Chapter 12, Problem 38CR

To determine

To prove: That the shaded area is equal to $\pi$ times the product of the radii of the unshaded semicircles.

Explanation of Solution

Given Information:

B is a point on the diameter AC of a semi-circle.

X and Y are the centres of the bigger and smaller circles

Formula used:

Area of a semi-circle $\left(A\right)=\frac{\pi r^2}{2}$

  $"r"$ is the radius of the circle

Let $C_1\text{and}{C}_2$ be the bigger and smaller semi-circles inside the main semi-circle $C\text{'}$

Let $r_1\text{and}{r}_2$ be the radii of $C_1\text{and}{C}_2$ respectively.

So, $d_1\text{and}{d}_2$ becomes diameters of $C_1\text{and}{C}_2$ respectively.

So, the diameter of main semi-circle will be $\left(d\right)=AB+BC=d_1+d_2$

  $\begin{array}{l}\therefore 2r=2r_1+2r_2\\ \therefore r=r_1+r_2\end{array}$

We know that, Area of a semi-circle $\left(A\right)=\frac{\pi r^2}{2}$

  $\therefore$ Area of main semi-circle $\left(A\text{'}\right)=\frac{\pi {\left(r_1+r_2\right)}^2}{2}$

  $\therefore A\text{'}=\frac{\pi \left(r_1{}^2+r_2{}^2+2r_1{r}_2\right)}{2}$

  $\therefore$ Area of semi-circle $C_1\text{is}\left(A_1\right)=\frac{\pi {\left(r_1\right)}^2}{2}$

  $\therefore$ Area of semi-circle $C_2\text{is}\left(A_2\right)=\frac{\pi {\left(r_2\right)}^2}{2}$

Area of the shaded region $\left(D\right)=A\text{'}-\left(A_1+A_2\right)$

  $\begin{array}{l}\therefore D=\frac{\pi \left(r_1{}^2+r_2{}^2+2r_1{r}_2\right)}{2}-\frac{\pi \left(r_1{}^2\right)}{2}-\frac{\pi \left(r_2{}^2\right)}{2}\\ \therefore D=\frac{\pi }{2}\left(r_1{}^2+r_2{}^2+2r_1{r}_2-r_1{}^2-r_2{}^2\right)\\ \therefore D=\pi \left(2r_1{r}_2\right)\end{array}$

Hence, the shaded area is equal to $\pi$ times the product of the radii of the unshaded semicircles.