Chapter 11.2, Problem 41WE

To determine

To Construct: An isosceles right triangle whose area is equal to the area of scalene triangle.

Explanation of Solution

Given: Scalene triangle.

Let the scalene triangle is

  

Area of scalene triangle

  $\begin{array}{l}b=a\\ h=h_1\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}a{h}_1\end{array}$

Now

We have to draw the isosceles right triangle

i.e.

Area of isosceles triangle

  $\begin{array}{l}b=x\\ h=x\\ A=\frac{1}{2}x(x)\\ A=\frac{1}{2}{x}^2\end{array}$

Now

  $\begin{array}{l}\frac{1}{2}{x}^2=\frac{1}{2}a{h}_1\\ x=\sqrt{a{h}_1}\end{array}$

Taking

  $\begin{array}{l}a=8,h_1=2\\ x=\sqrt{a{h}_1}\\ x=\sqrt{8(2)}\\ x=4\end{array}$

Thus the isosceles triangle shown below.