Question

Find the area of the quadrilateral whose vertices are (8,6) (5,11) (-5, 12) and (-4,3).​

Answer 1

Given

We have given four vertices of a quadrilateral

A(8,6) B( 5,11) ,C (-5,12) & D ( -4,3)

To find

We have to find area of quadrilateral

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

since Quadrilateral having two triangle inside or we can say that it is made of joining two triangle namely ∆ABC &∆ ACD

So,Area of Quadrilateral ABCD=Area of ∆ABC+Area of ∆ ACD

x₁ x₂ y₁ y₂ x₃ y₃

Formula for finding area of triangle=>

1/2[x₁(y₂ - y₃)+x₂ (y₃-y₁ )+x₃ (y₁ - y₂)]

A(8,6) B(5,11) & C (-5,12)

Area of ∆ABC= 1/2 [8(11-12)+5(12-6)-5(6-11)]

Area of ∆ABC= 1/2[8(-1)+5(6)-5(-5)]

Area of ∆ABC=1/2(-8+30+25)

Area of ∆ABC=1/2(47)= 47/2 sq.units

Area of ∆ACD

A(8,6) ,C(-5,12),D (-4,3)

Area of ACD= 1/2[8(12-3)-5(3-6)-4(6-12)]

Area of ∆ACD= 1/2[ 8(9)-5(-3)-4(-6)]

Area of ∆ACD= 1/2 ( 72+15+24)

Area of ∆ACD= 111/2 sq.units

Area of quadrilateral ABCD= Ar.of ∆ ABC+ Ar.of ∆ACD

Area of quadrilateral ABCD= 47/2+111/2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=158/2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=79

Hence,Area of Quadrilateral ABCD= 79 sq.units