I have a square of length $10$ and a circle of radius $1.$ The circle rolls along the outside perimeter of the square, and I want to find the area it sweeps out as it completes its path.

The corners are really troubling me. I seem to need to find the area of a $90$ degree cutout of a circle, when the angle starts on the perimeter. I hope that makes sense. Otherwise, I know I can just add the four rectangular parts after. Can I get some help finding the area of these corner parts?

  • 1
    $\begingroup$ Join the corner regions together to make a larger shape whose area should be easy to see. You actually get the same result with any convex polygon; it's based on the exterior angles adding up to 360°. $\endgroup$ Commented Nov 10, 2021 at 23:48
  • 1
    $\begingroup$ So is the larger area a circle of radius $2$? And would the final answer then be $4(2\times10)+4\pi=80+4\pi$? $\endgroup$
    – mauCheese
    Commented Nov 10, 2021 at 23:52
  • $\begingroup$ Right on the money! ⭐ $\endgroup$ Commented Nov 11, 2021 at 0:00

1 Answer 1


To Begin:
The shape created can be divided into $2$ distinct subsets of shapes. Firstly, the rectangles are the easy ones to deal with, while the circle simply rolls along the edge. The curves created at each corner of the square could be slightly harder to deal with, but they are still doable.

The Rectangles:
As the circle has a radius of $1$, its diameter is double that, or $2$. Therefore, each rectangle would have a width of $2$ units. Its length is the length of the square, which we know to be $10$. Each rectangle would then have an area of $2*10=20$, and all four combined would be $20*4=80$ units$^2$.

The Quarter-Circles:
Each corner of the square creates a curved shape as the circle rolls around it. As the distance from the square to the new outer edge is always the same, this curve is a quarter-circle with radius $2$. As there are four of these shapes, they combine to make a single full circle. Solving from here, we get $A=\pi 2^2=4\pi$ units$^2$.

This is a very simple process. We merely add $80$ and $4\pi$, which we can leave as $4\pi+80$ or evaluate fully to $92.5663706144...$ units$^2$.

  • $\begingroup$ Ironically, the OP got it 7 minutes earlier. $\endgroup$ Commented Nov 11, 2021 at 0:02
  • $\begingroup$ Truly so. I didn't notice as when I began writing this answer, the comment had not yet been posted. $\endgroup$
    – PiGuy314
    Commented Nov 11, 2021 at 0:04
  • $\begingroup$ What would be the proper way to proceed from here? $\endgroup$
    – PiGuy314
    Commented Nov 11, 2021 at 0:05
  • $\begingroup$ I would just leave it. The OP might accept your answer too. $\endgroup$ Commented Nov 11, 2021 at 0:09
  • $\begingroup$ Fair point. Thanks for the suggestion. : ) $\endgroup$
    – PiGuy314
    Commented Nov 11, 2021 at 0:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .