Question

Find the area of the quadrilateral those vertices are A (2,1) B(2,3) C(-2,2) D(-1,0)

Answer 1

Hi Mate !!


area of quadrilateral = area of ∆ABD + area of ∆BCD.


• In ∆ABD

x1 = 2 y1 = 1
x2 = 2 y2 = 3
x3 = ( - 1 ) y3 = 0

Area of ∆ABD =

$= \frac{1}{2}(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))$


$= \frac{1}{2} (2(3 - 0) + 2(0 - 1) - 1( 1 - 3)$

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


$= 3 {unit}^{2}$


• In ∆BCD

x1 = 2 y1 = 3

x2 = ( - 2 ). y2 = 2

x3 = ( - 1 ) y3 = 0

area of ∆BCD =

$= \frac{1}{2} (2(2 - 0) - 2(0 - 3) - 1(3 - 2))$


$= \frac{1}{2} (2 \times 2 - 2 \times ( - 3) - 1 \times 1)$

$= 4.5 {unit}^{2}$


• area of quadrilateral = 3 + 4.5 = 7.5unit²