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

$$\begin{array}{rl}& s+1=2r(1)\\ \\ & \text{Using Pythagoras theorem in the right grey triangle, we get}\\ \\ & r^2=(r-1{)}^2+{\left(\frac{s}{2}\right)}^2\\ \\ & r^2=r^2-2r+1+\frac{s^2}{4}\\ \\ & 1-2r+\frac{s^2}{4}=0\\ \\ & 4-8r+s^2=0\\ \\ & \text{Using (1), we get}\\ \\ & (2r-1{)}^2-8r+4=0\\ \\ & 4r^2-4r+1-8r+4=0\\ \\ & 4r^2-12r+5=0\\ \\ & 4(r-\frac{1}{2})(r-\frac{5}{2})=0\\ \\ & r=\frac{1}{2}Orr=\frac{5}{2}\\ \\ & r=\frac{s}{2}+\frac{1}{2}>\frac{1}{2}\\ \\ \Rightarrow & r=\frac{5}{2}\\ \\ \Rightarrow & \text{ Square’s area }=s^2=(2r-1{)}^2=16\end{array}$$

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