Chapter 1.7, Problem 17E

Solution

Write the area of the triangle as a function of the length of the base.

  $A=\frac{b^2\sqrt{7}}{4}$

Given:

The base of an isosceles triangle is half as long as the two equal sides.

Concept Used:

An Isosceles triangle is a triangle that has two equal sides. Also, the two angles opposite to the two equal sides are equal.

Calculation:

Let the equal sides of isosceles triangle = a

And base side of isosceles triangle = b

And height of isosceles triangle $h_b$

  

According to question −

Area of the isosceles triangle $A=\frac{b{h}_b}{2}$

  $\begin{array}{l}b=\frac{a}{2}\\ a=2b\end{array}$

  $\begin{array}{c}$ a^2=h_b{}^2+{(\frac{b}{2})}^2$2b^2=h_b{}^2+\frac{b^2}{4}\\ h_b{}^2=2b^2-\frac{b^2}{4}\\ h_b{}^2=\frac{8b^2-b^2}{4}\\ h_b{}^2=\frac{7b^2}{4}\\ h_b=\sqrt{\frac{7b^2}{4}}\\ h_b=\frac{\sqrt{7}b}{2}\end{array}$

Area of the isosceles triangle $A=\frac{b{h}_b}{2}$

  $\begin{array}{c}A=\frac{b{h}_b}{2}\\ =\frac{b(\frac{\sqrt{7}b}{2})}{2}\\ =\frac{b^2\sqrt{7}}{4}\end{array}$

So area $A=\frac{b^2\sqrt{7}}{4}$