## 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 \[S=a^{2}+b^{2}+\left(a+b\right)^{2}\] \[=a^{2}+b^{2}+a^{2}+b^{2}+2ab\] \[=2a^{2}+2b^{2}+2ab\] \[=2\left(a^{2}+b^{2}+ab\right)\] \[=2\left(a^{2}+4a^{2}+2a^{2}\right)\] \[=14a^{2}\] \[\large \Rightarrow S=14\]

Home -> Solved problems -> Areas three squares

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Home -> Solved problems -> Areas three squares