Find the area of quadrilateral formed by joining points (4, 2), (8, 2), (8, 14) and (4, 10).

Answer: Option D

Solution

Answer: Option D

Solution:

$$\begin{array}{rl}& \text{Area of }\Delta ABC\\ & =\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\\ & =\frac{1}{2}\left[4\left(2-14\right)+8\left(14-2\right)+8\left(2-2\right)\right]\\ & =\frac{1}{2}\left[4\times \left(-12\right)+8\left(12\right)+8\left(0\right)\right]\\ & =\frac{1}{2}\left[-48+96\right]\\ & =\frac{1}{2}\times 48\\ & =24\text{ sq}\text{. units}\\ & \text{Area of }\Delta ACD\\ & =\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\\ & =\frac{1}{2}\left[4\left(14-10\right)+8\left(10-2\right)+4\left(2-14\right)\right]\\ & =\frac{1}{2}\left[4\times 4+8\times 8+4\times \left(-12\right)\right]\\ & =\frac{1}{2}\left[16+64-48\right]\\ & =\frac{1}{2}\times 32\\ & =16\text{ sq}\text{. units}\\ & \text{Hence Area of Quadrilateral }ABCD\\ & =\text{Area of }\left(\Delta ABC+\Delta ACD\right)\\ & =24+16\\ & =40\text{ sq}\text{. units}\end{array}$$