system of equations.

Question

A right triangle has a hypotenouse of 50 cm and an area of 600cm². Find the lengths of the legs of the triangle.

Answer 1

Let the 2 legs have lengths A and B 

 

By the Pythagorean Theorem A² + B² = 50² = 2500 

 

The area of any triangle is (1/2)*Base*Height

In a right triangle the base is one leg and the height is the other leg so 

 

(1/2)*A*B = 600

A*B = 1200

A = 1200/B

 

Substitute this value for A into the 1st equation 

 

(1200/B)² + B² = 2500

1440000/B² + B² = 2500

1440000 + B⁴ = 2500*B²  

B⁴ - 2500*B² + 1440000 = 0 

 

This is a "quadratic in disguise".  Let X = B² so 

 

X² - 2500*X + 1440000 = 0  

 

Use the quadratic formula or factor to get 

 

X = [2500 ± √(2500² - 4*(1)*1440000)] / [2*(1)]

   = [2500 ± √(490000)] / 2 

   = [2500 ± 700] / 2

   = 1600  or  900  

 

Remember that X = B² so B = √X = 40  or  30  

 

If B = 40 then A = 1200/40 = 30 

If B = 30 then A = 1200/30 = 40 

 

so the two legs are the same in either case and are 30 cm and 40 cm 

 

Check:  30² + 40² = 900 + 1600 = 2500 = 50² = hypotenuse squared 

            (1/2)*30*40 = 600 cm² area