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

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Matheng
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

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