What is the area of the square

[american3141]

it is a circle and a square 1 unit apart Help???

 

[MarkFL]

Hello, and welcome to FMH!

I would let \(s>0\) be the side length of the square. Assuming the circle and square are tangent on the left side, we then know the diameter \(d\) of the circle is:

[MATH]d=s+1[/MATH]
In terms of the radius \(r\) of the circle, this is:

[MATH]2r=s+1[/MATH]
Using the Pythagorean theorem, we may state

[MATH](r-1)^2+\frac{s^2}{4}=r^2[/MATH]
Given that \(0<s\), what do you conclude?

 

[MarkFL]

Here's a diagram to help:

 

[MarkFL]

To follow up:

[MATH]r^2-2r+1+\frac{s^2}{4}=r^2[/MATH]
[MATH]-(2r-1)+\frac{s^2}{4}=0[/MATH]
[MATH]-s+\frac{s^2}{4}=0[/MATH]
[MATH]s^2-4s=0[/MATH]
[MATH]s(s-4)=0[/MATH]
The root \(s=0\) corresponds to a "degenerate" square of area 0 (a point on a circle of radius 1) and so we are left with:

[MATH]s=4[/MATH]
Thus, the area \(A\) of the square is:

[MATH]A=s^2=4^2=16[/MATH]

 

[american3141]

Thanks!