The important class 10 maths questions for chapter 12 i.e. areas related to circles provided here covers wide variations of questions from this chapter to help the students get completely acquainted with this chapter and thus, develop problem-solving skills. These questions from class 10 areas related to circles chapter from NCERT book can help students to score well in their CBSE class 10 maths exam.

**Also Check:**

- Important 2 Marks Questions for CBSE 10th Maths
- Important 3 Marks Questions for CBSE 10th Maths
- Important 4 Marks Questions for CBSE 10th Maths

## Important Areas Related to Circles Questions for Class 10- Chapter 12 (With Solutions)

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}

= 36/6 π 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} (self-assessment)

**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 cm^{2}

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.

### More Topics Related to Class 10 Areas Related to Circles

Revision Notes For CBSE Class 10 Maths Chapter 12 Areas Related to Circles |
NCERT Solutions For Class 10 Maths Chapter 12 – Areas Related to Circles |

### Extra Questions For Class 10 Chapter 12: Areas Related to Circles (NCERT)