Thread: [SOLVED] Triangle Question!

Hi, I'm a little wary regarding this problem. I'm pretty sure I'm doing it right but would appreciate it if someone verified my solution. The question is:

The perimeter of a right triangle is 60 cm. The length of the hypotenuse is 6 cm more than twice the length of one of the other sides. Find the lengths of all three sides.

So here' what i did.
Let one side of the triangle be x, let the other be y
60 - x - (2x+6) = y
(54-3x) = y

a^2 + b^2 = c^2
So X^2 + (54-3x)^2 = (2x+6)^2
After expanding i got
6x^2 - 348x + 2880 = 0
Using the quadratic formula, I got 10, 48.

Choosing 10 as the best answer for x and solving for y. I got the following at my answers:
10, 24, and 26 as the sides of the triangle. Is my approach correct?

THANKS!

Originally Posted by demo1

Hi, I'm a little wary regarding this problem. I'm pretty sure I'm doing it right but would appreciate it if someone verified my solution. The question is:

The perimeter of a right triangle is 60 cm. The length of the hypotenuse is 6 cm more than twice the length of one of the other sides. Find the lengths of all three sides.

So here' what i did.
Let one side of the triangle be x, let the other be y
60 - x - (2x+6) = y
(54-3x) = y

a^2 + b^2 = c^2
So X^2 + (54-3x)^2 = (2x+6)^2
After expanding i got
6x^2 - 348x + 2880 = 0
Using the quadratic formula, I got 10, 48.

Choosing 10 as the best answer for x and solving for y. I got the following at my answers:
10, 24, and 26 as the sides of the triangle. Is my approach correct?

THANKS!

Hi demo1,

You did good!

I'm so glad! Thanks for confirming my answer.