Chapter 13, Problem 12CR

To determine

To find: The following and determine its area.

Answer to Problem 12CR

The area of $△ADE=14.463\text{ square unit}$ .

Explanation of Solution

Given information:

  

Solution:

  

  $\begin{array}{l}\therefore \angle DAG={40}^{\circ }\\ \therefore \sin {40}^{\circ }=\frac{h}{5}\\ \Rightarrow h=5\times \sin {40}^{\circ }\\ \Rightarrow h=3.214\\ \therefore \tan {40}^{\circ }=\frac{h}{AG}\\ \Rightarrow AG=\frac{h}{\tan {40}^{\circ }}\\ \Rightarrow AG=\frac{3.214}{\tan {40}^{\circ }}\\ \therefore AG=3.83\\ GE=AE-AG=9-3.83=5.17\end{array}$

  $\therefore$ Area of $△ADG$

  $\begin{array}{l}\frac{1}{2}\times AG\times h\\ =\frac{1}{2}\times 3.83\times 3.214\\ =6.15481\text{ square unit}\end{array}$

  $\therefore$ Area of $△DGE$

  $\begin{array}{l}\frac{1}{2}\times DG\times GE\\ =\frac{1}{2}\times 3.214\times 5.17\\ =8.30819\text{ square unit}\end{array}$

  $\therefore$ Area of $△ADE=$ area of $△ADG+$ area of $△DGE$

  $\begin{array}{l}=\left(6.15481+8.30819\right)\\ =14.463\text{ square unit}\end{array}$

Hence, the area of $△ADE=14.463\text{ square unit}$ .