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?