Question

Given: ABCD is a trapezoid, m∠B = 90º, m∠ACD = 90º, AB = BC, AC = CD, AD = 12. Find: Area of ABCD.

Answer 1

in Δ ACD:
m∠ACD = 90º and AC = CD and AD = 12
Δ ACD is isosceles right triangle and its hypotenuse is AD
∴ AC = CD = AD/√2 = 12/√2 = 6√2

in Δ ABC:
m∠B = 90º and AB = BC and AC = 6√2
Δ ABC is isosceles right triangle and its hypotenuse is AC
∴ AB = BC = AC/√2 = 6√2/√2 = 6

Area of ABCD = 0.5 (AD + BC) * AB = 0.5 *(6 +12) * 6 = 54