Question

If $P(-5,-3),Q(-4,-6),R(2,-3)$ and $S(1,2)$ are the vertices of a quadrilateral $PQRS$, find its area.

Answer 1

he quadrilateral ABCD can be divided into triangles $PQS$ and $RSQ$ and hence the area of the quadrilateral is the sum of the areas of the two triangles.

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$

Area of quadrilateral PQRS = area of triangle PQS + area of triangle RSQ

area of $△PQS=\frac{1}{2}[-5(-6-2)+(-4\left)\right(2+3)+1(-3+6\left)\right]$

$=\frac{1}{2}\left[23\right]$sq. units

area of $△RSQ=\frac{1}{2}\left[2\right(2+6)+1(-6+3)+(-4\left)\right(-3-2\left)\right]$

$=\frac{1}{2}\left[33\right]$sq. units

Area of quadrilateral PQRS = area of triangle PQS + area of triangle RSQ

$=\frac{1}{2}\left[23\right]+\frac{1}{2}\left[33\right]$

$=\frac{56}{2}=28$sq.units