The question in full:
The triangle T has side lengths $a$, $b$ and $c$, whilst triangle U has side lengths $1/a$, $1/b$, and $1/c$. Both triangles are right-angled and T has twice the area of U. What is the perimeter of T ?
I can calculate what $b$ is, but cannot get any further. Any help would be appreciated.
How I calculate $b$: If we assume that $a$ is the shortest side of T, and that $b$ is the other side of the right-angle. Then $c$ must be the hypotenuse, and therefore longest side. Therefore, on U, $\dfrac{1}{c}$ must now be the shortest side, $\dfrac{1}{a}$ is the longest and $\dfrac{1}{b}$ is the "middle".
Lets call A the opposite angle of $a$ and F the opposite of $\dfrac{1}{c}$.
Using sohcahtoa we can then deduce that $$\sin{A} = \dfrac{a}{c}$$ and $$\sin {F} = \frac{\dfrac{1}{c}}{\dfrac{1}{a}} = \frac{a}{c}$$ therefore the triangles are similar. Also because we are given that $$A(T) = 2\cdot A(U)$$ we know that $a=\frac{1}{c}\sqrt{2}$ and $b=\frac{1}{b}\sqrt{2}$. the latter then allows us to calculate $b$, which is is the 4th root of 2.
Other worthy things to note: Using $$\text{Area of triangle} = \dfrac{1}{2} \:(\text{base} \times \text{height})$$ we get $$A(T)= \frac{1}{2}(ab)$$ and $$A(U) = \frac{1}{2bc}$$ therefore $$b^2 = \frac{2}{ac}$$ Also, since the triangles are similar, the ratio of $a/b$ must be the same as $(1/c) / (1/b)$ , thus $b^2 = ac$. Combining these gives $ac = 2 / ac$.
I have a feeling that I need to use the sine rule or cosine rule to get further, but can't quite get something that is solvable (other than $(a^4)/2 + (a^2)/\sqrt{2} = 1$)
Assume $a< b<c$ and start with the equations: $$ a^2+b^2=c^2,\quad \frac1{c^2}+\frac1{b^2}=\frac1{a^2}.\tag1 $$
Elimnation of $c$ results in the equation: $$ b^4-a^4=a^2b^2. $$ Resolving this quadratic (with respect to $a^2$) equation and using $c^2=a^2+b^2$ one obtains: $$ a^2=\frac{\sqrt5-1}2b^2,\quad c^2=\frac{\sqrt5+1}2b^2.\tag2 $$ Observe $a^2c^2=b^4$.
Let the ratio of the areas be $\lambda^4$ with $\lambda>0$. Then we have: $$ \lambda^4=ab^2c=b^4\implies b=\lambda. $$
Using $(2)$ to compute $a$ and $c$ one finally obtains for the perimeter the value: $$ \lambda\left[\sqrt{\frac{\sqrt5-1}2}+1+\sqrt{\frac{\sqrt5+1}2}\right] =\lambda\left[1+\sqrt{2+\sqrt5}\right]. $$
In your case $\lambda=2^\frac14$.