Question

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 6 cm and DA = 5 cm and diagonal AC = 5 cm.

Answer 1

Area of quadrilateral ABCD

= area of ∆ABC + area of ∆ACD

Since, sides of ∆ABC are Pythagorian triplet

So, area = \(\frac { 1 }{ 2 }\) x base x altitude

= \(\frac { 1 }{ 2 }\) x 3 x 4 = 6 cm²

Also ∆ACD is an isosceles triangle with equal side

(a) = 5 cm and third side (b) = 6 cm.

∴ Area of an isosceles triangle ACD