Diagonals, I need help asap

Question

The area of a rectangle is 10, and it’s diagonal is sqrt(101). Find it’s dimensions and perimeter.

Answer 1

d^2 = L^2 + W^2

d^2=101,

A=LW = 10,

L=10/W.

the diagonal is the hypotenuse of a right triangle with sides L and W

sqr(101)^2 = 101 =

101 = (10/W)^2 + W^2

multiply by W^2

101W^2 = 100 + W^4, Let X= W^2

X^2 -101X - 100 = 0

factor

(X-100)(X+1) = 0

X = 100 =W^2 (ignore the negative solution for X)

W = sqr100 = 10

L=10/10 = 1

dimensions are 10 x 1

Perimeter = 2(W+L) = 2(10+1) = 22