Math problem

Question

In a right triangle, the longer leg is 20ft longer than twice the length than the shorter leg. The hypotenuse is 10ft longer than the length of the longer leg. Find the lengths of all sides of the triangle.

Answer 1

 

Let X = shorter leg

Let 2X + 20 = longer leg

Let 2X + 30 = hypotenuse

 

Use pythagorean theorem to solve

 

(2x + 30)² = x² + (2x + 20)²

 

4x² + 120x + 900 = x² + 4x² + 80x + 400

 

Subtract 4x² from both sides

 

120x + 900 = x² + 80x + 400

 

0 = x² - 120x + 80x + 400 - 900

 

0 = x² - 40x - 500

 

0 = (x  - 50)(x + 10)

 

x - 50 = 0 .... x = 50

 

or x + 10 = 0 ... x = -10 ... discard this ... we cant have a negative side length

 

The side measures are:

 

50

 

2(50)+20 = 120

 

120 + 10 = 130

 

side measures are 50, 120, and 130