Chapter 13.4, Problem 9MRE

To determine

To find: The area of quad. ABCD.

Answer to Problem 9MRE

Area of quadrilateral ABCD $=25$

Explanation of Solution

Given information: The vertices of quad. ABCD are $A\left(2,0\right),B\left(7,0\right),C\left(7,5\right),D\left(2,5\right)$ .

Formula used: Area of triangle $=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$ .

Calculation: Area of quadrilateral ABCD $=\text{Area of }\Delta ABC+\text{Area of }\Delta ADC$

Area of triangleABC $=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$ .

Area of triangle ABC $=\frac{1}{2}\left[2\left(0-5\right)+7\left(5-0\right)+7\left(0-0\right)\right]$

Area of triangle ABC $=\frac{25}{2}$

Area of triangle ADC $=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$ .

Area of triangle ADC $=\frac{1}{2}\left[2\left(5-5\right)+2\left(5-0\right)+7\left(0-5\right)\right]$

Area of triangle ADC $=\frac{25}{2}$

Area of quadrilateral ABCD $=\text{Area of }\Delta ABC+\text{Area of }\Delta ADC$

Area of quadrilateral ABCD $=25$