Trig. problem

[Tinkermom]

I need help on the attached Trigonometry problem. I think I found the Area of the sector, but not the shaded area only. I see one mistake I think....Here is what I did.

A=1/2 r^2 (theta)

A = 1/2 (2(pi) - 3root3/12) (theta)

A = 12(4(pi) - 6root 3/12) (theta)

A= 4(pi) - 6root3/12 (theta)

A = pi - root3/6 (theta)

Oh boy....I need help.

 

[Deleted member 4993]

I need help on the attached Trigonometry problem. I think I found the Area of the sector, but not the shaded area only. I see one mistake I think....Here is what I did.

A=1/2 r^2 (theta) <<< How did you get that equation??
A = 1/2 (2(pi) - 3root3/12) (theta) <<< where did "r" go?? However, as defined in the problem, that should be "s".

A = 12(4(pi) - 6root 3/12) (theta)

A= 4(pi) - 6root3/12 (theta)

A = pi - root3/6 (theta)

Oh boy....I need help.

 

[soroban]

Hello, Tinkermom!

Your formula is correct.

\(\displaystyle \text{The area of a sector is: }\:A \:=\:\tfrac{1}{2}r^2\theta\)

. . \(\displaystyle \text{where }r\text{ is the radius and }\theta\text{ is the central angle in radians.}\)


\(\displaystyle \text{We have: \,radius }= s,\;\;\theta = 60^o = \tfrac{\pi}{3}\)

\(\displaystyle \text{Hence: }\:A(\text{sector}) \:=\:\tfrac{1}{2}s^2(\tfrac{\pi}{3}) \:=\:\tfrac{\pi}{6}s^2\)


\(\displaystyle \text{The area of an equilateral triangle of side }s\text{ is:}\)

. . \(\displaystyle A(\text{triangle}) \:=\:\tfrac{\sqrt{3}}{4}s^2\)


\(\displaystyle \text{Therefore, the area of the segment is:}\)

. . \(\displaystyle \frac{\pi}{6}s^2 - \frac{\sqrt{3}}{4}s^2 \;=\;s^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) \;=\;s^2\left(\frac{2\pi - 3\sqrt{3}}{12}\right)\)

 

[Tinkermom]

At first I didn't see how you got root3/4 s^2. I had to stare at it a while...but I got it.

Thank you for your help.