Chapter 10.7, Problem 24PSC

To determine

Find the ratio of the area of the inscribed circle to the area of the circumscribed circle.

Answer to Problem 24PSC

  $\frac{169}{16}$

Explanation of Solution

Given information:

A right triangle has legs measuring 5 and 12.

Calculation:

As per the given statement the figure has to be drawn as

  

  $AB$ is the diameter of the large circle,

  $\begin{array}{l}AB=\sqrt{5^2+{12}^2}\\ AB=\sqrt{25+144}\\ AB=\sqrt{169}=13\\ Radiusofl\arg ecircle,=\frac{AB}{2}=\frac{13}{2}\end{array}$

  $\begin{array}{l}Radiusofsmallcircle=5-r-12-r=13=5-r-12-r=13\\ 5-r+12-r=13\\ -2r+17=13\\ -2r=-4\\ r=2\end{array}$

  $\begin{array}{l}\text{Ratio of the area}\text{of }\text{circles,}\\ \frac{\pi R^2}{\pi r^2}=\frac{{\left(\frac{13}{2}\right)}^2}{2^2}=\frac{169/4}{4}=\frac{169}{16}\end{array}$

Hence, ratio of the area of the circumscribed circles is $\frac{169}{16}$