MCQ Questions for Class 7 Maths Perimeter And Area Quiz 8

Find the area of a parallelogram with a base of

$200$ cm and height of

$2.5$ cm.

0%
$500$ $cm^{2}$
0%
$510$ $cm^{2}$
0%
$520$ $cm^{2}$
0%
$300$ $cm^{2}$

Explanation

Area of a parallelogram

$=$ base

$\times$ height

$=200 \times 2.5$

$= 500$

$cm^{2}$

Find the base of parallelogram if its area is

$\displaystyle 80{ cm }^{ 2 }$ and altitude is

$10$ cm.

0%
$6$ cm
0%
$8$ cm
0%
$10$ cm
0%
None of the above

Explanation

Given, area of parallelogram

$=80\ cm^2$

Altitude,

$a=10\ cm$

Let, the base be

$b$

For a parallelogram,

$Area=Base\times Altitude$

$\Rightarrow 80=b\times 10$

$\Rightarrow b=\dfrac{80}{10}$

$\Rightarrow b=8$

Hence, the base of the parallelogram is

$8\ cm$

A circular lawn with a radius of

$5$ meters is surrounded by a circular walkway that is

$4$ meters wide (see figure). What is the area of the walkway?

0%
$48\ \pi \ m^2$ .
0%
$60\ \pi \ m^2$ .
0%
$56\ \pi \ m^2$ .
0%
$42\ \pi \ m^2$ .

Explanation

The area of the walkway is the area of the entire image (walkway + lawn) minus the area of the lawn. To find the area of each circle, use the formula:

Large circle:

$A = \pi r^2 = \pi (9)^2 = 81\pi$

Small circle:

$A = \pi r^2 = \pi(5)^2 =25\pi$

Thus, required area =

$81\pi-25\pi= 56\pi m^2$ .

Find the area of a parallelogram with a base of

$34$ meters and a height of

$8$ meters.

0%
262 $m^{2}$
0%
272 $m^{2}$
0%
282 $m^{2}$
0%
292 $m^{2}$

Explanation

Area of a parallelogram = base

$\times$ height
=

$34 \times 8$
=

$272$

$m^{2}$

Area

$= \dfrac{1}{2}\times base \times height$
Which of the following is true?

0%
Calculation of area needs attitude.
0%
Calculation of area by the given formula doesn't need attitude.
0%
Base can be $AB,\ BC$ or $AC$
0%
None of the above

Explanation

Area of

$\Delta ABC=\dfrac { 1 }{ 2 } \times base\times altitude$

To calculate the area we need the altitude which is not given but we can choose any side as a base from

$AB, BC$ or

$AC$ .

Therefore, statements

$A$ and

$C$ are true.

A parallelogram has an area of

$60$

$cm^{2}$ and a base of

$12cm$ . Find the height.

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

Explanation

Area of a parallelogram = base

$\times$ height

$60 = 12 \times height$
height

$=60 \div 12$
height

$=5$

$cm$

Calculate the area of a parallelogram with a base of

$12$ m and height of

$5$ m.

0%
59 $m^{2}$
0%
60 $m^{2}$
0%
61 $m^{2}$
0%
62 $m^{2}$

Explanation

Area of a parallelogram = base

$\times$ height
=

$12 \times 5$
=

$60$

$m^{2}$

If P represents the area and W represents the circumference of the circle, then P in terms of W is

0%
$\displaystyle \frac{2\pi }{W}$
0%
$\displaystyle \frac{4\pi^{2} }{W}$
0%
$\displaystyle \frac{2\pi^{2} }{W}$
0%
$\displaystyle \frac{W^{2} }{4\pi }$

Explanation

Since P is the area so P =

$\displaystyle \pi r^{2}$ ....(1)
W is the perimeter thus W =

$\displaystyle 2\pi r\Rightarrow r=\frac{W}{2\pi }$

Put the value of

$r$ in (1)

$\displaystyle \Rightarrow P=\pi \left (\frac{W}{2\pi } \right )^{2}$

$\displaystyle P=\pi \frac{W^{2}}{4\pi ^{2}}$

$\displaystyle \Rightarrow P= \frac{W^{2}}{4\pi }$

Find the perimeter of the figure outlined by solid line in terms of

$x$ .

0%
$5x+3\pi x$
0%
$5x+6\pi x$
0%
$10x+3\pi x$
0%
$10x+6\pi x$

Explanation

Perimeter of the figure

$=$ Length of slanting sides of triangle + Circumference of semicircle

Circumference of semicircle

$=\pi r = \dfrac {22}{7} \times 3x = \dfrac {66x}{7}= 3 \pi x$

For the two right angle triangle,

By Pythagoras theorem

$(Hyp)^2 = (4x)^2+ \left(\dfrac{6x}{2}\right)^2$

$=16x^2+9x^2$

$=25x^2$

Hyp

$= 5x$ units

Length of two slanting sides of triangle

$=2 \times 5x = 10x$

Perimeter of the figure

$=(3 \pi x + 10x)$ units

Find the area and inner circumference of the ring.

