Chapter 8.6, Problem 63CCR

To determine

Find the area of the figure.

Answer to Problem 63CCR

Total area of the figure ABCD is18 sq m

Explanation of Solution

Given:

The figure:

Given the Dimensions: AC = AE + EC = 3 + 3 = 6 m BD = DE + EB = 4 + 2 = 6 m

Concept Used:

Here are two right Triangle.

In the Right Triangle ABD: Base = BD = 6 m and height AE = 3 m

In the Right Triangle BDC: Base = BD = 6 m and height CE = 3 m

Rule: $\text{Area}\text{of}\text{Triangle}=\frac{1}{2}·\text{base}·\text{height}\text{.}$

Calculation:

In the Right Triangle ABD: Base = BD = 6 m and height AE = 3 m

In the Right Triangle BDC: Base = BD = 6 m and height CE = 3 m

Rule: $\text{Area}\text{of}\text{Triangle}=\frac{1}{2}·\text{base}·\text{height}\text{.}$

  $\text{Area}\text{of}\text{Triangle}\text{ABD}=\frac{1}{2}·\text{6}·\text{3= 9 sq m }$

  $\text{Area}\text{of}\text{Triangle}\text{BDC}=\frac{1}{2}·\text{6}·\text{3= 9 sq m }$

Total area of the figure ABCD = Area of Right Triangle ABD + Area of Right Triangle BDC Total area of the figure ABCD = 9 + 9 = 18 sq m

Thus, total area of the figure ABCD is18 sq m