# 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 $w$.

Thanks!

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 $w=4.29732800472$.

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,

$$\frac{5}{\sqrt{144-w^2}}+\frac{5}{\sqrt{100-w^2}}=1$$

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 $$4.29733$$.