0%
$9.48$ sq. cm, $3.14$ cm
0%
$9.42$ sq. cm, $6.48$ cm
0%
$9.42$ sq. cm, $2.14$ cm
0%
None of these

Explanation

Area of a circular ring =

$\pi(R^2-r^2)$

$=$

$3.14(2^2-1^2)$

$= 3.14 \times 3$

$= 9.42$ sq. cm
Inner circumference

$=$

$2\pi r$

$= 2 \times 3.14 \times 1$

$= 6.28$ cm

Calculate the area of a circular ring whose outer and inner radii are

$12$ and

$10\ cm.$

0%
$118.16$ $cm^2$
0%
$128.16$ $cm^2$
0%
$138.16$ $cm^2$
0%
$148.16$ $cm^2$

Explanation

Area of a circular ring

$=\pi(R^2-r^2)$

$=3.14(12^2-10^2)$

$= 3.14 \times 44$

$= 138.16$

$cm^2$

Find the width of the ring.

0%
$2\ cm$
0%
$3\ cm$
0%
$4\ cm$
0%
$5\ cm$

Explanation

Width of the ring,

$d =$ outer radius

$-$ inner radius

$d = R - r$

$=10 - 5 = 5\ cm$

The diameter of a circle is

$1$ . Calculate the area of the circle.

0%
$\dfrac {\pi}{8}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{2}$
0%
$\pi$
0%
$2\pi$

Explanation

formula to calculate area of a circle is

$\pi { r }^{ 2 }$ .
Given, diameter

$d=1$

Radius

$r=$

$\dfrac { d }{ 2 } =\dfrac { 1 }{ 2 }$

$\Rightarrow$ Area of circle

$=$

$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\dfrac { \pi }{ 4 }$ sq. units.

Find the area of the triangle given.

0%
$3$ sq units
0%
$6$ sq units
0%
$9$ sq units
0%
None of the above

Find the area of the shaded portion

0%
$4 cm^{2}$
0%
$5 cm^{2}$
0%
$6 cm^{2}$
0%
$7 cm^{2}$

Explanation

We know that the area of a triangle is given by the formula

$A=\dfrac{1}{2} bh$ , where b is the length of the base and h is the triangles height.

$ar (ADBE)=ar(ABC)-ar(DEC)$

$=\dfrac{1}{2}\times 2\times 6-\dfrac{1}{2}\times 1\times 2$

$=6-1=\boxed{5cm^2}$

1103374_515843_ans_21c6a340560043c8bef57325e72c3a91.png

Sonali pounded a stake into the ground, when she attached a rope to both the stake and her dog's collar, the dog could reach

$9$ feet from the stake in any direction. Find the approximate area of the lawn, in square feet, the dog could reach from the stake. (

$\displaystyle \pi$ =3.14)

0%
$28$
0%
$57$
0%
$113$
0%
$254$

Explanation

Dog can reach

$9$ feet from the stake in any direction, so radius of the circular ground will be

$9$ .
We know formula for area of circle is

$\pi { r }^{ 2 }$
Hence area will be

$3.14\times { 9 }^{ 2 }$ (value of pi is given)

$=3.14\times 81=254.34$

The radius of a circle whose circumference is

$\pi$ is

0%
$\dfrac{1}{2}$
0%
$1$
0%
$2$
0%
$\pi$
0%
$2 \pi$

Explanation

Given that the circumference of circle is

$\pi$ .

We know that the circumference of circle is

$2\pi r$ , where

$r$ is radius of circle.

$\therefore$ Circumference

$=\pi= 2 \pi r$

$\therefore r=\dfrac {1}{2}$

In the figure given above,

$\overline {AC}$ is a diameter of the large circle and B lies on

$\overline {AC}$ so that

$\overline {AB}$ is a diameter of the small circle. If

$AB = 1$ and

$BC = 2$ , Calculate the area of the shaded region.

0%
$\dfrac {\pi}{4}$
0%
$\pi$
0%
$2\pi$
0%
$\dfrac {9\pi}{4}$
0%
$9\pi$

Explanation

To find the area of the shaded region, we have to find the area of the bigger circle and the area of the smaller circle.
For large circle,

$AC$ is diameter measuring

$3$ .

$AC=AB+BC=1+2=3$
Area of larger circle is

$=$

$\pi { R }^{ 2 }$

$\pi { \left( \dfrac { 3 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 9 }{ 4 }$
Area of smaller circle is

$=\pi { r }^{ 2 }$

$\pi { \left( \dfrac { 1 }{ 2 } \right) }^{ 2 }=\pi \dfrac { 1 }{ 4 }$
Area of shaded region is

$=$ Area of larger circle

$-$ Area of smaller circle

$=\dfrac { 9\pi }{ 4 } -\dfrac { \pi }{ 4 }$

$=\dfrac { 9\pi -\pi }{ 4 } =\dfrac { 8\pi }{ 4 } =2\pi$

The area of the parallelogram

$ABCD$ is

0%
$20$
0%
$10$
0%
$5\sqrt{3}$
0%
$10\sqrt{3}$
0%
$18$

Explanation

From the given diagram, we conclude that the height of the parallelogram

$ABCD$ is

$4$ and its base is given by

$5$ .

