Question

Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).

Answer 1

By joining A and C, we get two triangles ABC and ACD

Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)

$(x_1=3,y_1=-1),(x_2=9,y_2=-5),(x_3=14,y_3=0),(x_4=9,y_4=19)$

Then

Area of triangle ABC

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

$=\frac{1}{2}\times \left(3\right(-5-0)+(9\left)\right(0+1)+(14\left)\right(-1+5\left)\right)$

$=\frac{1}{2}\times \left(3\right(-5)+9(1)+14(4\left)\right)$

$=\frac{1}{2}\times (-15+9+56)$

$=\frac{1}{2}\times \left(50\right)$

= 25 sq.units

Area of triangle ACD

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

$=\frac{1}{2}\times \left(3\right(0-19)+(14\left)\right(19+\left(1\right))+(9\left)\right(-1–0\left)\right))$

$=\frac{1}{2}\times \left(3\right(-19)+14(20)+9(-1\left)\right)$

$=\frac{1}{2}\times (-57+280–9)$

$=\frac{1}{2}\times \left(214\right)$

= 107 sq.units

So, the area of the quadrilateral is 25 + 107 = 132 sq. units