Question

Find the area of quadrilateral ABCD whose vertices are A(-3,-1), B(-2,-4), C(4,-1) and D(3, 4).

Answer 1

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

Let A (-3, -1), B (-2, -4) and C (4, -1) and D (3, 4)

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

Then

Area of triangle ABC

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

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

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

$=\frac{1}{2}\times (9+0+12)$

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

= 10.5 sq.units

Area of triangle ACD

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

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

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

$=\frac{1}{2}\times (15+20+0)$

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

= 17.5 sq.units

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