Find the dimensions of the rectangle

Question

The length of a rectangle is 31 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle.
w
2w+31
2w+33
width=? inches
length=? inches

Answer 1

w = w

l = 2w + 31

d = 2w + 33

 

The diagonal divides the rectangle into two congruent triangles.

Use Pythagorean Theorem, i.e. the hypotenuse squared is equal to the sum of the squares of the two legs.

 

d² = w² + l²

 

(2w + 33)² = w² + (2w + 31)²

 

4w² + 132w + 33² = w² + 4w² +124w + 31²

 

Simplify to get:

w² - 8w - 128 = 0

 

Factor to get:

(w - 16)(w + 8) = 0

 

So, w = 16, l = 63, and d = 65