A geometry problem by Swapnil Nikam

Let $AC = y$ and $AF = x$ . We are then wanting to find the value of $y^{2}$ . Now, since triangles $CAF$ and $FBE$ are similar right-angled triangles, we have that

(i) $x^{2} + y^{2} = 16$ , and

(ii) $\dfrac{4}{y} = \dfrac{3}{y - x} \Longrightarrow 4(y - x) = 3y \Longrightarrow x = \dfrac{y}{4}$ .

Substituting this last result into equation (i) yields

$(\dfrac{y}{4})^{2} + y^{2} = 16 \Longrightarrow y^{2} = \dfrac{16^2}{17} = \boxed{15.06}$ to $2$ decimal places.

$\quad \quad \quad \\ Let\quad \\ \qquad \angle ACF\quad =\quad \angle EFB\quad =\quad \theta \\ Given\\ \qquad AC\quad =\quad AF+FB\\ \\ \qquad 4cos\theta \quad =\quad 4sin\theta \quad +\quad 3cos\theta \\ \\ \qquad \theta \quad =\quad { tan }^{ -1 }(1/4)\quad =\quad 14.036\\ \\ Now\quad \\ \qquad AC\quad =\quad 4cos\theta \quad =4cos(14.036)\\ \qquad \qquad =\quad 3.8805\\ \\ Hence\quad Area\quad =\quad ACxAC\quad =\quad { 3.8805 }^{ 2 }\quad =\quad 15.05\\ \\ \\ \quad \quad$

This problem is solved using Pitagoras theorem all over the place and a system of equations.

First we obtain the hypotenuse of the inner triangle

$4^{2} + 3^{2} = 25$ , so CE=5

If we name the segments

a = AF b = FB c = BE d = ED

and if we name any of the square sides 's'

We have the following 5 equations (for those 5 variables)

1. s=a+b
2. s=c+d
3. $s^{2}+d^{2}=25$
4. $s^{2}+a^{2}=16$
5. $b^{2}+c^{2}=9$

This equation system will yield 4 possible sets of solutions, we'll discard those with negative values and when we pick the one in which all variables are positive (because these are segments of a square and a triangle and they can't possibly be negative)

the value of s for that set is $s=16/\sqrt{17}$

$s^{2} = 15.058$