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?

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Yan Peng
August 14, 2019 15:11 PM

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.

Yves Daoust
August 14, 2019 15:12 PM

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.

Parcly Taxel
August 14, 2019 15:13 PM