1
$\begingroup$

I'm doing some geometry exercises. The first thing I did was inscribe a circle within a square (gave them the same diameter) and found the ratio between the areas to be Circle/Square=$\pi$/4.

The second thing I did was inscribe a square within a circle. Now, the diameter of the circle is the diagonal of the square. Therefore, if we were to make a square out of that diagonal, it would have the same ratio as above to the circle. We also know that this square would have twice the area of the original square. Now, let's get into the math I had problems with.

Let: $d$=diagonal=diameter; $x$=sidelength of square; $r$=radius=$d/2$; $C$=area of circle; $x^2$= area of original square; $d^2$= area of diagonal square.

$d^2=2x^2$

$d=√(2x^2)=x√2=2r$

$r=(x√2)/2$

$C=πr^2=π([x√2]/2)^2$>

$C=π(x/2)^2 $

That last part there is my problem. I've looked over it multiple times and can't find where my math is wrong, but I know that it must be because the final line indicates that x/2=r when it does not. I've also tried it this way, plugging into the ratio:

$C/(d^2)=π/4 $

$C=(π[d^2])/4 $

$C=(π[{x√2}^2])/4 $

$C=(2πx^2)/4 $

$C=(πx^2)/2 $

This is the same answer given by a different method, which means that my error must be very basic. I'd like to say now that it has been a while since I've done algebra so I'd beg for mercy in criticising my mistake(s). Also, I'm new to this Exchange so I'm not sure how the formatting will turn out. Nonetheless, any help is greatly appreciated. Also, I'm going to be doing this with many different regular polygons to try finding patterns (although the struggle I've had with a simple square worries me). I'm sure it's been done before and I could probably just look it up, but it's all for the sake of practice. Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ This has nothing to do with algebraic geometry - please put geometry there instead $\endgroup$ – TMO Jul 1 '19 at 21:18
  • $\begingroup$ $\frac {\sqrt 2}2=\frac 1{\sqrt 2}\neq \frac 12$ $\endgroup$ – lulu Jul 1 '19 at 21:19
1
$\begingroup$

Just looking at your top calculation:

You are correct up to $$C=\pi r^2 = \pi \left (\frac {x\sqrt 2}2\right)^2$$

But then you somehow replace $\frac {\sqrt 2}2$ with $\frac 12$. If you correctly replace it with $\frac 1{\sqrt 2}$ you get the relation you expect.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot! Also, I should learn to format as you did haha much more readable. $\endgroup$ – Becausedefect Jul 1 '19 at 21:36
  • $\begingroup$ After looking over it again, I realized I skipped a step at that spot. For π([x√2]/2)^2 I got (2πx)/4, which then simplified to πx/2. I thought I caught onto my error but perhaps not, because I still feel like that is how the calculation should be. $\endgroup$ – Becausedefect Jul 1 '19 at 21:59
  • $\begingroup$ here is a good tutorial on formatting for this site. Also, if you click on "edit" you can see the syntax I used (but don't forget to "Cancel" afterwards). $\endgroup$ – lulu Jul 1 '19 at 23:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.