## Chapter 12 Areas Related to Circles Exercise 12.1

**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 2 circles.**

Answer:

Answer:

Let the radius of circle, whose radius is 9 cm be r and the radius of circle, whose radius is 19 cm be r'. Also let the radius of circle, whose circumference is equal to sum of circumferences of both circles having radii 19 cm and 9 cm respectively, be R.

Therefore, according to question,

2πR = 2πr + 2πr'

2 x 22/7 x R = 2π(r + r')

2 x 22/7 x R = 2 x 22/7 (9 + 19)

44/7 x R = 44/7 (28)

R = 28 cm (cancelling 44/7 on both sides)

Hence, radius of circle whose circumference is equal to sum of circumferences of both circles is 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.**

Answer:

Answer:

Given:

Radii of 2 circles i.e. 8 cm and 6 cm

Required to find:

Radius of circle having area equal to sum of the areas of the 2 circles

Now,

Let the radius of circle, whose radius is 6 cm be r and the radius of circle, whose radius is 8 cm be r'. Also let the radius of circle whose area is equal to sum of the areas of 2 circles, be R.

Therefore, according to question:

πR² = πr² + πr'²

= πR² = π(r² + r'²)

= R² = r² + r'²

= R = √[(6)² + (8)²]

= R = √[36 + 64]

= R = √100

= R = 10 cm

Hence, the radius of circle whose area is equal to sum of the areas of 2 circles is 10 cm.

**Question 3: Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre 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 10.5cm wide. Find the area of each of the five scoring regions.**

**Answer:**Diameter of gold circle = 21 cm

Therefore, the radius of gold circle = 21/2cm = 10.5 cm

Since the radius of scores (red, blue, black, white) after gold increases by 10.5 cm.

Therefore,

• Radius of red region = 10.5 + 10.5

= 21 cm

• Radius of blue region = 21 + 10.5

= 31.5 cm

• Radius of black region = 31.5 + 10.5

= 42 cm

• Radius of white region = 42 + 10.5

= 52.5 cm

Therefore,

Area of gold circle = πr²

= 22/7 x (10.5)²

= 346.5 cm²

Area of red region = Area of red region - area of gold circle

= πr² - 346.5

= 22/7 x (21)² - 346.5

= 1386 - 346.5

= 1039.5 cm²

Area of blue region = Area of blue region - (area of red region + area of gold circle)

= πr² - (1039.5 + 346.5)

= 22/7 x (31.5)² - 1386

= 3118.5 - 1386

= 1732.5 cm²

Area of black region = Area of black region - (area of blue region + area of red region + area of gold circle)

= πr² - (1735.5 + 1039.5 + 346.5)

= 22/7 x (42)² - 3118.5

= 5544 - 3118.5

= 2425.5 cm²

Area of white region = Area of white region - (area of black region + area of blue region + area of red region + area of gold circle)

= πr² - (2425.5 + 1732.5 + 1039.5 + 346.5)

= 22/7 x (52.5)² - 5544

= 8662.5 – 5544

= 3118.5 cm²

**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?**

Answer:

Answer:

Diameter of wheels, d = 80 cm

∴ Radius, r = d/2 ⇒ 80/2 ⇒ 40 cm

Circumference of wheel = 2πr

= 2 x π x 40

= 80π

Distance travelled by car in 60 minutes

= 66 x 1000

= 66000 m

Distance travelled by 1 minute

= 66000/60

= 1100 m

Distance travelled by 10 minutes

= 1100 x 10

= 11000 m or 1100000 cm

∴ Total number of revolutions = Distance travelled/Circumference of wheel

= (1100000)/(80 x 22/7)

= (1100000 x 7)/(80 x 22)

= 7700000/1760

= 4375

Hence the car takes 4375 revolutions

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

a) 2 units b) π units c) 4 units d) 7 units

Answer:

a) 2 units b) π units c) 4 units d) 7 units

Answer:

Since the area and perimeter of circle are equal

= πr² = 2πr

= r² = 2r (cancelling π and r on both sides)

= r = 2 cm

Hence, option a) is correct.

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