Inscribed Square in Inscribed Circle in Square

$\begingroup$

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.

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

$\endgroup$ 5

2 Answers

$\begingroup$

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 $\pi = 3.14$]

$\endgroup$ $\begingroup$

Hint

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: $(\sqrt{2}r)^2=2r^2$

$\endgroup$ 4

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy