# Homework Help: Area bounded inside the quarter-circles.

1. Jul 24, 2010

### kensaurus

Its a question that I had from a friend in the past.
I had tried solving it but to no avail.

Have tried integration and stuff like that, but I think there is an easier way to solve this question.

Question -> Square of 7cm, find the shaded area.
[PLAIN]http://img833.imageshack.us/img833/555/questionx.jpg [Broken]

Thanks for any help!

Last edited by a moderator: May 4, 2017
2. Jul 24, 2010

### hgfalling

Call the area in the center Z, the four pieces that border Z Y, and the flattish triangular pieces that don't border Z X.

Now write three equations based on the following:

1) The area of one of the quartercircles is pi/4 times the area of the square.
2) Consider the area if you take the two quartercircles at opposite ends of a diagonal. Together they are pi/2 * area of the square. But the entire square is included in both, with some excess.
3) Consider the "rounded" equilateral triangle made up of Z and the two Ys and X on one side of the square. Find the area of this "rounded" equilateral triangle, and set it equal to those three.

Once you do this you'll have three equations in three unknowns and voila!

3. Jul 24, 2010

### kensaurus

i just realized that the equations cant be solved based on your equations.

it still results in 2 equations and I AM UNABLE to find your third equation. any help please?

Last edited: Jul 24, 2010
4. Jul 25, 2010

### ehild

You can find the area of piece X by integration.

ehild

5. Jul 25, 2010

### kensaurus

so it all comes down to integration?
okay, thanks...

btw, this was a question from mensa, so i think there should be a brilliant way to solve it -.-

6. Jul 25, 2010

### Hurkyl

Staff Emeritus
It doesn't look to hard to split the center piece into 4 triangles and 4 circular segments.

But anyways, who's more brilliant? The person who solves it in a minute with integration, or the person who spends 5 minutes trying to come up with a "brilliant" solution?

7. Jul 25, 2010

### hgfalling

For a unit square, the three equations are:

$X + 2Y = 1 - \pi/4$
$Z + 2Y = \pi/2 - 1$
$X + 2Y + Z = \pi/3 - \sqrt{3}/4$

Solving these yields $Z = \pi/3 + 1 - \sqrt{3}$

8. Jul 25, 2010

### ehild

It is really brilliant! But isn't a typo in the first equation ? It should be $2X + Y = 1 - \pi/4$, isn't it?

ehild

9. Jul 25, 2010

### hgfalling

Oh, right, that's correct.

10. Jul 26, 2010