## Calculate the sum of areas of the three squares

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### Solution

We have a top part of a circle of diameter $$5$$ drawn with three squares. To start let $$a$$ be the side length of the blue square, and $$b$$ the side length of the green square
As we see in the previous figure the side of the grey square equals to $$a+b$$. Let’s take the next example of similar right triangles
We get these results: $a^{2}=d^{2}+h^{2}$ $b^{2}=e^{2}+h^{2}$ $\left(d+e\right)^{2}=a^{2}+b^{2}$ $d^{2}+e^{2}+2de=a^{2}+b^{2}$ $d^{2}+e^{2}+2de=d^{2}+h^{2}+e^{2}+h^{2}$ $2de=2h^{2}$ $h^{2}=de$
Now let’s return to our problem and apply the previous result to the next figure
Therefore we get, $b^{2}=\left(b-a\right)\left(2a+b\right)$ $=2ab+b^{2}-2a^{2}-ab$ $2a^{2}-ab=0$ $a\left(2a-b\right)=0$ $\color{blue} {a\neq0}$ $\Rightarrow2a=b$ Using another fact in the figure, we have $\left(b-a\right)+a+\left(a+b\right)=5$ $2b+a=5$ $2\left(2a\right)+a=5$ $5a=5$ $a=1$ Finally, let $$S$$ be the sum of areas of the three squares, thus $S=a^{2}+b^{2}+\left(a+b\right)^{2}$ $=a^{2}+b^{2}+a^{2}+b^{2}+2ab$ $=2a^{2}+2b^{2}+2ab$ $=2\left(a^{2}+b^{2}+ab\right)$ $=2\left(a^{2}+4a^{2}+2a^{2}\right)$ $=14a^{2}$ $\large \Rightarrow S=14$
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