(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: .$\displaystyle AB \:=\:54,\;\;CD = 2 $

Let the radius be $\displaystyle R.$

. . Then: .$\displaystyle OA = OB = OC = R$

. . And: .$\displaystyle OD = R - 2$

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

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

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

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

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

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

The area of sector $\displaystyle ACBO$ is:

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