Practice Questions: Areas 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:
Ans: (b)
Sol:
The length of an arc depends on whether the angle θ is measured in radians or degrees.
If θ is in radians, arc length l = rθ, which has the dimension of length.
If θ is in degrees, arc length l = (θ/360) × 2πr = (πrθ)/180, also a length.
Any expression that involves r² (area dimension) cannot represent an arc length. Hence the given statement is false.
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: 

Ans: (a)
Sol:
Circumference of a circle of radius x is 2πx.
Given: 2πR₁ + 2πR₂ = 2πR.
Divide both sides by 2π: R₁ + R₂ = R.
Therefore option (a) is correct.

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: 
Ans: (a)
Sol:
The radius of the upper semicircle is 7 units, so its diameter is 14 units. The given length 14 units is therefore the width (or side) of the rectangular part beneath the semicircle.
If the shaded region corresponds to the rectangle whose sides are 14 units each, its area = length × breadth = 14 × 14 = 196 square units.
Hence option (a) is correct.

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 =

Sol:
Arc length corresponding to angle θ (in degrees): l = (θ/360) × 2πr = (πrθ)/180.
If θ is given in radians, arc length is l = rθ.

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

Ans: (b)
Sol:
Radius r = diameter/2 = 10/2 = 5 cm.
Arc length = (θ/360) × 2πr = (144/360) × 2π × 5 = (2/5) × 10π = 4π cm.
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

Sol:
The area of the shaded region = Area of trapezium - Area of quadrant ABE.
Take π = 22/7.
Radius of the quadrant can be found from the given quadrant area used in calculation: here radius r = 3.5 cm (since (πr²)/4 = (22/7 × 3.5²)/4 = 9.625 cm²).
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.

Ans: True (enter 1)
Explanation:
Circumference = πd = 3.14 × 1.2 = 3.768 m.
Therefore the statement is true, so enter 1.

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

Sol:
Take π = 22/7 (standard value used in this set).
Given area = 154 cm² = πr² ⇒ r² = 154 × 7 / 22 = 49 ⇒ r = 7 cm.
Circumference = 2πr = 2 × (22/7) × 7 = 44 cm.

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

Ans: (c)
Sol:
Let the radius of the outer circle be r.
The square PQRS is inscribed in the outer circle, so its side length s = √2 · r and its mid-point square ABCD (whose vertices are the contact points of the inner circle with PQRS) has side length equal to r.
Thus perimeter of polygon ABCD = 4r.
Perimeter of outer circle = 2πr.
Ratio = (2πr) / (4r) = π/2.
Hence option (c) is correct.

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.

Ans: (a)
Sol:
Height of rectangle = h. Base (diameter of semicircle) = h/3 ⇒ radius r = (h/3)/2 = h/6.
Area of rectangle = base × height = (h/3) × h = h²/3.
Area of semicircle = (1/2) × πr² = (1/2) × π × (h/6)² = (πh²)/72.
Total area = h²/3 + (πh²)/72 = h²(24 + π)/72.
Hence option (a) is correct.