CBSE 10th Maths -  Areas Related to Circles- Exercise 12.1

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

circumference of the $1^{st}$ circle = $2 \pi ×19 = 38\pi$ cm

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

circumference of the $2^{nd}$ circle = $2 \pi ×9 = 18 \pi$ cm

So,

The sum of the circumference of two circles = $38 \pi +18 \pi = 56 \pi$ cm

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

the circumference of the $3^{rd}$ circle = $2 \pi R$

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

Hence, $56 \pi = 2 \pi$ R

Or, R = 28 cm.

The radius of $1^{st}$ circle = 8 cm (given)

Area of $1^{st}$ circle = $\pi (8)^2 = 64 \pi $

The radius of $2^{nd}$ circle = 6 cm (given)

Area of $2^{nd}$ circle = $\pi (6)^2 = 36 \pi $

So,

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

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

Area of the circle $3^{rd}$ circle = $\pi R^2$

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

Or, $\pi R^2 = 100 \pi \: cm^2$

$R^2 = 100 \: cm^2$

So, R = 10 cm

The radius of $1^{st}$ circle, $r_1 = \dfrac{21}{2}$ cm (as diameter D is given as 21 cm)

So, area of gold region = $\pi r{_1}^{2} = \pi (10.5)^2$ = 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, $r_2$ = 10.5 cm+10.5 cm = 21 cm

Thus,

area of red region = Area of $2^{nd}$ circle − Area of gold region = ($\pi r_{2}^{2}−346.5)$ $cm^2$

= ($\pi (21)^2 − 346.5)$ $cm^2$

= 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 the circle (n-1)

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

= $\pi (31.5)^2$ – 1386 $cm^2$

= 3118.5 – 1386 $cm^2$

= 1732.5 $cm^2$

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

= $\pi (42)^2$ – 1386 $cm^2$

= 5544 – 3118.5 $cm^2$

= 2425.5 $cm^2$

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

= $\pi (52.5)^2$ – 5544 $cm^2$

= 8662.5 – 5544 $cm^2$

= 3118.5 $cm^2$

The radius of car’s wheel = $\dfrac{80}{2}$ = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2 $\pi$ r = 80 $\pi$ cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 $\pi$ 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 = $\dfrac{(66×10^5×10)}{60}$ = 1100000 cm/s

distance covered by car = $11×10^5$ cm

Now, the no. of revolutions of the wheels =$(\dfrac{Distance \: covered \: by \: the \: car}{Circumference \: of \: the \: wheels})$

= $\dfrac{(11×10^5)}{80 \pi} $ = 4375.