Find the area of the square

Question

Knowing that the area of the square below can be written as an irreducible fraction, $\frac{p}{q}$ the alternative that the sum of the digits of the number that represents the result of the expression $\frac{p-1}{q}$ is:(Answer:$5$)

I try:

$\triangle_{SED} = \frac{l.x}{2}$

$\triangle_{AEB}=\frac{l.y}{2}$

$ \triangle _{AEG}: x^2+y^2= (\frac{3\sqrt2}{2})^2 \implies x^2+y^2 = \frac{9}{2}$

Answer 1

Solution found on the net (By:castelo)

Answer 2

Putting the thing into cartesian coordinates, we have $$\left\{\begin{array}* x^2+y^2=\frac{9}{2}\\ (a-x)^2+y^2=25\\ x^2+(a-y)^2=16\\ \end{array} \right.$$ Thus $a=\sqrt{\frac{65}{2}}$.
Fine, we can walk the steps $$\left\{\begin{array}* x^2+y^2=\frac{9}{2}\\ (a-x)^2+y^2=25\\ x^2+(a-y)^2=16\\ \end{array} \right.\Leftrightarrow$$ $$\left\{\begin{array}* x^2+y^2=\frac{9}{2}\\ a^2-2ax+x^2+y^2=25\\ x^2+a^2-2ay+y^2=16\\ \end{array} \right.\Leftrightarrow$$ $$\left\{\begin{array}* x^2+y^2=\frac{9}{2}\\ a^2-2ax=\frac{41}{2}\\ a^2-2ay=\frac{23}{2}\\ \end{array} \right.\Leftrightarrow$$ $$\left\{\begin{array}* x^2+y^2=\frac{9}{2}\\ 4ax=2a^2-41\\ 4ay=2a^2-23\\ \end{array} \right.\Rightarrow$$ $$16a^2\cdot\frac{9}{2}= \left(2a^2-41\right)^2+ \left(2a^2-23\right)^2$$ $$72s=(4s^2-164s+41^2)+(4s^2-92s+23^2)$$ $$8s^2-328s+2210 = 0\Rightarrow$$ $$\left[\begin{array}* s=\frac{17}{2}\\ s=\frac{65}{2}\\ \end{array} \right.$$ But $\sqrt{\frac{17}{2}}<5$ is too small for fitting a segment of length $5$ inside.
Thus $s=\frac{65}{2}$

Answer 3

$16+9/2-4(3\sqrt{2})\cos\theta_1=25+9/2-5(3\sqrt{2})\cos \theta_2=L^2$

$15\sqrt{2}\cos\theta_2-12\sqrt{2}\cos\theta_1=9$

$\cos\theta_2=3\sqrt{2}/(10)+(4/5)\cos\theta_1$

$\cos^2\theta_2=9/50+(16/25)\cos^2\theta_1+(12\sqrt{2}/25)\cos\theta_1$

$\sin^2\theta_2=41/50-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1$

$4^2+5^2-2\cdot 4\cdot 5\cos(2\pi-(\theta_1+\theta_2))=2L^2$

$41-40[\cos(\theta_1+\theta_2)]=2L^2$

$41-40(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)=2L^2$

$\cos\theta_1\cos\theta_2=3\sqrt{2}/10\cos\theta_1+(4/5)\cos^2\theta_1$

$\sin\theta_1\sin\theta_2=\sin\theta_1\sqrt{41/50-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1}$

$41-40([(3\sqrt{2}/10)\cos\theta_1]+(4/5)\cos^2\theta_1]-[\sin\theta_1\sqrt{41/50-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1}])=32+9-24\sqrt{2}\cos\theta_1$

$12\sqrt{2}\cos\theta_1+32\cos^2\theta_1+40\sin\theta_1\sqrt{41/50-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1}=24\sqrt{2}\cos\theta_1$

$-12\sqrt{2}\cos\theta_1+32\cos^2\theta_1=-40\sin\theta_1\sqrt{41/50-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1}$

$288\cos^2\theta_1+1024\cos^4\theta_1-24\cdot32\sqrt{2}\cos^3\theta_1=1600(1-\cos^2\theta_1)[(41/50)-(16/25)\cos^2\theta_1-(12\sqrt{2}/25)\cos\theta_1]$

$288\cos^2\theta_1+1024\cos^4\theta_1-24\cdot32\sqrt{2}\cos^3\theta_1=(1-\cos^2\theta_1)[32\cdot 41-1024\cos^2\theta_1-64\cdot 12\sqrt{2}\cos\theta_1]$

$9\cos^2\theta_1+32\cos^4\theta_1-24\sqrt{2}\cos^3\theta_1=(1-\cos^2\theta_1)[41-32\cos^2\theta_1-24\sqrt{2}\cos\theta_1]$

$9\cos^2\theta_1+32\cos^4\theta_1-24\sqrt{2}\cos^3\theta_1=41-32\cos^2\theta_1-24\sqrt{2}\cos\theta_1-41\cos^2\theta_1+32\cos^4\theta_1+24\sqrt{2}\cos^3\theta_1$

$0=41-82\cos^2\theta_1+48\sqrt{2}\cos^3\theta_1$

Now solve the cubic for $\cos\theta_1$. That will give you $\cos\theta_2$ and $L^2$.