Question

If $A(-7,5),B(-6,-7),C(-3,-8)$ and $D(2,3)$ are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Answer 1

Given: ABCD is a quadrilateral whose vertices are $A(-7,50,B(-6,-7),C(-3,-8)$ and $D(2,3)$.

By joining AC, we get two triangles ABC and ADC

We know that:

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

Area of triangle ABC:

$=\frac{1}{2}[-7(-7-(-8)]+(-6\left)\right[(-8)-5]+(-3\left)\right[5-(-7)]$

$=\frac{1}{2}[-7\times (1)+(-6)\times (-13)+(-3)\times 12]$

$=\frac{1}{2}[-7+78-36]$

$=\frac{1}{2}\times 35=\frac{35}{2}$ sq. units.

Area of triangle ADC.

$=\frac{1}{2}[-7(-8-3)+(-3\left)\right(3-5)+(2\left)\right(-5(-8)]$

$=\frac{1}{2}\left[\right(-7)\times (-11)+(-3)\times (-2)+2\times 13]$

$=\frac{1}{2}[77+6+26]=\frac{109}{2}$ sq. units.

Now, Area of quadrilateral PQRS$=$Area of triangle ABC$+$Area of triangle ADC

$=35/2+109/2$

$=72$ sq. units.