**Exercise: 12.1 (Page No: 230)**

**1. 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.**

**Solution:**

The radius of the 1^{st} circle = 19 cm (given)

∴ Circumference of the 1^{st} circle = 2π×19 = 38π cm

The radius of the 2^{nd} circle = 9 cm (given)

∴ Circumference of the 2^{nd} circle = 2π×9 = 18π cm

So,

The sum of the circumference of two circles = 38π+18π = 56π cm

Now, let the radius of the 3^{rd} circle = R

∴ The circumference of the 3^{rd} circle = 2πR

It is given that sum of the circumference of two circles = circumference of the 3^{rd} circle

Hence, 56π = 2πR

Or, R = 28 cm.

**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:**

Radius of 1^{st} circle = 8 cm (given)

∴ Area of 1^{st} circle = π(8)^{2} = 64π

Radius of 2^{nd} circle = 6 cm (given)

∴ Area of 2^{nd} circle = π(6)^{2} = 36π

So,

The sum of 1^{st} and 2^{nd} circle will be = 64π+36π = 100π

Now, assume that the radius of 3^{rd} circle = R

∴ Area of the circle 3^{rd} circle = πR^{2}

It is given that the area of the circle 3^{rd} circle = Area of 1^{st} circle + Area of 2^{nd} circle

Or, πR^{2} = 100πcm^{2}

R^{2} = 100cm^{2}

So, R = 10cm

**3. Fig. 12.3 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:**

The radius of 1^{st} circle, r_{1} = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = π r_{1}^{2 }= π(10.5)^{2 }= 346.5 cm^{2}

Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2^{nd} circle, r_{2} = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2^{nd} circle − Area of gold region = (πr_{2}^{2}−346.5) cm^{2}

= (π(21)^{2} − 346.5) cm^{2}

= 1386 − 346.5

= 1039.5 cm^{2}

Similarly,

The radius of 3^{rd} circle, r_{3} = 21 cm+10.5 cm = 31.5 cm

The radius of 4^{th} circle, r_{4} = 31.5 cm+10.5 cm = 42 cm

The Radius of 5^{th} circle, r_{5} = 42 cm+10.5 cm = 52.5 cm

For the area of n^{th }region,

A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle

= π(31.5)^{2} – 1386 cm^{2}

= 3118.5 – 1386 cm^{2}

= 1732.5 cm^{2}

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle

= π(42)^{2} – 1386 cm^{2}

= 5544 – 3118.5 cm^{2}

= 2425.5 cm^{2}

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

= π(52.5)^{2} – 5544 cm^{2}

= 8662.5 – 5544 cm^{2}

= 3118.5 cm^{2}

**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:**

The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km

Converting km into cm we get,

Distance covered by the car in 1hr = (66×10^{5}) cm

In 10 minutes, the distance covered will be = (66×10^{5}×10)/60 = 1100000 cm/s

∴ Distance covered by car = 11×10^{5} cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)

=( 11×10^{5})/80 π = 4375.

**5. Tick the correct Solution: 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:**

Since the perimeter of the circle = area of the circle,

2πr = πr^{2}

Or, r = 2

So, option (A) is correct i.e. the radius of the circle is 2 units.