Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,866

Problem was just posed by ganesh.

ganesh wrote:

Find the ratio of the areas of the incircle and circumcircle of a square.

I know there is a lot of ways to do this but supposing you did not have any idea how to solve this. Geogebra to the rescue!

1) Make a 4 sided regular polygon. ( a square )

2) Use the 3 point circle option to draw the outer circle using 3 of the vertices of the square.

3) Draw the diagonal line segments to get the center of the square.

4) Put a point F on the square. Make that line segment from the center to F parallel with the x axis.

5) use the point and radius circle option to make a point from the center of the square to f.

6) Get areas of both circles.

7) Take the ratio:

8) use one of the free vertices to expand the inner circle. Find the new areas. What do you deduce?

Looks like the ratio of the areas is 1 / 2. Not rigorous but definitely enough to go to war with!

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

hi bobbym

i solved this before,and it is very easy.

the radius of the inner circle is a/2 where a is the side of the square.so it's area is A1=pi*a^2/4

the radius of the outer circle is a/sqrt(2).it's area is A2=pi*a^2/2.

*Last edited by anonimnystefy (2011-08-19 03:33:26)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

Offline

Pages: **1**