Class 10 Maths Chapter 11 Practice Question Answers - Area Related to Circles

Q1. The length of an arc of a sector of a circle of radius r units and of centre angle θ is

(a) True
(b) False
(c) Neither
(d) Either
Ans: b
Sol: Length of arc subtending angle  Note that area will have the dimension of square of length. So,   cannot be the length.

Q2. If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then
(a) R₁ + R₂ = R
(b) R₁ + R₂ > r
(c) R1 + R2 < r
(d) Nothing definite can be said about the relation among R₁ ,R₂  and R.
Ans: (b)
Sol: 

The  circumference of circle with radius R₁= 2πR₁

and the circumference of circle with radius R₂ = 2πR₂

∴ The Sum of Circumferences, Sum=2π(R₁+ R₂) 

Again the circumference of circle with radius R=2πR.

∴ By given condition,

Q3. Find the area of shaded portion, where the length of it is 14 units and radius of the upper semicircle is 7 units.

(a) 196
(b) 98
(c) 200
(d) None
Ans: (a)
Sol: 
Requried area of figure =  Area of shaded portion
If we substract area of semi circle from area of rectangle & then add the area of upper semi circle, we get the net area or alternatively, if we calculate the area of rectangle, we get the net area = l x b
= 14 x 14
=196

Q4. Length of an arc of a sector of a circle with radius r and angle with degree measure θ is
m is
Sol: 
The whole circle of 360º  makes arc length of 2πr
Thus angle of θ will make a length =

Q5. If the sector of a circle of diameter 10 cm subtends an angle of 144º  at the centre, then the length of the arc of the sector is
(a) 2π cm
(b) 4π cm
(c) 5π cm
(d) 6π cm
Ans: (b)
Sol:  Given, diameter =10 cm, θ=144º
Length of an arc of a circle

Hence, option B is correct.

Q6. ABCD is a  trapezium  of area  24.5 sq. cm. In it, AD∥BC,∠DAB=90º, AD=10 cm and BC=4 cm. If ABE is a quadrant of a circle, find the area of the shaded region.
(Take π = 22/7)

Sol:
Given:
AD=10 cm
BC=4 cm
Area of trapezium =24.5 cm²


The area of shaded region = Area of trapezium – Area of given quadrant

The area of shaded region =24.5–9.625=14.875 cm²

Q7. Take π=3.14, the circumference of a circle with diameter = 1.2 m is 3.768 m. If true then enter 1 and if false then enter 0.
Sol:
Given that,
Diameter of a circle is 1.2 m.
The circumference of the circle is 3.768 m.
We know that, circumference =2πr or πd.
=3.14×1.2[ given π=3.14 ]
=3.768 m
So, it is true that the circumference of the circle is 3.768 m.
Hence, 1 is the answer.

Q8. If the area of a circle is 154 cm² , then its circumference in cm is
Sol:
Area of the circle =πr²

Circumference of the circle

Q9. The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point B,C,D and A. The ratio of the perimeter of the outer circle to that of polygon ABCD is

(a) π/4
(b) 3π/4
(c) π/2
(d) π
Ans: (c)
Sol: 
Let the radius of the outer circle be r.
∴ perimeter of outer circle = 2πr
But OQ=BC=r       [diagonals of the square BQCO]
∴ Perimeter of Square ABCD =4r

Q10. The figure is shown in a rectangle topped with a semicircle. The base of the rectangle is one-third its height. If h is the height of the rectangle, what is the area of the figure?
(a)
(b)
(c)
(d)
Ans: (a)
Sol:
Given that:
The height of the rectangle is equal to h.
The base of the rectangle = h/3
The base of the rectangle = Diameter of the semicircle
∴ Radius of the circle = h/6
Now,
Area of the figure = Area of the rectangular part + Area of the semicircular part

Hence, option A is correct.