# Thread: Circle Geometry - area of segment.

1. ## Circle Geometry - area of segment.

Iam having trouble getting the correct ans of 7.16m

2. 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.

3. 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!

4. 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.

5. Originally Posted by heatly
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.
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.

6. ok ,yes i was assuming that-oops.

7. Originally Posted by heatly
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.
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.

8. Thanks for that

9. 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