Solving a quadratic with Geometry.

Can someone please tell me if I managed to do this correctly?

"The length of one leg of a right triangle is 2 ft longer than the other leg. If the hypotenuse is 10 ft, find the perimeter of the triangle."

My answer was

P (perimeter) = 12+2y

If this is wrong someone please tell me. I don't understand how to do these types of problems at all.

Re: Solving a quadratic with Geometry.

I would let x be the length of the shorter leg, then x + 2 is the length of the longer leg. By Pythagoras we then have:

$\displaystyle x^2+(x+2)^2=10^2$

$\displaystyle x^2+x^2+4x+4=100$

Now collect like terms, simplify, and factor to find x.

Then the perimeter P will be:

$\displaystyle P=x+x+2+10=2(x+6)$

Re: Solving a quadratic with Geometry.

Quote:

Originally Posted by

**WannaBSmart** Can someone please tell me if I managed to do this correctly?

"The length of one leg of a right triangle is 2 ft longer than the other leg. If the hypotenuse is 10 ft, find the perimeter of the triangle."

My answer was

P (perimeter) = 12+2y

If this is wrong someone please tell me. I don't understand how to do these types of problems at all.

Let $\displaystyle x$ be the shorter leg. Therefore the longer leg has length $\displaystyle x+2$.

The perimeter is given by $\displaystyle P=2(l+w)$ where $\displaystyle l$ and $\displaystyle w$ are length and width respectively. In your case $\displaystyle l = x$ and $\displaystyle w = x+2$ and so $\displaystyle P = 2(x+x+2) = 2(2x+2) = 4x+4$

To find the length x use Pythagoras since you are given the hypotenuse: $\displaystyle x^2 + (x+2)^2 = 10^2 \ \ ,\ \ x > 0$

Re: Solving a quadratic with Geometry.

Hi,

Let the length of one side be 'x'. We now have a triangle such that we have

one side has length 'x'

the other side is 'x+2'

and the hypotenuse is '10'

so your perimeter = 10+(x+2)+x which simplifies too 12+2x

Now in order to solve that you need to use r^2= a^2+b^2

r = 10

a = x+2

b = x

Solve for x and plug into the perimeter equation and you will have an answer

I agree with MarkFL2.

e^(i*pi), i think you solved using a rectangle instead of a triangle :P

Also how do i use math symbols n stuff like you guys did? :S

Re: Solving a quadratic with Geometry.

Quote:

Originally Posted by

**e^(i*pi)** Let $\displaystyle x$ be the shorter leg. Therefore the longer leg has length $\displaystyle x+2$.

The perimeter is given by $\displaystyle P=2(l+w)$ where $\displaystyle l$ and $\displaystyle w$ are length and width respectively. In your case $\displaystyle l = x$ and $\displaystyle w = x+2$ and so $\displaystyle P = 2(x+x+2) = 2(2x+2) = 4x+4$

To find the length x use Pythagoras since you are given the hypotenuse: $\displaystyle x^2 + (x+2)^2 = 10^2 \ \ ,\ \ x > 0$

Hmm, I don't think I understand. I don't think that is the formula for perimeter of a right triangle.

Since I have X on the shorter leg, X+2 on the longer leg, and 10ft as the hypotenuse, shouldn't I just group common variables and get

P= 2x+12 OR P= 2(x+6) as my answer?

Re: Solving a quadratic with Geometry.

Quote:

Originally Posted by

**WannaBSmart** Hmm, I don't think I understand. I don't think that is the formula for perimeter of a right triangle.

Since I have X on the shorter leg, X+2 on the longer leg, and 10ft as the hypotenuse, shouldn't I just group common variables and get

P= 2x+12 OR P= 2(x+6) as my answer?

That is right, but is not the complete answer. Solve for 'x' using pythagorean theorem!

Re: Solving a quadratic with Geometry.

Quote:

Originally Posted by

**Crzyrio** That is right, but is not the complete answer. Solve for 'x' using pythagorean theorem!

Hmm, I'm not following. I thought the purpose of Pythagorean theorem is to find the hypotenuse? I don't quite understand how I can use Pythagorean theorem to solve for 'x'.

Re: Solving a quadratic with Geometry.

Quote:

Originally Posted by

**WannaBSmart** Hmm, I'm not following. I thought the purpose of Pythagorean theorem is to find the hypotenuse? I don't quite understand how I can use Pythagorean theorem to solve for 'x'.

Ohhhh haha. Never mind. As soon as I wrote this I realized what you were telling me.

Thank you very much for all the help!