Question

If $A(-5,7),B(-4,-5),C(-1,-6)$ and $D(4,5)$ are the vertices of a quadrilateral, find the area of the quadrilateral $ABCD$.

Answer 1

Let the vertices of the quadrilateral be $A(-5,7),B(-4,-5),C(-1,-6),D(4,5)$.

Joining $AC$ there are two triangles $ABC,ADC$

Therefore area of quadrilateral $ABCD=$Area of $\Delta ABC+$Area of $\Delta ADC$

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

Here $x_1=-5,y_1=7,x_2=-4,y_2=-5,x_3=-1,y_3=-6$

Area of $\Delta ABC=\frac{1}{2}[-5(-5+6)-4(-6-7)-1(7+5\left)\right]=\frac{1}{2}\left(35\right)$ square units.

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

Here $x_1=-5,y_1=7,x_2=4,y_2=5,x_3=-1,y_3=-6$

Area of $\Delta ADC=\frac{1}{2}[-5(5+6)+4(-6-7)-1(7-5\left)\right]=\frac{1}{2}(-109)=\frac{1}{2}\left(109\right)$ square units. (since area cannot be negative)

Therefore area of quadrilateral $ABCD=$Area of $\Delta ABC+$Area of $\Delta ADC$

$=\frac{1}{2}\left(35\right)+\frac{1}{2}109$

$=\frac{1}{2}\left(144\right)$

$=72$

Therefore area of quadrilateral $ABCD=72$ square units