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 $\displaystyle \begin{align*} a \end{align*}$.
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?
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
AD=x DC = 8-x
ABD is similar to BDC Equal angles
x/2 =2/8-x
x^2-8x+4=0
x=0.54
AB^2= 2^2 +(0.54)^2
AB = 2.07 short leg
BC = 7.73 long leg
AC = 8.00 hypothenuse
perimeter = 17.8 units