Geometry puzzle

**Rheghead** · 13-Sep-05, 12:56

This one has got me stumped.

1 Draw a 10cm square.

2 Draw 4 10cm 90 degree radius arcs from each corner of the square, from one corner inside the square to the other.

What is the area of the the middle shape which is a wonky sided circle/square thing?

My niece has asked me this but it has got me. I think it has to be answered by integration rather than trigonometry.

DrSzin this must appeal to you?

**Whitewater** · 13-Sep-05, 15:12

Hi Rhegers, I havn't worked it out but should it not be the difference between the area of the 10cm circle and the area of the 10cm square, (you have four quarters of the circle)

**Whitewater** · 13-Sep-05, 16:01

Hi Rhegers (again) after having sketched out your puzzle what I've assumed is rubbish. sorry!!

**Rheghead** · 13-Sep-05, 17:25

You should have a square with a four point star in it

**jimag** · 13-Sep-05, 17:35

Try 31.514674 square centimeters for starters.

**Rheghead** · 13-Sep-05, 17:37

Originally Posted by jimag

Try 31.514674 square centimeters for starters.

How did you work it out?

**jimag** · 13-Sep-05, 17:38

I must have drawn it wrongly because I do not have a star. (Dont feel one either)

**jimag** · 13-Sep-05, 17:39

Rheghead

You tell me how you are going about it.

**Rheghead** · 13-Sep-05, 17:46

I tried deducting (pi r^2)/4from 100 then I took that answer from (pi r^2)/4, but that just gave me an area for the pointy oval shape.

**Tristan** · 13-Sep-05, 19:27

The answer should be 14.15926536 cm2

The area of the square is 100 cm2

If a whole circle was drawn the area would be Pi*100 = 314.1592654cm2
1/4 of the circle is 78.53981634cm2
subtract this from the area of the square 100-78.5398 = 21.46018366cm2
This is the area that is between the outside of the circle but still within the square
There are 4 of these so 4*21.46018366 = 85.84073464cm2 (the area outside all 4 aches
subtract this from 100cm2 and you have the area of the inner bit
100-85.84073464 =
14.15926cm2

**George Brims** · 13-Sep-05, 19:44

The area is 31.5147 square cm as jimag said. I drew the figure in AutoCAD and used the area tool. Now somewhere around 1968 I could have worked it out from Geometry but if my employer is going to buy me $5000 worth of software and a nice computer, why should I trouble my ever-declining brain?

**hereboy** · 13-Sep-05, 19:45

or so you would think...

but the four areas which are between each quarter circle and the rest of the square intersect and overlap each other... so the area outside the shape is not 85 but less than that, which would make the area inside the shape greater than 14...

try again....

PS. my reply was to Tristan.

**Rheghead** · 13-Sep-05, 20:23

As always when we are back to school, the teacher wants to see the working out.

It has got me beat!!

**DrSzin** · 13-Sep-05, 20:28

Originally Posted by jimag

Try 31.514674 square centimeters for starters.

That's what I get too! Did I really get it right first time? Amazing!

A quick calculation gives, for a square of side R:

Area = (Pi/3 + 1 - sqrt(3)) R^2

where Pi = 3.14159..., sqrt means square root, and R^2 is R-squared.

I haven't checked my calculation but it surely can't be a coincidence that my answer agrees with jimag's to 8 significant figures.

I'll provide working for teacher if he really wants it. I just integrated over the plane area as he suggested. I suspect there's a sneaky geometrical solution, but I'm hopeless at geometry.

**Rheghead** · 13-Sep-05, 20:34

jeepers, how did you come up with the solution?

**George Brims** · 13-Sep-05, 20:36

Where did you get that formula DrSzin? I am not sure what you mean by "a square of side R".

Anyway I just wasted a good half hour drawing extra lines on that thing and putting in angles, and I did finally manage to work it out instead of having the computer do it. It's not as hard as I at first thought. Bit hard to explain here in words without the benefit of pictures though.

**DrSzin** · 14-Sep-05, 00:27

Originally Posted by George Brims

Where did you get that formula DrSzin? I am not sure what you mean by "a square of side R".

I meant a square whose side has length R (R = 10cm in the original version).

Originally Posted by George Brims

Anyway I just wasted a good half hour drawing extra lines on that thing and putting in angles, and I did finally manage to work it out instead of having the computer do it. It's not as hard as I at first thought. Bit hard to explain here in words without the benefit of pictures though.

You're right, it is indeed easy to work out using simple geometry. But at least my solution above was correct, even if I did use a sledgehammer to crack a nut.

And it's too late at night to explain the simple solution in words...

**scotsboy** · 14-Sep-05, 09:17

I had a totally different shape as I drew the "arcs" on the outside of the square

It wasn't till Rheghead posted the wee graphic that I realised they should be inside the box - a picture is worth a thousand words......or at least a few hundred

**jimag** · 14-Sep-05, 09:53

DrSzin

Sledge hammer or not you are to be commended for your effort, but where did you get the formula. Would you like to show us how it was derived. I, like our friend across the pond, took the easy route and drew it out using CAD.