The Length of a rectangle is 1 inch less than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle

Answer 1

8 inches by 15 inches are the dimensions, with a diagonal of 17 inches

Let L=length, W=width

L=2W-1

The diagonal is the hypotenuse of a right triangle with L and W as the other two sides

The diagonal equals the square root of L^2 + W^2

=the square root of [W^2 + (2W-1)^2] = square root of [W^2 + 4W^2-4W+1]

=square root of [5W^2-4W+1] =2+L=2+2W-1=2W+1

square both sides to get

5W^2-4W+1=(2W+1)^2 = 4W^2+4W+1 Subtract 4W^2+4W+1 from both sides

W^2-8W=0

W(W-8)=0

W=8

L=2W-1=16-1=15

The diagonal is L+2=15+2=17

8 squared plus 15 squared =64=225=289 = 17 squared

Dimensions of the rectangle are 8x15 inches

There was a 2nd possible answer, with W=0, but that makes L=-1, and a negative length does not make sense. If you allowed it, then the diagonal would be 1, which is L+2, and is the square root of the sum of the squares of 0 and -1.