal divides a parallelogram into two triangles of equal area.

$\Rightarrow$

$ar(ABC)=\dfrac{1}{2}\times ar(ABCD)$

$=\dfrac{1}{2}\times 32$

$=16\,cm^2$

Say true or false.

It is possible to have a regular polygon with the measure of each exterior angle as

$22^{0}.$

0%
True
0%
False

Explanation

Here, let the number of sides of the polygon be

$n.$

So, the number of exterior angles is

$n$ .

Also, it is a regular one. Therefore, the total number of exterior angles=

$n$ and they are equal to each other.

$\therefore$ The sum of the exterior angles =

${ 360 }^{ o }$

$\therefore$ Each angle=

$\theta =\dfrac { { 360 }^{ o } }{ n } \Longrightarrow n=\dfrac { { 360 }^{ o } }{ \theta }$ when

$n \epsilon N$

i.e

$\theta$ should be a complete divisor of

${ 360 }^{ o }.$

Here,

$\theta ={ 22 }^{ o }$ which is not a complete divisor of

${ 360 }^{ o }.$

So any regular polygon can not have exterior angle =

${ 22 }^{ o }$

In Fig.8.2,

$ABCD$ and

$AEFG$ are two parallelograms. If

$\angle C = 55^o$ , determine

$\angle F$ .

0%
$27.5^o$
0%
$45^o$
0%
$55^o$
0%
$65^o$

Explanation

$ABCD$ and

$AEFG$ are two parallelograms sharing two sides

$AD$ (or

$AG$ ) and

$AE$ (or

$AE$ ) and the included angle.

$\angle C={ 55 }^{ o }$ .

To find out-

$\angle F$

Solution-

We have

$\angle F=\angle A$ ......(opposite angles of a parallelogram).........(i)

Again

$\angle A=\angle C$ ......(opposite angles of a parallelogram)...........(ii)

$\therefore \angle F=\angle C$ ......[From (i) and (ii)]

And

$\angle C={ 55 }^{ o }$ .......(given)

$\Rightarrow \angle F={ 55 }^{ o }$

Opposite angles of a quadrilateral

$ABCD$ are equal. If

$AB = 4$ cm, determine

$CD$ .

0%
$3$ cm
0%
$4$ cm
0%
$5$ cm
0%
$6$ cm

Explanation

Let

$ABCD$ be a quadrilateral.

Given, opposite angles of a quadrilateral

$ABCD$ are equal.

So, opposite sides of the quadrilateral

$ABCD$ are equal.

$\Rightarrow AB=CD$

Given,

$AB=4$ cm

Therefore,

$CD=4$ cm.

The diagonals

$AC$ and

$BD$ of a parallelogram

$ABCD$ intersect each other at the point

$O$ .

$\angle DAC = 32^o$ and

$\angle AOB = 70^o$ , then

$\angle DBC$ is equal to?

0%
$24^o$
0%
$86^o$
0%
$38^o$
0%
$32^o$

Explanation

In given figure,

Quadrilateral ABCD is a parallelogram.

So, AD

$\left|\right|$ BC

$\therefore$

$\angle$ DAC =

$\angle$ ACB --- ( Alternate angle)

$\therefore$

$\angle$ ACB = 32

$^\circ$

$\angle$ AOB +

$\angle$ BOC = 180

$^\circ$ --- (straight angle)

$\Rightarrow$ 70

$^\circ$ +

$\angle$ BOC = 180

$^\circ$

$\therefore$

$\angle$ BOC = 110

$^\circ$

$\triangle$ BOC,

$\angle$ OBC +

$\angle$ BOC +

$\angle$ OCB = 180

$^\circ$

$\Rightarrow$

$\angle$ OBC + 110

$^\circ$ + 32

$^\circ$ = 180

$^\circ$

$\Rightarrow$

$\angle$ OBC = 38

$^\circ$

$\therefore$

$\angle$ DBC = 38

$^\circ$

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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