nandinigaur wrote:

as its a rectangle and we know that the diagonal is 10... cant we consider the rectangle to be 2 triangles of 45:45:90... and assume sides to be in ratio 1:1:root 2??? that ways 1st equation is sufficient.

If you assume rectangle to be 2 triangles of 45:45:90, then it makes a square and the sides will be 10/\(\sqrt{2}\) :10/\(\sqrt{2}\): 10

Also it is not given that the sides are integers so we can have \(x^2+y^2=10^2\). Now x=\(\sqrt{99}\) and y= 1 then perimeter, p= 2(\(\sqrt{99}\) +1)

Now if x=6,y=8 then perimeter, p= 2*(14)= 28.

2 different answers are possible using St1 and therefore it is not alone

From St 2 we have xy=48 (Many combinations possible such as x=4,y=12, p=32 or x=6,y=8, p=28). Not sufficient

Combining we get, only 1 possible value of x=6and y=8 or vice versa meet both conditions.

Ans is C

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