NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.1

NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.1 are designed to help students understand how to calculate areas related to circles, including sectors and segments. The exercise focuses on solving problems involving the area of a circle, sector, and segment using trigonometric concepts and basic geometry.

Exercise 12.1 helps students practice:

NCERT Solutions Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.1

Area Related to Circles

Medium

Q.

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.

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Area Related to Circles

Medium

Q.

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

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Area Related to Circles

Medium

Q.

The following figure depicts an archery target marked with its five scoring areas 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.

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Area Related to Circles

Medium

Q.

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?

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Area Related to Circles

Medium

Q.

Tick the correct answer in the following and justify your choice: If the perimeter and the 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

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• Area of a circle:

  $A = \pi r^2$A=πr2

• Area of a sector:

  $A = \frac{\theta}{360^\circ} \times \pi r^2$A=360∘θ​×πr2 (where

  $\theta$θ is the angle in the sector)

• Area of a segment: Area of the sector minus the area of the triangle formed by the center and the endpoints of the chord.

The solutions provided are clear, step-by-step, making it easier for students to understand how to apply these formulas to solve real-life problems and prepare for their Class 10 exams.

Q1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Difficulty Level: Easy
What is the known/given?

• Radii of two circles.

What is unknown?

• Radius of the 3rd circle.

Reasoning:

• Using the formula for the circumference of a circle, we find the radius of the circle.

Solution:
Radius of the 1st circle = 19 cm
Radius of the 2nd circle = 9 cm

Let the radius of the 3rd circle be

$r$

r.
Circumference of the 1st circle =

$2\pi \times 19 = 38\pi$

2π×19=38π
Circumference of the 2nd circle =

$2\pi \times 9 = 18\pi$

2π×9=18π
Circumference of the 3rd circle =

$2\pi r$

2πr

Given that:
Circumference of the 3rd circle = Circumference of the 1st circle + Circumference of the 2nd circle

$$2\pi r = 38\pi + 18\pi = 56\pi$$

2πr=38π+18π=56π

$$r = \frac{56\pi}{2\pi} = 28$$

r=2π56π​=28

Therefore, the radius of the circle which has a circumference equal to the sum of the circumferences of the two given circles is 28 cm.

Q2. 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.

Difficulty Level: Easy
What is the known/given?

• Radii of two circles.

What is unknown?

• Radius of the 3rd circle.

Reasoning:

• Using the formula for the area of a circle, we find the radius of the circle.

Solution:
Radius of the 1st circle = 8 cm
Radius of the 2nd circle = 6 cm

Let the radius of the 3rd circle be

$r$

r.
Area of the 1st circle =

$\pi \times 8^2 = 64\pi$

π×82=64π
Area of the 2nd circle =

$\pi \times 6^2 = 36\pi$

π×62=36π
Area of the 3rd circle =

$\pi r^2$

πr2

Given that:
Area of the 3rd circle = Area of the 1st circle + Area of the 2nd circle

$$\pi r^2 = 64\pi + 36\pi = 100\pi$$

πr2=64π+36π=100π

$$r^2 = 100$$

r2=100

$$r = 10 \, \text{cm}$$

r=10cm

Therefore, the radius of the circle having an area equal to the sum of the areas of the two given circles is 10 cm.

Q3. Given figure depicts an archery target marked with its five scoring areas from the center outwards as Gold, Red, Blue, Black, and White. The diameter of the region representing the 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.

Difficulty Level: Medium
What is the known/given?

• Diameter of the gold region and width of the other regions.

What is unknown?

• Area of each scoring region.

Reasoning:

• The area of the region between two concentric circles is given by

  $\pi (r_2^2 - r_1^2)$π(r22​−r12​).

Solution:
Radius of the gold region (1st circle) =

$\frac{21}{2} = 10.5 \, \text{cm}$

221​=10.5cm
Given that each circle is 10.5 cm wider than the previous one.
Therefore:
Radius of the 2nd circle =

$10.5 + 10.5 = 21 \, \text{cm}$

10.5+10.5=21cm
Radius of the 3rd circle =

$21 + 10.5 = 31.5 \, \text{cm}$

21+10.5=31.5cm
Radius of the 4th circle =

$31.5 + 10.5 = 42 \, \text{cm}$

31.5+10.5=42cm
Radius of the 5th circle =

$42 + 10.5 = 52.5 \, \text{cm}$

42+10.5=52.5cm

Now, calculate the areas of each region:

