Question : In a trapezium ABCD, AB and DC are parallel to each other with a perpendicular distance of 8 m between them. Also, AD = BC = 10 m, and AB = 15 m < DC. What is the perimeter (in m) of the trapezium ABCD?
Option 1: 50
Option 2: 66
Option 3: 62
Option 4: 58
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Correct Answer: 62
Solution : ⇒ AD = BC = 10 m ⇒ AB = EF = 15 m ⇒ AE = BF = 8 m In $\triangle$ADE, By Pythagoras theorem, AD 2 = AE 2 + DE 2 DE = $\sqrt{10^2-8^2}$ = $\sqrt{100-64}$ = $\sqrt{36}$ = 6 m ⇒ DE = CF = 6m Perimeter of trapezium ABCD = AB + BC + CF + FE + ED + DA = 15 + 10 + 6 + 15 + 6 + 10 = 62 m Hence, the correct answer is 62 m.
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