Question 262110
find the perimeter of a rectangle ABCD with a side of 8 inches and a diagonal of 10 inches?

Let the other side be x
Side = 8 inches
Diagonal = 10 inches

10^2 = 8^2 + x^2  pythagoras 

100-64=x^2
36=x^2
X=6

Perimeter = 2*(l+b)
=2(6+8)
=28 inches

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