a leg of a right triangle is 17 inches longer than the other leg. The hypotenuse is 25 inches long. how long is the shorter of the legs?

Question

algebra 2

Answer 1

a^2 + b^2=c^2 c^2=625, b^2=(a-17)^2 equation then becomes 2a^2-34a+289=625 Solving you get a=-7 or 24. Since a length must be positive a must be 24.

Since be is the shorter length which is a-17 b=24-17=7 inches long.

 

Answer 2

Hi Savana, let x represent one leg of the triangle. Then the other leg is represented by x + 17.

 

Using the Pythagorean Theorem:

 

x² + (x + 17)² = 25²

 

x² + x² + 34x + 289 = 625

 

Combine terms and subtract 625 from both sides:

 

2x² + 34x - 336 = 0

 

Divide all terms by 2:

 

x² + 17x - 168 = 0

 

At this point you can use the Quadratic Formula or find factors that will fit the equation.

 

The factors are 7 and 24 giving:

 

(x - 7) * (x + 24) = 0

 

Using the zero product rule:

 

x + 24 = 0

 

Subtract 24 from both sides:

x = -24, this solution and be discarded since the leg of a triangle cannot be negative.

 

x - 7 = 0

 

Add 7 to both sides:

 

x = 7

 

One leg of the triangle is 7 inches. The other leg is: 7 + 17 or 24 inches long

 

Check:

 

7² + 24² = 25²

 

49 + 576 = 625

 

625 = 625

 

Values check

 

Questions?