anamikarai wrote:

In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possible area of the rectangle ABCD?

A - 1000

B - 1200

C - 1250

D - 1500

E - 1600

In rectangle ABCD, lengths of AB=CD and AD=BC

Hence, AB+BC+CD=100 => 2AB+CD=100

AB cannot be equal to 50, since it will not be a rectangle then as 2AB=100 => CD=0. Not possible.

AB<50.

If AB=49, CD=100-2x49=2

Area of rectangle=49x2=98

Going on like this, if AB=25, CD=100-2x25=50

Area=25x50=1250

If AB=24, CD=100-2x24=52

Area=24x52=1248. Observe that the area is decreasing from 1250 obtained earlier.

Hence max. area can be 1250.

Answer C.

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