Question

Find the area of quadrilateral PQRS whose vertices are P(-5, -3), Q(-4,-6), R(2,-3) and S(1,2).

Answer 1

By joining P and R, we get two triangles PQR and PRS

Let P (-5, -3), Q (- 4, – 6) and R (2, -3) and S (1, 2)

$(x_1=-5,y_1=-3),(x_2=–4,y_2=–6),(x_3=2,y_3=-3),(x_4=1,y_4=2)$

Then

Area of triangle PQR

$=\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

$=\frac{1}{2}\times (-5(-6+3)-4(-3+3)+2(-3+6\left)\right)$

$=\frac{1}{2}\times (-5(-3)–4(0)+2(3\left)\right)$

$=\frac{1}{2}\times (15–0+6)$

$=\frac{1}{2}\times \left(21\right)$

= 10.5 sq.units

Area of triangle PRS

$\frac{1}{2}\times \left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

$=\frac{1}{2}\times (-5(-3-2)+2(2+3)+1(-3+3\left)\right)$

$=\frac{1}{2}\times (-5(-5)+2(5)+1(0\left)\right)$

$=\frac{1}{2}\times (25+10+0)$

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

= 17.5 sq.units

So, the area of the quadrilateral is 10.5 + 17.5 = 28 sq. units