# Thread: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

1. ## Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Hi everyone!

I've been struggling with this problem recently, and don't really understand how I'm supposed to solve it:

A hypotenuse of a right triangle has length 8. The area of the triangle is also 8 units squared. What is the perimeter of the triangle?

Any help would be appreciated!

2. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

You have \displaystyle \begin{align*} \frac{1}{2}\,a\,b = 8 \end{align*} and \displaystyle \begin{align*} a^2 + b^2 = 64 \end{align*}. Rearranging the first equation we have \displaystyle \begin{align*} b = \frac{16}{a} \end{align*}, so substituting into the second equation gives \displaystyle \begin{align*} a^2 + \left( \frac{16}{a} \right) ^2 = 64 \end{align*}. Expand and multiply both sides by \displaystyle \begin{align*} a^2 \end{align*}, giving you a quadratic equation you should be able to solve for \displaystyle \begin{align*} a^2 \end{align*} and thus solve for \displaystyle \begin{align*} a \end{align*}.

3. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Alright, so I've tried 1/2ab=8^2 and solved for b, getting b = 128/a. Then I tried plugging that into the equation a^2 + b^2 = 64, like so:

a^2 + (128/a)^2 = 64

a^ 2 + (16384/a^2) = 64

16384/a^2 = 64 - a^2

16384 = 64a^2 -a^4

-a^4 + 64a^2 - 16384 = 0

-1(a^4 - 64a^2 + 16384) = 0

a^4 - 64a^2 + 16384 = 0

Then, when I tried to solve for that with the quadratic formula:

-(-64) +/- sqrt[(-64)^2 - 4(1)(16384)]/2(1)

I got an imaginary answer, since 4096 - 65536 = a negative number, which you can't find a real square root of.

I'm still stuck, so can someone shed some light on what I did wrong?

4. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Originally Posted by Lexielai
Alright, so I've tried 1/2ab=8^2 and solved for b, getting b = 128/a. Then I tried plugging that into the equation a^2 + b^2 = 64, like so:

a^2 + (128/a)^2 = 64

a^ 2 + (16384/a^2) = 64

16384/a^2 = 64 - a^2

16384 = 64a^2 -a^4

-a^4 + 64a^2 - 16384 = 0

-1(a^4 - 64a^2 + 16384) = 0

a^4 - 64a^2 + 16384 = 0

Then, when I tried to solve for that with the quadratic formula:

-(-64) +/- sqrt[(-64)^2 - 4(1)(16384)]/2(1)

I got an imaginary answer, since 4096 - 65536 = a negative number, which you can't find a real square root of.

I'm still stuck, so can someone shed some light on what I did wrong?
The question is flawed.Max area of a rt triangle with hypothenuse of 8 = 16 sq units

5. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Originally Posted by Lexielai
Alright, so I've tried 1/2ab=8^2
NO! You have been told the area of the triangle is \displaystyle \begin{align*} 8\,\textrm{units}^2 \end{align*}, NOT 64. Read my previous post properly!

6. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Oh, that makes much more sense now! I was totally thinking about this the wrong way. Thanks.

7. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

Well, for practice, try this one:

right triangle ABC with sides a,b,c (a<b, c = hypotenuse).

area of ABC = 78a = 30c

Calculate a,b,c.

8. ## Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

I misinterpreted area as 64 sq units.For an area of 8 sq units
Draw a semicircle radius 4 Mark diameter ends A and C. On the arc where altitude is 2 mark B on arc.D is the base of alt from B to AC.
ABC is a right angle. 1/2*180. Area triangle ABC = 1/2 * 2*8