need help from a math wiz!!!

[ruck'n'reefer]

I need to find the area of...part of a circle.
It has a 2.5 inch diameter. I need to know the area of the 1/2 inch outer cross section.
I hope this makes sense.
If I was to cut off 1/2 an inch of the circle how much would I have?

Thanx muchly

[DJ88]

HI rob.. ok from what you are wanting I presume you mean by the 1/2 inch as a cirlce within the circle right?

1/2" trimmed off the outside? If so the area of the 1.5 inch diameter circle(center piece) will be 1.767 inch^3.

The original 2.5 inch diameter circle has an area of 4.908inches^3.


The 1/2 inch piece you trimmed off is basically pi.. 3.141592654inches^3 etc etc etc..

Does that help?

[ 22 October 2001: Message edited by: DJ88 ]

[StirCrazy]

ok what you have to do is figure out the area of 2 circles and then subtract them.. so your outer circle would be 2 x pi x r(squared)
then draw your smaller cirle repersenting the part you don't want and figure that one out.. then subtract them

hope this helps
Steve

[DJ88]

Is it 2 pi r^2? or pi r^2? damn..

It's pi r^2.. 2 pi r is the circumference and pi r^2 is the area..

FYI ^2 means squared.. [img]images/smiles/icon_wink.gif[/img]

[ 22 October 2001: Message edited by: DJ88 ]

[FlameAngel]

Pir^2 = area
2Pir = circumference

FlameAngel

P/S Why isn't there a 'Pi' on the keyboard ARGRRGHH

[reefburnaby]

Hello,

Let's define some equations.
A = pi * R * R
D = 2 * R

so

A = pi * D * D / 4

where pi = 3.14159..., D = diameter, R is radius.

The area of the whole circle is

AW = pi * 2.5 * 2.5 / 4
AW = pi * 25/16 square inchs.

The area of the circle that is trimmed off is

AT = pi * 1.5 * 1.5 /4
AT = pi * 9/16 square inches

The remaining "donut" that is 1/2 thick and 2.5" diameter (outside) is

AD = AW - AT
AD = pi (25/16 - 9/16)
AD = pi square inches.

So...Darren is correct if it is a donut. On the other hand, if it is a 1/2" slot (basically a semicircle)...then this is not the correct solution. Hope that helps.

- Victor.

[ruck'n'reefer]

Thank you for the input!!!!!
I don't think I explained well enough!!!

Here is the full circle O
here is the other piece C
I need to know the area of the piece left over. This piece is .5". The diameter of the circle is 2.5". The "C" piece is 2". The other piece is .5"

I hope this helps.
Rob

[reefburnaby]

Hello,

You mean this,
where you are trying to find the area of A. Diameter of the circle is 2.5" and the maximum height of the semicircle is 0.5".

- Victor.

[ 23 October 2001: Message edited by: reefburnaby ]

[ 23 October 2001: Message edited by: reefburnaby ]

[ruck'n'reefer]

Yes!!!!
Thanx for the graphic Victor!!!
That is it!

[reefburnaby]

Hello,

Here we go....

Were R is the radius, AC is perpendicular to BO. We are trying to find the area in semicircle ABCD.
------------
First, find the area of the pie ABCO.

R = AO = BO = CO = 1.25"
BD = 0.5"

DO = BO - BD = 1.25" - 0.5" = 0.75"

Angle DOC = inv cos (DO / CO)
Angle DOC = 0.9273 rad or 53.13 degrees.

Since AD = DC, then angle DOC = angle DOA.

Therefore Angle AOC = 2 * angle DOC = 106.26 deg.

A circle has 360 degrees, so the pie ABCO is 106.26 / 360 = 29.52% of the circle.

Area of ABCO pie = pi * R * R * pie's percentage of the circle.
Area of ABCO pie = 1.4489 square inches.

-------------
Next, find area of triangle ADCO.

DC = AD = sqrt (CO * CO - DO * DO)
DC = AD = 1"

Area of ADCO = AC * DO / 2 = 0.75 square inches.

--------------
Area of the semicircle.

Semicircle ABCD = Pie ABCO - Triangle ADCO
Semicircle ABCD = 0.69889 square inches.

- Victor.

[ 23 October 2001: Message edited by: reefburnaby ]