NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.1

NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.1 helps in understanding the areas and perimeters related to circles. The chapter explains concepts of sectors, segments, and sector angles; however, the exercise mainly focuses on the areas and perimeters of circles. These formulas and concepts are not only useful for comprehension of other higher-level topics in mathematics but also crucial for the solution of real-world problems. Let us start exploring this key Class 10 topic, which is crucial for both higher-level mathematics and exams.

1.0Download NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.1 : Free PDF 

2.0Introduction to Circles: 

Before diving into the formulas and concepts of area and perimeter, let’s take a quick look at circles. A circle is a two-dimensional geometric figure with all the points at equal distances from a constant point known as the centre. These equal distances from the centre to any point at the edge of the circle are known as the radius of the circle. The line passing through the centre of the circle, doubling the radius, is known as the diameter of the circle. Circles are one of the most common geometric figures, with applications in daily life, such as wheels and circular tables. 

3.0Exercise 11.1 Overview: Key Concepts

Perimeter and Area of a Circle — A Review 

1. The Perimeter of a Circle: 

First, let us recall a fundamental property of a circle: its perimeter. The perimeter of a circle is also called the circumference, which is the total distance around the circle. In the exercise, you will be solving many questions which use the formula for the circumference of the circle. The formula for the perimeter or circumference can be expressed as: 

$Circumference=2\pi r$

Here, r is the radius of the circle and is a constant value equal to $\frac{22}{7}$ or 3.14. 

2. Area of a Circle:

The area is the total space enclosed within the boundary of a geometric figure. Similarly, the area of a circle is the total space enclosed within the circular boundaries of the circle. The area of a circle is directly proportional to the square of the radius of the circle; in simple terms, the area of a circle is related to the size of the circle. The formula for the area of a circle is primarily used to solve most of the questions of exercise, which can be expressed as: 

$\text{Area of a Circle}=\pi r^2$

Here also, r is the radius of the circle and is a constant value equal to $\frac{22}{7}$ or 3.14. The unit of area of a circle is the square of the unit of radius; for example, if the unit of radius is given in meter (m), then the area will be measured in meter square (m²). 

When Area and Perimeter are Numerically Equal:

In the exercise, the relation between area and perimeter being equal is also used in some questions. This intriguing problem pushes the boundaries of properties for geometric figures in such a way that they work in a balanced proportion. This simply gives rise to the following mathematical relations, which is given as: 

$2\pi r=\pi r^2$

Interestingly, when the area and perimeter of a circle are numerically equal, the radius of the circle is 2 units. This answer holds true for every circle with any unit with a radius of 2 units. 

To increase your score in the exercise, start practising today from the NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.1 from the right place.    

4.0NCERT Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.1 : Detailed Solutions

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution:

The circumference of the circle having radius 19 cm = 2π × 19 cm = 38π cm

The circumference of the circle having radius 9 cm = 2π × 9 cm = 18π cm

Sum of the circumferences of the two circles = (38π + 18π) cm = 56π cm

Therefore, if r cm be the radius of the circle which has circumference equal to the sum of the circumference of the two given circles, then

2πr = 56π => r = 28

Hence, the required radius is 28 cm.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Solution:

We have,

Radius of circle-I, r1 = 8 cm

Radius of circle-II, r2 = 6 cm

∴ Area of circle-I = πr1² = π(8)² cm²

Area of circle-II = πr2² = π(6)² cm²

Let the radius of the circle-III be R

∴ Area of circle-III = πR²

=> π(8)² + π(6)² = πR²

=> π(8² + 6²) = πR²

=> 8² + 6² = R²

=> 64 + 36 = R²

=> 100 = R²

=> 10² = R² => R = 10 cm

Thus, the radius of the new circle = 10 cm.

3. The fig. depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:

Radius of the Gold scoring region = 21/2 cm = 10.5 cm (Since Diameter = 21 cm)

Therefore, the area of the Gold scoring region (circle) = π × (21/2)² cm² = (22/7) × (21/2) × (21/2) cm²

= (33 × 21)/2 cm² = 693/2 cm² = 346.5 cm²

Radius of the combined circular region scoring Gold and Red = radius of Gold scoring region + width of the Red scoring band = 10.5 cm + 10.5 cm = 21 cm

The area of the Red scoring region = Combined area of Gold and Red scoring region - area of Gold scoring region.

= {π × (21)² - π × (21/2)²} cm²

= π × {(21)² - (21/2)²} cm²

= (22/7) × {(21/2 × 2)² - (21/2)²} cm²

= (22/7) × (21/2)² × (4 - 1) cm²

= (22/7) × (21/2) × (21/2) × 3 cm²

= (693/2) × 3 cm² = 2079/2 cm² = 1039.5 cm²

Radius of the combined circular region scoring Gold, Red and Blue = 63/2 cm

Then, area of the Blue scoring region = {Combined area of Gold, Red and Blue scoring regions} - {Combined area of Gold and Red scoring regions}

= π × (63/2)² - π × (21)²

= π × (21)² × (9/4 - 1)

= (22/7) × 21 × 21 × (5/4) cm² = (11 × 63 × 5)/2 cm²

= 1732.5 cm²

Similarly, we find the area of the Black scoring region

= {π × (31.5 + 10.5)² - π × (31.5)²} cm²

= {π × (42)² - π × (63/2)²} cm²

= π × (21)² × {4 - 9/4} cm²

= (22/7) × 21 × 21 × (7/4) cm²

= (231 × 21)/2 cm² = 4851/2 cm² = 2425.5 cm²

Now, the area of the White scoring region

= {π × (42 + 10.5)² - π × (42)²} cm²

= {π × (105/2)² - π × (42)²} cm²

= π × (21)² × {25/4 - 4} cm²

= (22/7) × 21 × 21 × (9/4) cm²

= 6237/2 cm² = 3118.5 cm²

4. The wheels of a car are 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:

Diameter of the wheel of the car = 80 cm

Then, the radius of the wheel of the car = 40 cm = 0.4 m

Distance travelled by the car when the wheel of the car completes one revolution = 2π × (0.4) m = (4π/5) m

Let us suppose the wheel of the car completes n revolutions in 10 minutes when travelling at the speed of 66 km per hour. Then distance travelled by making a complete revolution of the wheel in 10 minutes = (4π/5 × n) m

Also, distance travelled in 60 minutes = 66 km = 66 × 1000 m

Then the distance travelled in 10 minutes = (66 × 1000 / 60) × 10 m = 11000 m

Therefore, we have (4π/5) × n = 11000 (∵ Distance travelled in 10 minutes is same)

=> (4/5) × (22/7) × n = 11000

=> n = (11000 × 5 × 7) / (4 × 22) = 125 × 35 = 4375

Hence, the number of complete revolutions made by the wheel in 10 minutes is 4375.

5. Tick the correct answer in the following and justify your choice. If the perimeter and area of a circle are numerically equal, then the radius of the circle is

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

Solution:

(A) We have

πr² = 2πr

=> πr² - 2πr = 0

=> r² - 2r = 0

=> r(r - 2) = 0

=> r = 0 or r = 2

But r cannot be zero

∴ r = 2 units.

Thus, the radius of the circle is 2 units.

5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.1

• Helps understand the formulas for the area and circumference of circles, sectors, and segments.
• Covers important board exam questions for scoring well in geometry-related problems.
• Helps in the preparation of NTSE, Olympiads, and entrance exams that involve mensuration.
• Summarized solutions and key formulas make last-minute revision easier before exams.