Question
Draw a quadrilateral in the cartesian plane, whose vertices are $(-4,5),(0,7),(5,-5)$ and $(-4,-2)$. Also find its area.
✓
Answer
Let the given points be $A(-4,5), B(0,7), C(5,-5)$ and $D(-4,-2) . A, B, C, D$ are plotted on a graph paper. $A$, $B ; B, C ; C, D$ and $D, A$ are joined.
Now, quadrilateral $A B C D$ is divided in two triangles, $\triangle A B D$ and $\triangle B C D$.
Area of $\triangle A B D$
$\begin{array}{l}
=\frac{1}{2}|\{-4(7+2)+0(-2-5)-4(5-7)\}| \\
=\frac{1}{2}|(-36+8)| \\
=14 \text { sq. units }
\end{array}$
area of $\triangle B C D$
$\begin{array}{l}
=\frac{1}{2}|\{0(-5+2)+5(-2-7)-4(7+5)\}| \\
=\frac{1}{2}|-45-48| \\
=\frac{93}{2} \text { sq.units }
\end{array}$
Area of quadrilateral $A B C D=$ Area of $\triangle A B C+$ $\triangle B C D$
$\begin{array}{l}=14+\frac{93}{2} \\ =\frac{121}{2} \\ =60.5 \text { sq. units. }\end{array}$
Now, quadrilateral $A B C D$ is divided in two triangles, $\triangle A B D$ and $\triangle B C D$.
Area of $\triangle A B D$
$\begin{array}{l}
=\frac{1}{2}|\{-4(7+2)+0(-2-5)-4(5-7)\}| \\
=\frac{1}{2}|(-36+8)| \\
=14 \text { sq. units }
\end{array}$
area of $\triangle B C D$
$\begin{array}{l}
=\frac{1}{2}|\{0(-5+2)+5(-2-7)-4(7+5)\}| \\
=\frac{1}{2}|-45-48| \\
=\frac{93}{2} \text { sq.units }
\end{array}$
Area of quadrilateral $A B C D=$ Area of $\triangle A B C+$ $\triangle B C D$
$\begin{array}{l}=14+\frac{93}{2} \\ =\frac{121}{2} \\ =60.5 \text { sq. units. }\end{array}$
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