Important questions for Class 10 Maths Chapter 12 Areas Related to Circles are given here based on the new pattern of CBSE for 2019-2020. Students who are preparing for the **board exams of 2020** can practice these questions of Areas Related to Circles For Class 10 to score full marks for the questions from this chapter.

This chapter contains many formulas and concepts thus, it is important for students from the examination perspective since most of the questions and objective-based questions will come for the exam from this chapter will be formula based. Students can also refer to these important questions as a part of their exam preparation. NCERT Solutions are also available at BYJU’S to help the students in scoring full marks.

**Read more:**

- Area Perimeter Formula
- Important 4 Marks Question For Cbse Class 10 Maths
- Maths Formulas For Class 10

## Important Questions & Answers For Class 10 Maths Chapter 12 Areas Related to Circles

Some of the important questions for areas related to circles for class 10 chapter 12 are given below. The questions include both short-answer type questions and long answer type questions along with HOTS questions.

### Short Answer Type Questions

**1. If the radius of a circle is 4.2 cm, compute its area and circumference.**

**Solution:**

Area of a circle = πr^{2}

So, area = π(4.2)^{2} = 55.44 cm^{2}

Circumference of a circle = 2πr

So, circumference = 2π(4.2) = 26.4 cm

**2. What is the area of a circle whose circumference is 44 cm?**

**Solution:**

Circumference of a circle = 2πr

From the question,

2πr = 44

Or, r = 22/π

Now, area of circle = πr^{2} = π × (22/π)^{2}

So, area of circle = (22×22)/π = 22 × 7 = 154 cm^{2}

**3. Calculate the area of a sector of angle 60°. Given, the circle is having a radius of 6 cm.**

**Solution:**

Given,

The angle of the sector = 60°

Using the formula,

The area of sector = (θ/360°)×π r^{2}

= (60°/360°) × π r^{2} cm^{2}

Or, area of the sector = 6 × 22/7 cm^{2} = 132/7 cm^{2}

**4. A chord subtends an angle of 90°at the centre of a circle whose chord is 20 cm. Compute the area of the corresponding major segment of the circle.**

**Solution:**

Point to note:

Area of the sector = θ/360 × π × r^{2}

Base and height of the triangle formed will be = radius of the circle

Area of the minor segment = area of the sector – area of the triangle formed

Area of the major segment = area of the circle – area of the minor segment

Now,

Radius of circle = r = 20 cm and

Angle subtended = θ = 90°

Area of the sector = θ/360 × π × r^{2} = 90/360 × 22/7 × 20^{2}

Or, area of the sector = 314.2 cm^{2}

Area of the triangle = ½ × base × height = ½ × 20 × 20 = 200 cm^{2}

Area of the minor segment = 314.2 – 200 = 114.2 cm^{2}

Area of the circle = π × r^{2}

Area of the major segment = π × r^{2} – 114.2 = 1142 .94 cm^{2}

So, the area of the corresponding major segment of the circle = 1142 .94 cm^{2}

**5. A square is inscribed in a circle. Calculate the ratio of the area of the circle and the square.**

**Solution:**

As the square is inscribed in a circle, a diagonal of the square will be = the diameter of the circle.

Let “r” be the radius of the circle and “d” be the length of each diagonal of the square.

We know,

Length of the diagonal of a square = side (s) × √2

So,

d = 2r

And, s × √2 = 2r

Or, s = √2r

We know, the area of the square = s^{2}

Thus, the area of the square = (√2r)^{2} = 2r^{2}

Now, the area of the circle = π × r^{2}

∴ Area of the circle : area of the square = π × r^{2} : 2r^{2} = π : 2

So, the ratio of the area of the circle and the square is π : 2.

**6. Find the area of the sector of a circle with radius 4cm and of angle 30°. Also, find the area of the corresponding major sector.**

**Solution:**

46.05 cm^{2} (Try yourself)

**7. A vehicle is moving at a speed of 66 km/hr. Find the number of complete revolutions made by each wheel in 10 minutes if the diameter of each wheel is 80 cm.**

**Solution:**

Here, the circumference of the wheel (circle) will give the distance covered in one revolution.

No. of complete revolutions in 10 minutes = Distance covered in 10 mins/Circumference of wheel

It is given that the speed of the vehicle is 66 km/ hr = (66 × 5/18) m/s = 55/3 m/s

So, distance covered in 10 mins = (55/3 × 10 × 60)m = 11,000 m

Circumference of the wheel = 2πr = 2π(80/2) = (80π) cm = (0.8π) m

Now, the number of complete revolutions in 10 minutes = 11,000/0.8π = 4376.76

**8. Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm ^{2}**

**Solution:**

Here, as the equilateral triangle 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 × r^{2} = 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) × x^{2}

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

So, r = [x^{2}(√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.

### Long Answer Type Questions

**Q.1:** **The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 per m ^{2}. Find the cost of ploughing the field (Take π = 22/7).**

Solution:

Length of the fence (in metres) = Total cost/Rate = 5280/24 = 220

So, the circumference of the field = 220 m

If r metres is the radius of the field, then 2πr = 220

2 × (22/7) × r = 220

r = (220 × 7)/ (2 × 22)

r = 35

Hence, the radius of the field = 35 m

Area of the field = πr^{2}

= (22/7) × 35 × 35

= 22 × 5 × 35 m^{2}

= 3850 sq.m.

