Question

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

Answer 1

Given,

Coordinates of the four vertices of a quadrilateral are (8, 6), (5, 11), (-5, 12) and (-4, 3)

To find,

The area of the quadrilateral.

Solution,

We can easily solve this mathematical problem by using the following mathematical process.

Point A = (8,6)

Point B = (5,11)

Point C = (-5,12)

Point D = (-4,3)

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆CDA

For, ∆ABC

Point A = (8,6) = X1,Y1

Point B = (5,11) = X2,Y2

Point C = (-5,12) = X3,X3

Area of ∆ABC = ½ [X1(Y2-Y3) + X2 (Y3-Y1) + X3 (Y1-Y2)]

= ½ [8(11-12) + 5(12-6) -5(6-11)]

= ½ (-8+30+25)

= ½ × 47

= 23.5 sq. units

For, ∆CDA

Point C = (-5,12) = X1,Y1

Point D = (-4,3) = X2,Y2

Point A = (8,6) = X3,Y3

Area of ∆CDA = ½ [X1(Y2-Y3) + X2 (Y3-Y1) + X3 (Y1-Y2)]

= ½ [-5(3-6)-4(6-12)+8(12-3)]

= ½ (15+24+72)

= ½ × 111

= 55.5 sq. units

Total area = (23.5+55.5) = 79 sq. units

Hence, the area of the given quadrilateral is 79 unit².