Chapter 9.4, Problem 5PSA

To determine

To find: The value of perpendicular $JG$ in an isosceles triangle

Answer to Problem 5PSA

The value of perpendicular $JG$ in an isosceles triangle is 9 units.

Explanation of Solution

Given information :

Triangle is isosceles.

One of the two equal sides is 15 units.

Base of the triangle = 24 units

  $FG=\frac{1}{2}FH=\frac{1}{2}\times 24=12\text{ units}$

  $JF=15\text{ units and }\angle JGF{\text{ = 90}}^{\text{o}}$

Formula used :

Altitude bisects the base of isosceles triangle at a right angle.

Calculation :

On using Pythagoras theorem in $\Delta JGF$ , we have

  $\begin{array}{l}J{F}^2=F{G}^2+J{G}^2\\ \Rightarrow {15}^2={12}^2+J{G}^2\\ \Rightarrow J{G}^2={15}^2-{12}^2=225-144=81\\ \Rightarrow JG=9\text{ units}\end{array}$