Chapter 12.4, Problem 34WE

To determine

To find: The area of the circle that is formed by the interaction.

Answer to Problem 34WE

Area of the circle formed will be $144\pi {\text{ cm}}^2$ .

Explanation of Solution

Given information:

Two intersecting sphere with radii 15 cm and 20 cm respectively. The centres of the spheres are 25cm apart.

Calculation:

The two intersecting circle is sketched below:

  

From the above figure, ABO is a right angles triangle. Therefore, by Pythagoras theorem

  $\begin{array}{l}O{A}^2=O{B}^2+A{B}^2\\ {20}^2=x^2+y^2\\ y^2={20}^2-x^2\end{array}$ …… (1)

ABO’ is also a right angled triangle. Therefore,

  $\begin{array}{l}O\text{'}{A}^2=O\text{'}{B}^2+A{B}^2\\ {15}^2={\left(25-x\right)}^2+y^2\\ y^2={15}^2-{\left(25-x\right)}^2\end{array}$ …… (2)

From equation (1) and (2)

  $\begin{array}{l}{20}^2-x^2={15}^2-{\left(25-x\right)}^2\\ 400-x^2=225-\left(625-50x+x^2\right)\\ 400-x^2=225-625+50x-x^2\\ 400-x^2+x^2-225+625=50x\\ 800=50x\\ x=\frac{800}{50}\\ x=16\end{array}$

Substitute $x=16$ in equation (1) and solve for y:

  $\begin{array}{l}{y}^2={20}^2-{16}^2\\ y^2=400-256\\ y^2=144\\ y=12\end{array}$

Thus, radius of the circle formed will be 12 cm.

Therefore, the area of the circle formed will be

  $\begin{array}{l}A=\pi y^2\\ A=\pi {\left(12\right)}^2\\ A=144\pi {\text{ cm}}^2\end{array}$

Hence,

Area of the circle formed will be $144\pi {\text{ cm}}^2$ .