In this page we have *Important Questions for Mensurations Class 8 Maths Chapter 11 * . Hope you like them and do not forget to like , social share
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Find the circumference and Area of the circle whose radius are gives below

(a) 21 cm

(b) 6.3 cm

(c) 14 mm

(d) 28 cm

(e) 49 cm (f) 77 mm

Take π = 22/7 in all the above questions

$C= 2 \pi r$ and $A= \pi r^2$

(a) r=21 cm , $C=2 \pi r= 132 \ cm$ , $A= \pi r^2= 1386 \ cm^2$

(b) r=6.3 cm , $C=2 \pi r= 39.6 \ cm$ , $A= \pi r^2= 124.74 \ cm^2$

(c)r=14 mm , $C=2 \pi r= 88 \ mm$ , $A= \pi r^2= 616 \ m^2$

(d) r=28 cm , $C=2 \pi r= 176 \ cm$ , $A= \pi r^2= 2464 \ cm^2$

(e) r=49 cm , $C=2 \pi r= 308 \ cm$ , $A= \pi r^2= 7546 \ cm^2$

(f) r=77 mm , $C=2 \pi r= 484 \ mm$ , $A= \pi r^2= 18634 \ mm^2$

If the circumference of a circular sheet is 176 m, find its diameter and area.

Here C=176 m or $2 \pi r =176$ or r=28 m

D= 2r= 56 cm

$A= \pi r^2= 2464 \ cm^2$

The area of a circle is 616 cm

Here A=616 cm^{2}

or

$\pi r^2= 616$ or r=14 cm

D=2r = 28 cm

C=88 cm

From a circular sheet of a radius 5 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.

Area of remaining sheet is the difference in the area of 5 cm circle and 3 cm circle

$A= \pi (5)^2 - \pi (3)^= 50.28 \ cm^2$

Find the perimeter semicircle whose radius is 5 cm including the diameter.

Perimeter = Half Circumference of circle of radius 5 cm + Diameter of the circle $=\pi \times 5 + 10=25.7 \ cm$

The diameter of a wheel is 70 cm. How many times the wheel will revolve in order to cover a distance of 110 m?

Circumference of wheel = 2 π r = 2 π (35) = 220 cm

Now Wheel moves 220 cm in 1 round

then it will move 110m in (11000/220)=50 rounds

The ratio of the radii of two wheels is 3 : 2. Find the ratio of their circumference.

We have $\frac {r_1}{r_2} = \frac {3}{2}$

$\frac {C_1}{C_2} = \frac {2 \pi r_1}{2 \pi r_2} = \frac {r_1}{r_2} = \frac {3}{2}$

A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 616 cm, find the width of the parapet.

Radius of Outer Circle=Circumference/2π= 616/2 π = 98 cm

Radius of the inner circle= Diameter/2 = 150/2= 75 cm

The width of the parapet= 98-75=23 cm

A thin wire is in the form of an equilateral triangle of side 11 cm. Find the area of a circle whose circumference is equal to the length of the wire.

Length of thin wire= Perimeter of equilateral triangle= 3 × 11= 33 cm

Now

Radius of the circle= Circumference/2π = 33/2 π = 5.25 cm

Area of circle = π r^{2} = 86.625 cm^{2}

Find the area of a circle whose circumference is same as the perimeter of square of side 22 cm.

Perimeter of Square = 88 cm

Now

Radius of the circle= Circumference/2π = 88/2 π = 14 cm

Area of circle = π r^{2} = 616 cm^{2}

Two circles have areas in the ratio 16 : 121. Find the ratio of their circumference and diameter

$\frac {A_1}{A_2} = \frac {16}{121}$

$\frac { \pi r_1^2}{\pi r_2^2}= \frac {16}{121}$

$\frac {r_1}{r_2} = \frac {4}{11}$

Now

$\frac {C_1}{C_2} = \frac {2 \pi r_1}{2 \pi r_2}=\frac {r_1}{r_2} = \frac {4}{11}$

$\frac {D_1}{D_2} = \frac {2 r_1}{2 r_2}=\frac {r_1}{r_2} = \frac {4}{11}$

Find the circumference of a wheel whose radius is 49 cm. Find the distance covered in 120 seconds, if it revolves 5 times per second.

circumference of a wheel= 2 π r = 2 π (49) = 308 cm

Distance covered in 1 round= circumference of a wheel

Wheel covers 5 revolutions in 1 sec

So , it does 600 revolutions in 120 sec

Now Distance covered in 1 round is 308 cm

Then Distance covered in 600 round will be 308 × 600 = 184800 cm = 1848 m

The radius of a wheel is 63 cm. Find the number of turns required to cover a distance of 1584 m.

Circumference=2 π r = 2 π (63) = 396 cm

Distance covered in 1 round= circumference of a wheel

So total turns required = 158400/396=400 turns

This Important Questions for Mensurations Class 8 Maths is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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