• Jan 10th 2012, 03:44 PM
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."

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.
• Jan 10th 2012, 03:51 PM
MarkFL
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)\$
• Jan 10th 2012, 03:51 PM
e^(i*pi)
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."

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\$
• Jan 10th 2012, 03:54 PM
Crzyrio
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
• Jan 10th 2012, 04:25 PM
WannaBSmart
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?
• Jan 10th 2012, 04:33 PM
Crzyrio
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!
• Jan 10th 2012, 04:46 PM
WannaBSmart
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'.
• Jan 10th 2012, 04:47 PM
WannaBSmart
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!