by Oliver
Last Updated August 14, 2019 15:20 PM

Find the area of the blue shaded region of the square in the following image:

How to solve this question?

**Hint:**

Consider a unit square with the bottom left corner at the origin and the point $(t,1)$.

The oblique from the bottom left corner has equation

$$y=\frac{x}{t}$$

and the other oblique is perpendicular, from the bottom right corner, hence

$$y=-t(x-1).$$

The intersection point is

$$\left(\frac{t^2}{t^2+1},\frac{t}{t^2+1}\right).$$

To determine $t$, we express the value of the ratio of the known sides,

$$\frac{4.8}{6}=\frac{\left\|\left(\dfrac{t^2}{t^2+1},\dfrac{t}{t^2+1}\right)-(t,1)\right\|}{\left\|\left(\dfrac{t^2}{t^2+1},\dfrac{t}{t^2+1}\right)-(1,0)\right\|}.$$

All the rest will follow.

The "extra information" required is down there in fine print: the outer shape is square.

Without loss of generality let us scale the figure by $\frac56$ so that the given lengths become $4$ and $5$. Now let the square's side length be $x$. We can set up an equation for it (using multiple invocations of Pythagoras's theorem) as
$$\sqrt{41-x^2}+\sqrt{(4+\sqrt{x^2-25})^2-x^2}=x$$
Solving this using SymPy I found *two* possible values for $x$, both of which give admissible diagrams:
$$x=\sqrt{\frac{65\pm5\sqrt5}2}$$
Now, given this $x$, the triangle on the left has an area of
$$\frac{\sqrt{x^2-25}(\sqrt{x^2-25}-1)}2$$
and that on the right has an area of
$$\frac{x\sqrt{41-x^2}}2$$
Substituting the values of $x$ into the sum of these arsas reveals that both choices give **the same area**, namely $10$. Remember that we scaled down beforehand, so we multiply by $\left(\frac65\right)^2$ and obtain the final answer of $14.4$ square centimetres.

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