The length of a rectangle is 20 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle.

Question

I need to find the width and length of the rectangle in inches.

Answer 1

When a diagonal is drawn across a rectangle, two congruent right triangles are formed.  The sides of the rectangle are the legs of the triangle and the diagonal is the hypotenuse.

 

Let width = x

Let length = 2x + 20

diagonal = (2x + 20) + 2 = 2x + 22

 

Because we have a right triangle, we can use the Pythagorean theorem.

 

x² + (2x + 20)² = (2x + 22)²

 

x² + (2x + 20)(2x + 20) = (2x + 22)(2x + 22)

 

x² + 4x² + 80x + 400 = 4x² + 88x + 484

 

 

Bring all the terms to the left side of equation to make the right side equal to zero.

 

x² - 8x - 84 = 0

 

 

We try to factor using FOIL.

 

(x - 14)(x + 6) = 0

 

x = 14        and         x = -6

 

 

Since we cannot have a negative value for the sides, we accept the positive value of x:

 

x = 14

 

 

Now we substitute this value of x into the width and length.

 

width = 14

 

length = 2(14) + 20

          = 28 + 20

          = 48