We know that, Area of parallelogram

$=\text{base} \times \text{height}$

$\therefore$ Area of parallelogram

$ABCD$ is given by

$\text{base}\times \text{height}=4\times 5=20$

Hence, the answer is

$20$ .

A circle of radius

$x$ has an area twice that of a square of side

$a.$ The equation used to find the radius of the circle is

0%
$\pi a^2 = 2x^2$
0%
$\pi x^2 = 2a^2$
0%
$\pi x^2 = 4a^2$
0%
$\pi a^2 = 4x^2$
0%
$4\pi x^2 = a^2$

Explanation

The area of square whose side length is

$a$ is

${a}^{2}$
So the area of circle is

$2{a}^{2}$

Since, the radius of circle be

$x$ , then the area is

$\pi {x}^{2} = 2{a}^{2}$

The figure shows a rectangle

$ABCD$ .
Calculate the unshaded area.

0%
$4x+3y$
0%
$8x+5y$
0%
$12xy$
0%
$22xy$

Explanation

For the bigger rectangle,

Length of the rectangle

$AD=BC =6x$ cm

and breadth of the rectangle

$AB=CD=4y$ cm

So, area of rectangle

$=6x\times 4y\ sq.cm$

$=24xy\ sq.cm$

For the shaded rectangle

Length of shaded rectangle

$=2x$ cm

and breadth of shaded rectangle

$=y$ cm

So, area of shaded rectangle

$=2x\times y \ sq.cm$

$=2xy\ sq.cm$

Hence, the area of unshaded part

$=(24 xy-2 xy)\ sq.cm=22 xy\ sq.cm$

In figure, what is the approximate area of parallelogram DAWN?

0%
11.57
0%
13.64
0%
14.63
0%
17.25
0%
20.00

Explanation

area of parallelogram

$=ab\sin \theta$

$AD\times DN \sin 47^o$

$4\times 5\times \sin 47^o$

$20\times 0.731$

$14.63$

2114827_537247_ans_e7c13f4e42cb4786b885e1cbd2bdf110.png

What will it cost to carpet a room with indoor/outdoor carpet if the room is 10 feet wide and 12 feet long? The carpet costs 12.51 per square yard.

0%
$166.80
0%
$175.90
0%
$184.30
0%
$189.90

A lawn

$30$ m long and

$16$ m wide is surrounded by a path

$2$ m wide. What is the area of the path?

0%
$200$ sq. m
0%
$280$ sq. m
0%
$300$ sq. m
0%
$320$ sq. m

Explanation

Given, length of lawn

$=16$ m, width

$=2$ m

We need to find the area of the path.

Figure is shown according to question having sides

$=16+2+2 , 30+2+2 = 20m, 34m$

Thus area of path

$= (34\times 20 - 30\times 16)m^{2}$

$= (680 - 480)$

$= 200$ sq. m

What is the area of a circle with a diameter of

$16$ ?

0%
$8$ $\pi$
0%
$16$ $\pi$
0%
$64$ $\pi$
0%
$128$ $\pi$
0%
$256$ $\pi$

Explanation

Given, diameter

$=d=16$

We need to find area of circle.

Area of circle

$=\pi r^2$

Therefore, radius

$=r= \dfrac {16}{2} = 8$
Area of the circle

$= \pi r^2 = \pi (8)^2 = 64\pi$

Two similar parallelograms have corresponding sides in the ratio

$1:k$ . What is the ratio of their areas?

0%
$1:3{k}^{2}$
0%
$1:4{k}^{2}$
0%
$1:{k}^{2}$
0%
$1:2{k}^{2}$

The area of MCM college is

$213$ hectares. Find its area in meters.

0%
$2130$
0%
$21300$
0%
$213000$
0%
$2130000$

Convert

$200 cm^2$ in

$mm^2$ .

0%
$2000$ $mm^2$
0%
$20000$ $mm^2$
0%
$200000$ $mm^2$
0%
$2000000$ $mm^2$

A wire of length 36 cm is bent in the form of a semicircle. What is the radius of the semicircle?

0%
$9\ cm$
0%
$8\ cm$
0%
$7\ cm$
0%
$6\ cm$

Explanation

Length of wire is in the form of semicircle.

Given, length of wire

$=36\ cm$

The length of the wire will cover the diameter as well as the circumference of the semicircle.

$\therefore \$ Length of the wire

$= \dfrac {1}{2} (2\pi r) + 2r\\$

$\Rightarrow 36 = (\pi + 2)r\\$

$\Rightarrow 36 = \left (\dfrac {22}{7} + 2\right ) r = \dfrac {36}{7}\times r\\$

$\Rightarrow r = 7\ cm$ .

Hence, the radius of the semicircle is

$7\ cm$ .

In the following figure, what is the area of the shaded circle inside of the square?

0%
512
0%
256
0%
16
0%
50.24
0%
12.57

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 8 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 8 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.