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
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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