Unit Test (Solutions): Areas Related to Circles | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: If the perimeter of the circle and square are equal, then the ratio of their areas will be equal to: (1 Mark)

(a) 14:11
(b) 22:7
(c) 7:22
(d) 11:14

Ans: (a)

Given,

The perimeter of circle = perimeter of the square

2πr = 4a

a=πr/2

Area of square = a² = (πr/2)²

A circle/A square = πr²/(πr/2)²

= 14/11

Q2: The area of the circle that can be inscribed in a square of side 8 cm is (1 Mark)

(a) 36 π cm²
(b) 16 π cm²
(c) 12 π cm²
(d) 9 π cm²

Ans: (b)

Given,

Side of square = 8 cm

Diameter of a circle = side of square = 8 cm

Therefore, Radius of circle = 4 cm

Area of circle

= π(r)²

= π (4)²

= 16π cm² 

Q3: The area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.  (1 Mark)

(a) 142/7
(b) 152/7
(c) 132/7
(d) 122/7

Ans: (c)

Angle of the sector is 60°

Area of sector = (θ/360°) × π r²

∴ Area of the sector with angle 60° = (60°/360°) × π r² cm²

= (36/6) π cm²

= 6 × (22/7) cm²

= 132/7 cm²  

Q4: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The area of the sector formed by the arc is:  (1 Mark)

(a) 200 cm²
(b) 220 cm²
(c) 231 cm²
(d) 250 cm²

Ans: (c)

The angle subtended by the arc = 60°

So, area of the sector = (60°/360°) × π r² cm²

= (441/6) × (22/7) cm²

= 231 cm²   

Q5: If the area of a circle is 154 cm2, then its perimeter is  (1 Mark)

(a) 11 cm
(b) 22 cm
(c) 44 cm
(d) 55 cm

Ans: (c) 

Given,

Area of a circle = 154 cm2

πr² = 154

(22/7) × r² = 154

r² = (154 × 7)/22

r² = 7 × 7

r = 7 cm

Perimeter of circle = 2πr = 2 × (22/7) × 7 = 44 cm

Q6: What is the area of a circle whose circumference is 44 cm?  (2 Marks)

Ans: Circumference of a circle = 2πr

From the question,

2πr = 44

Or, r = 22/π

Now, area of circle = πr² = π × (22/π)²

So, area of circle = (22×22)/π = 154 cm² 

Q7: Find the area of the sector of a circle with a radius of 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14) (2 Marks)

Ans: Let OAPB be the sector.

Area of the major sector = [(360 – θ)/ 360] × πr2

= [(360 – 30)/360] × 3.14 × 4 × 4

= (330/360) × 3.14 × 16

= 46.05 cm2

= 46.1 cm2 (approx)

Q8: Calculate the area of a sector of angle 60°. Given, the circle has a radius of 6 cm. (2 Marks)

Ans: Given,

The angle of the sector = 60°

Using the formula,

The area of sector = (θ/360°) ×π r²

= (60°/360°) × π r² cm²

Or, area of the sector = 6 × 22/7 cm² = 132/7 cm² 

Q9: Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm²  (3 Marks)  

Ans: Here, as the equilateral triangle is inscribed in a circle, the circle is an incircle.

Now, the radius of the incircle is given by,

r = Area of triangle/semi-perimeter

In the question, it is given that area of the incircle = 154 cm²

So, π × r² = 154

Or, r = 7 cm

Now, assume the length of each arm of the equilateral triangle to be “x” cm

So, the semi-perimeter of the equilateral triangle = (3x/2) cm

And, the area of the equilateral triangle = (√3/4) × x²

We know, r = Area of triangle/semi-perimeter

So, r = [x²(√3/4)/ (3x/2)]

=> 7 = √3x/6

Or, x = 42/√3

Multiply both numerator and denominator by √3

So, x = 42√3/3 = 14√3 cm

Now, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm

Q10: Find the Radius of a circle whose Circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm. (3 Marks)

Ans:

Radius of the first circle = r1 = 15 cm

Radius of the second circle = r2 = 18 cm

∴ Circumference of the first circle = 2πr1 = 30π cm

Circumference of the second circle = 2πr2 = 36π cm

assume, the radius of the circle = R

As per the question,

Circumference of circle = Circumference of the first circle + Circumference of the second circle

2πR= 2πr1+ 2πr2

⇒ 2πR = 30π + 36π

⇒ 66π ⇒ R = 33

⇒ Radius = 33 cm

Hence, the required radius of a circle is 33 cm.

Q11: Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm² (3 Marks)

Ans: 

Here, as the equilateral triangle is inscribed in a circle, the circle is an incircle.

Now, the radius of the incircle is given by,

r = Area of triangle/semi-perimeter

In the question, it is given that area of the incircle = 154 cm²

So, π × r² = 154

Or, r = 7 cm

Now, assume the length of each arm of the equilateral triangle to be “x” cm

So, the semi-perimeter of the equilateral triangle = (3x/2) cm

And the area of the equilateral triangle = (√3/4) × x²

We know, r = Area of triangle/semi-perimeter

So, r = [x²(√3/4)/ (3x/2)]

⇒ 7 = √3x/6

Or x = 42/√3

Multiply both numerator and denominator by √3

So, x = 42√3/3 = 14√3 cm

Now, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm

Q12: A chord of a circle of Radius 10 cm subtends a right angle at the center. Find the Area of the corresponding: (5 Marks)

(i) minor segment

(ii) major sector. (Use π = 3.14)

 Ans:

AB is the chord that subtends an angle of 90° at the center O.

It is seen that the radius (r) of the circle = 10 cm

(i) Area of the minor sector = (90/360°) × πr²

= (1/4) × (22/7) × 10²

Or, Area of the minor sector = 78.5 cm2

Also, the Area of the ΔAOB = 1/2 × OB × OA

Where, OB and OA are the radii of the circle i.e., = 10 cm

Thus, the Area of ΔAOB = 1/2×10×10

= 50 cm²

Then, Area of minor segment = area of the minor sector – Area of ΔAOB

= 78.5 – 50

= 28.5 cm²

(ii) Area of the major sector = Area of circle – Area of minor sector

= (3.14×102) - 78.5

= 235.5 cm²   

Q13: A chord of the circle of Radius 15 cm makes the angle of 60° at the center. Find out the areas for the corresponding minor and the major segments of the circle. (Use π = 3.14 and √3 = 1.73) (5 Marks)

Ans:

We know that

Radius = 15 cm

θ = 60°.

Hence,

The Area of the sector OAPB = (60°/360°) × πr² cm²

= 225/6 πcm².

Then, ΔAOB is equilateral as two sides are the circle’s radii and, 

Therefore, equal, and one angle is 60°.

So, the Area of ΔAOB = (√3/4) ×a²

And (√3/4) ×15²

∴ Area of ΔAOB = 97.31 cm².

Then, the Area of minor segment APB = Area of OAPB – Area of ΔAOB

And, area of minor segment APB = ((225/6)π – 97.31) cm² = 20.43 cm²

Or

The Area of major segment = Area of the circle – Area of segment APB

And area of major segment = (π×152) – 20.4 = 686.06 cm²