JEE Math Practice Question
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)$$^2$$ is equal to :
- A.64
- B.80
- C.60
- D.72
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