Solving two right-angled triangles with the same hypotenuse

Question

How can I use Pythagoras Theorem on the two right-angled triangles to get relationships between $W, D, L, x, R$ and hence find $D$ in terms of $W, L,$ and $R$ (eliminating $x$)? I have tried relating equating $R^2$ for both triangles, but I can't seem to find it in terms of $D$.

Answer 1

there are actually three right-angled triangle, but we'll need only two

On the first triangle bounced by $R$, $W+x$ and $L$, Using Pythagoras here gives us

$R^2 = (W+x)^2+L^2$

On the second triangle bounded by $R$, $x$ and $L+D$, Using Pythagoras here gives us

$R^2 = x^2+(L+D)^2$

Since the aim here is to put $D = f(W,L,R)$

$(L+D)^2 = R^2-x^2$, $L+D = \sqrt{R^2-x^2}$, $D = -L ± \sqrt{R^2-x^2}$

But $(W+x)^2 = R^2-L^2$, $x = -W ± \sqrt{R^2-L^2}$

Therefore $D = -L ± \sqrt{R^2 - (-W±\sqrt{R^2-L^2})^2}$, $D = -L ± \sqrt{R^2-W^2±2*W*\sqrt{R^2-L^2}+R^2-L^2}$

$D = -L ± \sqrt{2*R^2-W^2-L^2±2*W*\sqrt{R^2-L^2}}$

Answer 2

Hint:

Observing the lower right-angled triangle has a side equal to $x$, the other side is equal to $L+D$. Expanding Pythagoras for each right-angled triangle, $x^2$ is eliminated, and you should obtain the relations $$D^2+LD=W^2+2xW=R^2.$$