Find the length of $w$ using a numerical method I need to find the value of $w$ using a numerical method for finding function's root (like Newton's method).
By the Pythagorean theorem we know that $BF = \sqrt {100 -w^2}$ and $AD = \sqrt {144 - w^2}$. 
I'm not so sure how to utilize $OE$ to create a function so that it's root is the answer for $w$.
I'd be glad for help
Thanks!

 A: Put the origin at $D$, $A=(0,a)$, $B=(w,b)$. Then the diagonal lines have the equations $y=a·(1-\frac xw)$ and $y=b·\frac xw$. They intersect at $x=w·\frac{a}{a+b}$ at height $y=\frac{ab}{a+b}$.
The three unknowns $a,b,w$ are thus connected to the 3 given lengths via
$$
a^2+w^2=144\\
b^2+w^2=100\\
ab=5(a+b)
$$
leading to
$$
\sqrt{(144-w^2)(100-w^2)}=5(\sqrt{144-w^2}+\sqrt{100-w^2}).
$$
This can now be solved using your favorite univariate solver, like bisection, secant method, Newton method and others to find the root close to $w=4.29732800472$.
A: From triangle $ADF$ and triangle $ODF$:
$$\frac{AD}{DF}=\frac{OE}{EF}$$
$$\frac{AD}{w}=\frac{5}{EF}$$
$$EF=\frac{5w}{AD}$$
From triangle $BFD$ and $OED$:
$$\frac{BF}{DF}=\frac{OE}{DE}$$
$$\frac{BF}{w}=\frac{5}{DE}$$
$$DE=\frac{5w}{BF}$$
Since $DE+EF=w$
$$\frac{5}{AD}+\frac{5}{BF}=1$$
Hence,
$$\frac{5}{\sqrt{144-w^2}}+\frac{5}{\sqrt{100-w^2}}=1$$
A: Starting from @Siong Thye Goh's answer,
$$\frac{5}{\sqrt{144-w^2}}+\frac{5}{\sqrt{100-w^2}}=1$$ let $x=w^2$ and consider that you look for the zero of function
$$f(x)=\frac{5}{\sqrt{144-x}}+\frac{5}{\sqrt{100-x}}-1$$ If you plot it, it looks very linear and this is very good for a root-finding algorithm.
Being very lazy, using $x_0=0$, the first iteration of Newton, Halley and Householder methods will be
$$\frac{7200}{341} \qquad \frac{89280}{4823}\qquad \frac{173628000}{9399763}$$ Converted to decimals, th first estimate of $w$ is $4.59504$, $4.30248$ and  $4.29785$ while the exact solution is $4.29733$.
