Question

The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is [CBSE 2012]
(a) 20 (b) 12 (c) 6 (d) 16

Answer 1

Let A(x₁ = 5, y₁ = 0), B(x₂ = 8, y₂ = 0) and C(x₃ = 8, y₃ = 4) be the vertices of the triangle. Then
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[5\left(0-4\right)+8\left(4-0\right)+8\left(0-0\right)\right] \\ =\frac{1}{2}\left[-20+32+0\right] \\ =6\mathrm{sq}.\mathrm{units} \end{gathered}$$
Hence, the correct answer is option (c).