Question

find the area of the quadrilateral whose vertices are at (-9,-2),(-8,-4),(2,2),(1,-3).​

Answer 1

Given:

The vertices of quadrilateral are $(-9,-2), (-8,-4), (2,2), (1,-3)$.

Find:

The area of quadrilateral.

Solution:

Consider the given quadrilateral ABCD.

There is a attachment of the quadrilateral ABCD.

Find the area of triangle ABD.

Area of triangle ABD $=\frac{1}{2} \left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

Here,

${x_1} = - 9$ , ${x_2} = - 8$, ${x_3} = 1$, ${y_1} = - 2$, ${y_2} = - 4$, ${y_3} = - 3$

Substitute the values in the above formula, we get

Area of triangle ABD

$=\frac{1}{2} \left[ { - 9\left( { - 4 - \left( { - 3} \right)} \right) + \left( { - 8} \right)\left( { - 3 - \left( { - 2} \right)} \right) + 1\left( { - 2 - \left( { - 4} \right)} \right)} \right] \\=\frac{1}{2} \left[ { - 9\left( { - 4 + 3} \right) - 8\left( { - 3 + 2} \right) + 1\left( { - 2 + 4} \right)} \right] \\=\frac{1}{2} \left[ { - 9\left( { - 1} \right) - 8\left( { - 1} \right) + 1\left( 2 \right)} \right] \\=\frac{1}{2} \left[ {9 + 8 + 2} \right] \\=\frac{19}{2}$

Find the area of triangle CBD.

Area of triangle CBD $=\frac{1}{2} \left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

Here,

${x_1} = 2$ , ${x_2} = - 8$, ${x_3} = 1$, ${y_1} = 2$, ${y_2} = - 4$, ${y_3} = - 3$

Substitute the values in the above formula, we get

Area of triangle CBD

$=\frac{1}{2} \left[ {2\left( { - 4 + 3} \right) - 8\left( { - 3 - 2} \right) + 1\left( {2 + 4} \right)} \right] \\=\frac{1}{2} \left[ {2\left( { - 1} \right) - 8\left( { - 5} \right) + 1\left( 6 \right)} \right] \\=\frac{1}{2} \left[ { - 2 + 40 + 6} \right] \\=\frac{44}{2} \\=22$

Area of quadrilateral  ABCD = Area of triangle ABD + Area of triangle CBD$\frac{19}{2} +22=\frac{63}{2} =31.5$

So, the area of quadrilateral will be $31.5$ sqaure unit.