A square has an inscribed circle which has an inscribed square in, as shown.
Show that the area of the white bit is approximately 28.5% of the whole shape.Inscribed shapes

I started by calling the circle's radius r, so the length and width of the blue square is r and the area is $r^2$. The area of the circle would be $r^2* π$. I then searched my formula book about areas of inscribed shapes but I didn't find any. Can anyone help me?

P.S. I 'm just a Year 7 and this is Year 9 work so please explain clearly how you found the solution

  • $\begingroup$ Hint: the diameter of your circle is the diagonal of the inner square. $\endgroup$ – lulu Mar 21 '17 at 11:59
  • $\begingroup$ @lulu Doesn't that mean that I also need the length of the red square in order to solve it using Pythagoras' theorem $\endgroup$ – bio Mar 21 '17 at 12:04
  • $\begingroup$ Take any number you want, or use a variable $L$. The percent is scale invariant which means that this parameter will cancel out in the end. $\endgroup$ – lulu Mar 21 '17 at 12:07
  • $\begingroup$ Please clarify: the white bit is approximately 28.5% of what? (Also, to be precise, it's the area of the white bit that is approximately 28.5% of the area of something else.) $\endgroup$ – Barry Cipra Mar 21 '17 at 15:13
  • $\begingroup$ @BarryCipra Sorry I'll clarify $\endgroup$ – bio Mar 22 '17 at 7:54

Having called the radius of the circle as $r$, you can find out that the side of the outer square is $2r$. So, the area of the outer square comes out to be $4r^2$.

Now, if you notice carefully, some of the diameters of the circle also act as diagonals of the inner square. So, taking the side of the inner square as $a$, the length of diagonal ie $\sqrt2 a$ is the same as the diameter of the the circle ie $2r$. $$\implies \sqrt2 a = 2r \implies a = r\sqrt2$$..... (1)

Now, dividing the circle into symmetric halves by a diameter which is also the diagonal of the inner square, you can calculate the area of the white portion inside that half. This area will be the difference of the area of the semicircle and the area of the right triangle, which is a part of the inner square.

So, area of half white portion $= \frac{\pi r^2}{2} - \frac{1}{2}*a^2$ $$= \frac{\pi r^2}{2} - r^2 = \frac{\pi r^2 - 2r^2}{2}$$ [From (1)]

So, area of whole white portion $=2*\frac{\pi r^2 - 2r^2}{2} = \pi r^2 - 2r^2$

Now, calculating what percentage of the total area is the area of the white portion found above, we have;

$$\frac{\pi r^2 - 2r^2}{4r^2} * 100 = \frac{\pi - 2}{4} * 100 $$ $$= \frac{1.14}{4} * 100 = 28.5\%$$ [Taking $\pi = 3.14$]



If you called the circle's radius $r$, then the length of blue square is $2r$ (the diameter) and the diagonal of the red square is also $2r$ (the diameter), so its length is $\frac{2r}{\sqrt{2}}=\sqrt{2}r$.

So, we have the areas:

  • Blue square: $(2r)^2=4r^2$

  • Circle: $\pi r^2$

  • Red square: $(\sqrt{2}r)^2=2r^2$

  • $\begingroup$ Does this mean i can substitute r for a number and see if it makes the white bits 28.5% $\endgroup$ – bio Mar 21 '17 at 12:10
  • $\begingroup$ @bio: I don't need to do that. When you take the ratio, the factor $r^2$ will vanish. $\endgroup$ – Arnaldo Mar 21 '17 at 12:12
  • $\begingroup$ @bio: Just see that the white area is given by: circle's area - red square's area $\endgroup$ – Arnaldo Mar 21 '17 at 12:14
  • $\begingroup$ doesn't that make it π-$r^2$ $\endgroup$ – bio Mar 21 '17 at 12:16

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