In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 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:

We have, r_{1} = 19 cm and r_{2} = 9 cm

∴ Circumference of circle – I = 2πr_{1} = 2π(19) cm

and circumference of circle – II = 2πr_{2} = 2π(9) cm

Sum of the circumferences of circle-I and circle-II

= 2π(19) cm + 2π(9) cm

= 2π(19 + 9) cm

= 2π(28) cm

Let R be the radius of the circle – III.

∴ Circumference of circle – III = 2πR

According to the condition, 2πR = 2π(28)

⇒ R = \(\frac{2 \pi(28)}{2 \pi}\) = 28 cm

Thus, the radius of the new circle = 28 cm.

Question 2.

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.

Solution:

We have,

Radius of circle – I, r_{1} = 8 cm

Radius of circle – II, r_{2} = 6 cm

∴ Area of circle – I = πr_{1}^{2} = π(8)^{2} cm^{2}

Area of circle-II = πr_{2}^{2} = π(6)^{2} cm^{2}

Let the radius of the circle – III be R cm.

∴ Area of circle-III = πR^{2}

Now, according to the condition,

πr_{1}^{2} + πr_{2}^{2} = πR^{2}

⇒ π(8)^{2} + π(6)^{2} = πR^{2}

⇒ π(8^{2} + 6^{2}) = πR^{2}

⇒ 8^{2} + 6^{2} = R^{2}

⇒ 64 + 36 = R^{2}

⇒ 100 = R^{2}

⇒ 10^{2} = R^{2} ⇒R = 10

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

Question 3.

The given figure 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:

Diameter of the innermost (Gold scoring) region = 21 cm

Radius of the innermost (Gold scoring) region = \(\frac{21}{2}\) = 10.5 cm

∴ Area of Gold scoring region = π(10.5)^{2} cm^{2}

= \(\frac{22}{7} \times\left(\frac{105}{10}\right)^{2} \mathrm{cm}^{2}=\frac{22}{7} \times \frac{105}{10} \times \frac{105}{10} \mathrm{cm}^{2}\)

= \(\frac{22 \times 15 \times 105}{100}\) cm^{2} = 346.50 cm^{2}

Since, each band is 10.5 cm wide.

∴ Radius of Red scoring region

= 10.5 cm + 10.5 cm

= 21 cm

Area of Red scoring region

= π(10.5 + 10.5)^{2} cm^{2} – π(10.5)^{2} cm^{2}

= [π(21)^{2} – π(10.5)^{2}] cm^{2}

= π[(21)^{2} – (10.5)^{2}] cm^{2}

Area of White scoring region

= π[(42 + 10.5)^{2} – (42)^{2}] cm^{2}

= π[(52.5)^{2} – (42)^{2}] cm^{2}

= π[(52.5 + 42)(52.5 – 42)] cm^{2}

= \(\frac{22}{7}\) × 94.5 × 10.5 = 22 × \(\frac{945}{10} \times \frac{15}{10}\)

= 3118.5 cm^{2}

Question 4.

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:

Diameter of a wheel = 80 cm

∴ Radius of the wheel = \(\frac{80}{2}\) cm = 40 cm

So, circumference of the wheel

= 2πr = 2 × \(\frac{22}{7}\) × 40 cm

⇒ Distance covered by a wheel in one revolution = \(\frac{2 \times 22 \times 40}{7}\) cm

Distance travelled by the car in 1 hour (i.e., in 60 mins)

= 66 km = 66 × 1000 × 100 cm

∴ Distance travelled in 10 minutes

Thus, the required number of revolutions = 4375

Question 5.

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

Solution:

(A): We have

Numerical area of the circle

= Numerical circumference of the circle

⇒ πr^{2} = 2πr

⇒ πr^{2} – 2πr = 0

⇒ r^{2} – 2r = 0

⇒ r(r – 2) = 0

⇒ r = 0 or r = 2

But r cannot be zero

∴ r = 2

Thus, the radius of circle is 2 units.