Cost of ploughing 1 m^{2} of the field = Rs. 0.50

So, total cost of ploughing the field = 3850 × Rs. 0.50 = Rs. 1925

**Q.2: The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?**

**Solution: **

The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km

Converting km into cm we get,

Distance covered by the car in 1hr = (66 × 10^{5}) cm

In 10 minutes, the distance covered will be = (66 × 10^{5} × 10)/60 = 1100000 cm/s

∴ Distance covered by car = 11 × 10^{5} cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) = 11 × 105 /80 π = 4375.

**Q.3: Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14)**

**Solution: **

Let OAPB be the sector.

Area of the major sector = [(360 – θ)/ 360] × πr^{2}

=[(360 – 30)/360] × 3.14 × 4 × 4

= (330/360) × 3.14 × 16

= 46.05 cm^{2}

= 46.1 cm^{2} (approx)

**Q.4: Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and ∠ AOB = 120°. (Use π = 22/7).**

**Solution: **

Area of the segment AYB = Area of sector OAYB – Area of ∆ OAB …..(1)

Area of the sector OAYB = (120/360) × (22/7) × 21 × 21 = 462 cm^{2} ……(2)

Draw OM ⊥ AB.

OA = OB (radius)

Therefore, by RHS congruence, ∆ AMO ≅ ∆ BMO.

M is the mid-point of AB and ∠ AOM = ∠ BOM = (1/2) × 120° = 60°

Let OM = x cm

In triangle OMA,

OM/OA = cos 60°

x/21 = ½

x = 21/2

OM = 21/2 cm

Similarly,

AM/OA = sin 60°

AM/21 = √3/2

AM = 21√3/2 cm

AB = 2 AM = 2 (21√3/2) = 21√3 cm

Area of triangle OAB = (½) × AB × OM

= (½) × 21√3 × (21/2)

= (441/4)√3 cm^{2} ………..(3)

From (1), (2) and (3),

Area of the segment AYB = [462 – (441/4)√3] = (21/4) [88 – 21√3] cm^{2}

**Q.5: Find the area of the shaded design in given figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14).**

**Solution: **

Let us assign I, II, III and IV for the unshaded regions.

Give that, side of square ABCD = 10 cm

Sides of a square are also the diameters of semicircles.

The radius of semicircle = 10/2 = 5 cm

Now, area of the region I +III = Area of square ABCD – Area of two semicircles of radius 5 cm

= (10)^{2} – 2 × (½) π ×(5)^{2}

= 100 – 3.14 × 25

= 100 – 78.5

= 21.5 cm^{2}

Similarly,

Area of the region II + Iv = 21.5 cm^{2}

Area of the shaded region = Area of square ABCD – Area of the region (I + II + III + IV)

= 100 – 2× 21.5

= 100 – 43

= 57 cm^{2}

**Q.6: A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm ^{2}. (Use 3 = 1.7)**

**Solution:**

Total number of equal designs = 6

∠AOB = 360°/6 = 60°

The radius of the cover = 28 cm

Cost of making design = Rs. 0.35 per cm^{2}

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be √3/4 × a^{2}

Here, a = OA

∴ Area of equilateral ΔAOB = √3/4 × 282 = 333.2 cm^{2}

Area of sector ACB = (60°/360°) × π r^{2} cm^{2}

= 410.66 cm^{2}

Area of a single design = area of sector ACB – Area of ΔAOB

= 410.66 cm^{2} – 333.2 cm^{2} = 77.46 cm^{2}

∴ Area of 6 designs = 6 × 77.46 cm^{2} = 464.76 cm^{2}

So, the total cost of making design = 464.76 cm^{2} × Rs. 0.35 per cm^{2}

= Rs. 162.66

**Q.7: In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.**

**Solution:**

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm^{2}

Area of larger circle = πR^{2} = 22/7 × 72 = 154 cm^{2}

Area of larger semicircle = 154/2 cm2 = 77 cm^{2}

Area of smaller circle = πr^{2} = 22/7 × 7/2 × 7/2 = 77/2 cm^{2}

Area of the shaded region = Area of the larger circle – Area of a triangle – Area of larger semicircle + Area of the smaller circle

Area of the shaded region = (154 – 49 – 77 + 77/2) cm^{2}

= 133/2 cm^{2} = 66.5 cm^{2}

**Q.8: Area of the largest triangle that can be inscribed in a semi-circle of radius r units is**

**(A) r ^{2} sq. units (B) ½ r^{2} sq. units**

**(C) 2 r ^{2} sq. units (D) √2 r^{2} sq. units**

**Solution:**

The largest triangle that can be inscribed in a semi-circle of radius r units is the triangle having its base as the diameter of the semi-circle and the two other sides are taken by considering a point C on the circumference of the semi-circle and joining it by the endpoints of diameter A and B.

∴ ∠ C = 90° (by the properties of the circle)

So, ΔABC is a right-angled triangle with a base as diameter AB of the circle and height be CD.

Height of the triangle = r

∴ Area of largest ΔABC = (1/2)× Base × Height = (1/2)× AB × CD

= (1/2)× 2r × r = r^{2} sq. units

∴ Option A is correct.

### Practice Questions For Class 10 Maths Chapter 12 Areas Related to Circles

- A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in the area of the grassy lawn in which the calf can graze.
- 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.
- In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

- Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.