• Area of the gold region:

$$\text{Area of 1st circle} = \pi \times (10.5)^2 = 346.5 \, \text{cm}^2$$

Area of 1st circle=π×(10.5)2=346.5cm2

• Area of the red region:

$$\text{Area of 2nd circle} = \pi \times (21)^2 = 441\pi \, \text{cm}^2$$

Area of 2nd circle=π×(21)2=441πcm2

$$\text{Area of red region} = 441\pi - 346.5\pi = 330.75\pi \, \text{cm}^2$$

Area of red region=441π−346.5π=330.75πcm2

• Area of the blue region:

$$\text{Area of 3rd circle} = \pi \times (31.5)^2 = 992.25\pi \, \text{cm}^2$$

Area of 3rd circle=π×(31.5)2=992.25πcm2

$$\text{Area of blue region} = 992.25\pi - 441\pi = 551.25\pi \, \text{cm}^2$$

Area of blue region=992.25π−441π=551.25πcm2

• Area of the black region:

$$\text{Area of 4th circle} = \pi \times (42)^2 = 1764\pi \, \text{cm}^2$$

Area of 4th circle=π×(42)2=1764πcm2

$$\text{Area of black region} = 1764\pi - 992.25\pi = 771.75\pi \, \text{cm}^2$$

Area of black region=1764π−992.25π=771.75πcm2

• Area of the white region:

$$\text{Area of 5th circle} = \pi \times (52.5)^2 = 2756.25\pi \, \text{cm}^2$$

Area of 5th circle=π×(52.5)2=2756.25πcm2

$$\text{Area of white region} = 2756.25\pi - 1764\pi = 992.25\pi \, \text{cm}^2$$

Area of white region=2756.25π−1764π=992.25πcm2

Thus, the areas of the gold, red, blue, black, and white regions are approximately:
346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm², and 3118.5 cm².

Q4. 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 traveling at a speed of 66 km per hour?

Difficulty Level: Medium
What is the known/given?

• Diameter of the wheel of the car and the speed of the car.

What is unknown?

• Revolutions made by each wheel.

Reasoning:

• The distance travelled by the wheel in one revolution is equal to the circumference of the wheel itself.

Solution:
Diameter of the wheel of the car = 80 cm
Radius (r) of the wheel of the car =

$\frac{80}{2} = 40 \, \text{cm}$

280​=40cm
Distance travelled in 1 revolution = Circumference of wheel

$$\text{Circumference of wheel} = 2\pi r = 2\pi \times 40 = 80\pi \, \text{cm}$$

Circumference of wheel=2πr=2π×40=80πcm

Speed of the car = 66 km/hour
Convert this to cm/min:

$$66 \, \text{km/hour} = 66 \times 1000 \times 100 \, \text{cm/hour} = 6600000 \, \text{cm/hour}$$

66km/hour=66×1000×100cm/hour=6600000cm/hour

$$\text{Speed in cm/min} = \frac{6600000}{60} = 110000 \, \text{cm/min}$$

Speed in cm/min=606600000​=110000cm/min

Distance travelled by the car in 10 minutes:

$$\text{Distance} = 110000 \times 10 = 1100000 \, \text{cm}$$

Distance=110000×10=1100000cm

Let the number of revolutions of the wheel of the car be

$n$

n.

$$n \times 80\pi = 1100000$$

n×80π=1100000

$$n = \frac{1100000}{80\pi} = 4375 \, \text{revolutions}$$

n=80π1100000​=4375revolutions

Therefore, each wheel of the car will make 4375 revolutions.

Q5. Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is (A) 2 units (B)

$\pi$

π units (C) 4 units (D) 7 units

Difficulty Level: Easy
What is the known/given?

• Perimeter and area of the circle are numerically equal.

What is unknown?

• Radius of the circle.

Reasoning:

• Given that perimeter and area of the circle are numerically equal, we get

  $2\pi r = \pi r^2$2πr=πr2. Using this relation, we find the radius.

Solution:
Let the radius of the circle =

$r$

r.
Circumference of circle =

$2\pi r$

2πr
Area of circle =

$\pi r^2$

πr2

Given that the circumference of the circle and the area of the circle are equal,

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

2πr=πr2

$$2r = r^2$$

2r=r2

$$r = 2$$

r=2

Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.1

Q1. What is the focus of Exercise 12.1?
Answer:
Exercise 12.1 focuses on calculating the area of circles, sectors, and segments. It involves applying the formulas for the area of a circle, sector, and segment to solve problems.

Q2. How do I calculate the area of a circle?
Answer:
The area of a circle is given by the formula:

$$A = \pi r^2$$

A=πr2

where

$r$

r is the radius of the circle.

Q3. How is the area of a sector of a circle calculated?
Answer:
The area of a sector is calculated using the formula:

$$A = \frac{\theta}{360^\circ} \times \pi r^2$$

A=360∘θ​×πr2

where

$\theta$

θ is the central angle of the sector in degrees, and

$r$

r is the radius of the circle.

Q4. How do I find the area of a segment of a circle?
Answer:
To find the area of a segment:

1. Calculate the area of the sector using the formula:

    

   $$A = \frac{\theta}{360^\circ} \times \pi r^2$$A=360∘θ​×πr2

2. Subtract the area of the triangle formed by the radius and the chord from the area of the sector.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, detailed steps for solving problems involving the area of a circle, sector, and segment. By practicing these problems, students can improve their understanding of these concepts and apply them confidently in board exams.