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.

${\displaystyle \text{let} A(x_1,y_1)=A(-3,-1),B(x_2,y_2)=B(-2,-4),C(x_3,y_3)=C(4,-1)\text{and} Then D(x_4,y_4)=D(3,4)}$

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

${\displaystyle =\frac{1}{2}[-3(-4+1)-2(-1+1)+4(-1+4)]}$

${\displaystyle =\frac{1}{2}[9-0+12]=\frac{21}{2}}$ sq. units

${\displaystyle \text{Area of} \Delta ACD=\frac{1}{2}[x_1(y_3-y_4)+x_3(y_4-y_1)+x_4(y_1-y_3)]}$

${\displaystyle =\frac{1}{2}[-3(-1-4)+4(4+1)+3(-1+1)]}$

${\displaystyle =\frac{1}{2}[15+20+0]=\frac{35}{2}}$ sq. units

So, the area of the quadrilateral ABCD is ${\displaystyle \frac{21}{2}+\frac{35}{2}=28}$.sq units sq units