Chapter 16, Problem 6RP

To determine

To find: The length of side AH.

Answer to Problem 6RP

The length of side AH is $\frac{5\sqrt{35}+\sqrt{119}}{6}\text{units}$ .

Explanation of Solution

Given information:

A quadrilateral RHOA is inscribed in a circle.

Consider the figure of the quadrilateral RHOA as,

  

In this quadrilateral $\angle RAO$ is a right angle, so apply the Pythagorean Theorem to find the value of $AO$ as,

  $\begin{array}{l}{\left(AO\right)}^2={\left(RO\right)}^2-{\left(AR\right)}^2\\ {\left(AO\right)}^2={\left(12\right)}^2-{\left(5\right)}^2\\ AO=\sqrt{119}\text{units}\end{array}$

Similarly, calculate the value of $HO$ as,

  $\begin{array}{l}{\left(HO\right)}^2={\left(RO\right)}^2-{\left(HR\right)}^2\\ {\left(HO\right)}^2={\left(12\right)}^2-{\left(2\right)}^2\\ HO=\sqrt{140}\\ HO=2\sqrt{35}\text{units}\end{array}$

The length of side AH is calculated by using Ptolemy’s Theorem as,

  $\begin{array}{l}12\left(AH\right)=5\left(HO\right)+2\left(AO\right)\\ 12\left(AH\right)=5\left(2\sqrt{35}\right)+2\left(\sqrt{119}\right)\\ AH=\frac{5\sqrt{35}+\sqrt{119}}{6}\text{units}\end{array}$

Thus, the length of side AH is $\frac{5\sqrt{35}+\sqrt{119}}{6}\text{units}$ .