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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Question 1
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
Solution
$C= 2 \pi r$ 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 , $C=2 \pi r= 484 \ mm$ , $A= \pi r^2= 18634 \ mm^2$
Question 2
If the circumference of a circular sheet is 176 m, find its diameter and area.
Solution
Here C=176 m or $2 \pi r =176$ or r=28 m
D= 2r= 56 cm
$A= \pi r^2= 2464 \ cm^2$
Question 3
The area of a circle is 616 cm
2. Find its diameter and circumference.
Solution
Here A=616 cm2
or
$\pi r^2= 616$ or r=14 cm
D=2r = 28 cm
C=88 cm
Question 4
From a circular sheet of a radius 5 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Solution
Area of remaining sheet is the difference in the area of 5 cm circle and 3 cm circle
$A= \pi (5)^2 - \pi (3)^= 50.28 \ cm^2$
Question 5
Find the perimeter semicircle whose radius is 5 cm including the diameter.
Solution
Perimeter = Half Circumference of circle of radius 5 cm + Diameter of the circle
$=\pi \times 5 + 10=25.7 \ cm$
Question 6
The diameter of a wheel is 70 cm. How many times the wheel will revolve in order to cover a distance of 110 m?
Solution
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
Question 7
The ratio of the radii of two wheels is 3 : 2. Find the ratio of their circumference.
Solution
We have $\frac {r_1}{r_2} = \frac {3}{2}$
$\frac {C_1}{C_2} = \frac {2 \pi r_1}{2 \pi r_2} = \frac {r_1}{r_2} = \frac {3}{2}$
Question 8
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.
Solution
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
Question 9
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.
Solution
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 r2 = 86.625 cm2
Question 10.
Find the area of a circle whose circumference is same as the perimeter of square of side 22 cm.
Solution
Perimeter of Square = 88 cm
Now
Radius of the circle= Circumference/2π = 88/2 π = 14 cm
Area of circle = π r2 = 616 cm2
Question 11.
Two circles have areas in the ratio 16 : 121. Find the ratio of their circumference and diameter
Solution
$\frac {A_1}{A_2} = \frac {16}{121}$
$\frac { \pi r_1^2}{\pi r_2^2}= \frac {16}{121}$
$\frac {r_1}{r_2} = \frac {4}{11}$
Now
$\frac {C_1}{C_2} = \frac {2 \pi r_1}{2 \pi r_2}=\frac {r_1}{r_2} = \frac {4}{11}$
$\frac {D_1}{D_2} = \frac {2 r_1}{2 r_2}=\frac {r_1}{r_2} = \frac {4}{11}$
Question 12.
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.
Solution
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
Question 13.
The radius of a wheel is 63 cm. Find the number of turns required to cover a distance of 1584 m.
Solution
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
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