Chapter 11.9, Problem 44WE

To determine

To find: Expression for the area of triangle.

Answer to Problem 44WE

The expression for area of triangle is $\frac{32+22\sqrt{5}}{20}$ .

Explanation of Solution

Given information:

Base of triangle is $\frac{4\sqrt{5}-2}{5}$ .

Height of triangle is $\frac{\sqrt{5}+6}{2}$

Formula used:

  $Areaoftriangle=\frac{1}{2}\times Base\times Height$

As we know, the area of triangle is given by:

  $Areaoftriangle=\frac{1}{2}\times Base\times Height$

Given that, Base of triangle is $\frac{4\sqrt{5}-2}{5}$ .

Height of triangle is $\frac{\sqrt{5}+6}{2}$

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\times Base\times Height\\ Areaoftriangle=\frac{1}{2}\times \frac{4\sqrt{5}-2}{5}\times \frac{\sqrt{5}+6}{2}\\ Areaoftriangle=\frac{\left(4\sqrt{5}-2\right)\left(\sqrt{5}+6\right)}{2\cdot 5\cdot 2}\\ Areaoftriangle=\frac{4\sqrt{5}\left(\sqrt{5}+6\right)-2\left(\sqrt{5}+6\right)}{20}\end{array}$

  $\begin{array}{l}Areaoftriangle=\frac{4\sqrt{5}\cdot \sqrt{5}+4\sqrt{5}\cdot 6-2\cdot \sqrt{5}+2\cdot 6}{20}\\ Areaoftriangle=\frac{4\cdot 5+24\sqrt{5}-2\sqrt{5}+12}{20}\\ Areaoftriangle=\frac{20+24\sqrt{5}-2\sqrt{5}+12}{20}\\ Areaoftriangle=\frac{32+22\sqrt{5}}{20}\end{array}$

The expression for area of triangle is $\frac{32+22\sqrt{5}}{20}$ .