Thread: Finding the area of a circle segment with very limited information.

1. Finding the area of a circle segment with very limited information.

I am trying to help my grandfather who is a making a model sail boat. The sails are isosceles triangles with a curve along the largest edge. He needs the area in square inches of the sails. I have found the area of the triangles themselves, but i cant seem to come up with a method of getting the area of the curved part, which is basically a circle segment. The only information i have on them is the chords and the distance between the chord and the arc.

Length of chord: 54 in.
Distance to arc: 2 in.

If i can see the method i can use it on the other sail. But after searching various data banks i can not find anything. I have found the area using radius of the circle and degree of the angle made from the two points of the chord, but i do not have either.

Any help would be much appreciated!

2. (Sleepy)Hello, Gedalas!

I am trying to help my grandfather who is a making a model sailboat.
The sails are isosceles triangles with a curve along the largest edge.
He needs the area in square inches of the sails.
I have found the area of the triangles themselves, but i can't seem to come up with
a method of getting the area of the curved part, which is basically a circle segment.

The only information i have on them is the length of the chord
. . and the distance between the chord and the arc.

. . Length of chord: 54 in.
. . Distance to arc: 2 in.

I will assume that the circular arc is outside the isosceles triangle.

Code:
                    C
*  *  *
*       |2      *
*           |           *
A * - - - - - - + - - - - - - * B
\     27     |D    27     /
\           |           /
\          |          /
\         |         /
\        |        /
R \       |R-2    / R
\      |      /
\     |     /
\    |    /
\   |   /
\  |  /
\ | /
\|/
*
O

We are given: . $AB \:=\:54,\;\;CD = 2$

Let the radius be $R.$
. . Then: . $OA = OB = OC = R$
. . And: . $OD = R - 2$

In right triangle $BDO:\;\;(R-2)^2 + 27^2 \:=\:R^2$

. . $R^2-4R + 4 + 729 \;=\;R^2 \quad\Rightarrow\quad 4R \:=\:733$

Hence, the radius is: . $R \;=\;\dfrac{733}{4} \;=\;183.25\text{ in}$

In $\Delta AOB$, apply the Law of Cosines:

. . $\cos(\angle AOB) \;=\;\dfrac{183.25^2 + 183.25^2 - 54^2}{2(183.25)^2} \;=\;0.956582026$

. . $\angle AOB \:=\:16.9455792 \;\approx\;17^o$

The area of sector $ACBO$ is:

. . $\dfrac{17^o}{360^o}\pi(183.25)^2 \;\approx\;4981.78\text{ in}^2 \;\approx\;34.60\text{ ft}^2$

3. First off, my apologies on saying isosceles, it was a slip of the fingers, it is an acute triangle, but that does not affect your solution which does indeed help me. Thank you! You gave me the area of the sector, while i needed the area of the segment, anyway, i was able to use your radius and interior angle answers and plugged them into the A= (R^2/2)(theta-sin(theta)) for the circle segment. If anyone is interested the answer is approx. 72.77 in. sq.