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 :
- 64
- 80
- 60
- 72
Answer
(4) 72
Solution
Let AB = 4 make angle with PQ:
[figure]
PQ = 4 + 2 , QR = 4 + 2
Area = (4 + 2 )(2 + 4 )
= 8( 2 + 2 ) + 20 = 8 + 10 2
∵ maximum when 2 = 1, i.e. = 45
a + b = 6 45 + 6 45 = 6 2
(a + b)2 = 72
Answer: (4)