Question

If $A(0,0),B(12,0),C(12,2),D(6,7)$ and $E(0,5)$ are the vertices of the pentagon $ABCDE$, then its area (in sq units) is

• A. $58$
• B. $60$
• C. $61$
• D. $62$
• E. $63$

Answer 1

The correct option is E

$63$

Explanation for the correct option:

Finding the area:

Let $(x_1,y_1)=(0,0),(x_2,y_2)=(12,0),(x_3,y_3)=(12,2),(x_4,y_4)=(6,7),(x_5,y_5)=(0,5)$

The area of pentagon $\mathbf{=}\mathbf{}\mathbf{½}\mathbf{}\mathbf{\left[}\mathbf{\right(}{\mathit{x}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{2}}\mathbf{+}\mathbf{}{\mathit{x}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{x}}_{\mathbf{3}}{\mathit{y}}_{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{x}}_{\mathbf{4}}{\mathit{y}}_{\mathbf{5}}\mathbf{+}\mathbf{}{\mathit{x}}_{\mathbf{5}}{\mathit{y}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{(}{\mathit{y}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathbf{}{\mathit{y}}_{\mathbf{2}}{\mathit{x}}_{\mathbf{3}}\mathbf{+}\mathbf{}{\mathit{y}}_{\mathbf{3}}{\mathit{x}}_{\mathbf{4}}\mathbf{+}{\mathit{y}}_{\mathbf{4}}{\mathit{x}}_{\mathbf{5}}\mathbf{+}{\mathit{y}}_{\mathbf{5}}{\mathit{x}}_{\mathbf{1}}\mathbf{)}$

Now, substitute the values, we get

$$\begin{gathered} =(½)\left[0\right(0)+12(2)+12(7)+6(5)+0(0)-0+0+2\left(6\right)+7\left(0\right)+5\left(0\right)] \\ =(½)(24+84+30-12) \\ =63 \end{gathered}$$

Therefore, the correct option is option $\left(E\right)$.