Q1.Find the area of the quadrilateral whose vertices are (8,6) (5,11) (-5, 12) and (-4,3).
Answers
Explanation:
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
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
