Iam having trouble getting the correct ans of 7.16m

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- Mar 18th 2011, 06:04 PMheatlyCircle Geometry - area of segment.
Iam having trouble getting the correct ans of 7.16m

- Mar 18th 2011, 06:45 PMProve It
http://i22.photobucket.com/albums/b3...cleproblem.jpg

Here is a picture of the way I would try this problem. I would evaluate the area of the segment, and subtract the area of the triangle. Of course, to do this, you need to know the size of the angle. - Mar 18th 2011, 06:55 PMQuacky
With your shorthand notation, I can't even comprehend the question, but why did you use $\displaystyle A=\pi~r^2$ to find the area of A1? A1 isn't a circle!

- Mar 18th 2011, 07:02 PMheatly
basically the area i have shaded is 7.87m^2.

the vertical height of the shaded area has to be half the height of the radius to the circle.

I need to find the radius of the circle hence then the diameter. - Mar 18th 2011, 07:19 PMQuacky
I understand your method, but the quoted $\displaystyle A=\pi~r^2$ formula only applies to full circles. It cannot be used to find an area of a segment, sector, etc. Or am I misinterpreting your logic? It seems to me that you are assuming that you can use the '0.5R' as a radius and the given area of the segment of 7.87 to find the radius of the whole circle.

- Mar 18th 2011, 07:25 PMheatly
ok ,yes i was assuming that-oops.

- Mar 18th 2011, 11:59 PMearboth
1. You probably found the correct solution already. If not, here are the steps:

a) Calculate the value of the central angle of the sector:

$\displaystyle \cos\left(\frac12 \alpha\right=\dfrac{\frac12 R}{R}=\frac12~\implies~\frac12 \alpha = 60^\circ$. So the complete central angle is 120°.

b) The shaded area is a segment.

$\displaystyle area_{segment}=area_{sector} - area_{triangle}$

$\displaystyle 7.87 = \frac13 \cdot \pi R^2 - \frac12 \cdot R \cdot \underbrace{R \cdot \sin(120^\circ)}_{\text{height of triangle}}$

$\displaystyle 7.87 = \left(\frac13 \pi - \frac14 \sqrt{3} \right)R^2$

Solve for R.

c) the diameter is 2R. - Mar 19th 2011, 12:27 AMheatly
Thanks for that

- Mar 19th 2011, 12:37 AMCaptainBlack
If you want anymore help with this create a new thread and this time if you must post a PDF with a scan of the question

**MAKE SURE IT IS THE RIGHT WAY UP**

CB