Chapter 9.4, Problem 26PSC

To determine

To calculate: The length of a leg of an isosceles triangle.

Answer to Problem 26PSC

The length of a leg of an isosceles triangle is 10.

Explanation of Solution

Given information:

Perimeter $=32,$

Side $AD=8$ .

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

An isosceles triangle has two legs congruent.

  $\overline{AB}\cong \overline{AC}$

Two triangles are congruent under ASA congruence rule.

  $\Delta ABD\cong \Delta ACD$

  $\begin{array}{l}BC=32-2x\\ BD=DC=16-x\end{array}$

Side x can be calculated by applying Pythagoras Theorem in triangle ADB .

In right angled triangle ADB , we get

  $\begin{array}{l}{\left(AB\right)}^2={\left(AD\right)}^2+{\left(BD\right)}^2\\ {\left(x\right)}^2={\left(8\right)}^2+{\left(16-x\right)}^2\\ {\left(x\right)}^2=64+256-32x+x^2\\ 320-32x+x^2=x^2\\ 320=32x\\ \therefore x=10\end{array}$