Home -> Solved problems -> What is the area of the square?
Solution
To start, let \(s\) be the side length of the square and \(r\) the radius of the circle. Let’s simplify the problem using the next figure.
The square’s area is equal to \(s^{2}\), let’s find the side length of the square \(s\). Using the previous figure, we get
\[\begin{aligned}
&s+1=2 r\;\;\;\;\;\;\;(1)\\\\
&\text {Using Pythagoras theorem in the right grey triangle, we get}\\\\
&r^{2} =(r-1)^{2}+\left(\frac{s}{2}\right)^{2} \\\\
&r^{2} =r^{2}-2 r+1+\frac{s^{2}}{4} \\\\
&1-2 r+\frac{s^{2}}{4} =0 \\\\
&4-8 r+s^{2} =0\\\\
&\text {Using (1), we get}\\\\
&(2 r-1)^{2}-8 r+4=0 \\\\
& 4 r^{2}-4 r+1-8 r+4=0 \\\\
& 4 r^{2}-12 r+5=0 \\\\
& 4\left(r-\frac{1}{2}\right)\left(r-\frac{5}{2}\right)=0 \\\\
&r=\frac{1}{2} \quad Or \quad r=\frac{5}{2} \\\\
&r=\frac{s}{2}+\frac{1}{2}>\frac{1}{2} \\\\
\Rightarrow & r=\frac{5}{2} \\\\
\Rightarrow & \text { Square’s area }=s^{2}=(2 r-1)^{2}=16
\end{aligned}
\]
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Home -> Solved problems -> What is the area of the square?
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Home -> Solved problems -> What is the area of the square?