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If $A (-3, 5), B (-2, -7), C (1, -8)$ and $D (6, 3)$ are the vertices of a quadrilateral $ABCD$, find its area.
Given:
$A (-3, 5), B (-2, -7), C (1, -8)$ and $D (6, 3)$ are the vertices of a quadrilateral $ABCD$.
To do:
We have to find the area of the quadrilateral.
Solution:
Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.
This implies,
Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[-3(-7+8)+(-2)(-8-5)+1(5+7)] \)
\( =\frac{1}{2}[-3(1)+(-2)(-13)+1(12)] \)
\( =\frac{1}{2}[-3+26+12] \)
\( =\frac{1}{2} \times (35) \)
\( =\frac{35}{2} \) sq. units.
Area of triangle \( ADC=\frac{1}{2}[-3(3+8)+6(-8-5)+1(5-3)] \)
\( =\frac{1}{2}[-3(11)+6(-13)+1(2)] \)
\( =\frac{1}{2}[-33-78+2] \)
\( =\frac{1}{2} \times (-109) \)
\( =\frac{109}{2} \) sq. units.
Therefore,
The area of the quadrilateral $ABCD=\frac{35}{2}+\frac{109}{2}=\frac{35+109}{2}=72$ sq. units.
The area of the quadrilateral $ABCD$ is $72$ sq. units.
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