Chapter 10.1, Problem 14PSB

To determine

Radius of other circle.

Answer to Problem 14PSB

  $20\text{ cm}$

Explanation of Solution

Given information :

  $\begin{array}{l}\text{Two circles intersect and have a common chord 24 cm long}\text{. The centres of the circles are}\\ \text{21 cm apart}\text{. Radius of one circle is 13 cm}\text{.}\end{array}$

  $\begin{array}{l}\text{AB = 21 cm, D = 24 cm}\\ \text{Join AC, AC = 13 cm}\\ \text{Join BC}\\ \text{And CE is half of CD}\\ \text{it means CE = }\frac{\text{24}}{\text{2}}=12\text{ cm}\\ \text{In triangle}\text{ACE,}\\ A{C}^2=A{E}^2+C{E}^2\text{(Using Pythagoras Theorem)}\\ {\text{13}}^{\text{2}}=A{E}^2+{12}^2\\ 169=A{E}^2+144\\ A{E}^2=169-144=25\\ AE=5\text{ cm, Therefore}\\ AB=AE+EB\\ 21=5+EB\Rightarrow EB=16\text{cm}\\ \text{In triangle BEC, (Using Pythagoras Theorem)}\\ B{C}^2=B{E}^2+C{E}^2\\ r^2={16}^2+{12}^2\\ r^2=256+144=400\\ r=\text{20 cm}\\ \text{Radius of other circle is 20 cm}\text{.}\end{array}$