Sector of a circle

[1141]

Can someone help with the attached questions, please?

For i.) I'm not sure how to start it? How do I find AX?

For ii.) I know I need to find the area of the sector, which I found:

1/2r[sup:1ohvsnib]2[/sup:1ohvsnib]?
= 1/2 . 12[sup:1ohvsnib]2[/sup:1ohvsnib] . 1/3?
= 24?

But to carry on, I first need to find AX, but since I'm not sure how to do i.) I can't carry on.

 

[soroban]

Hello, 1141!

For (i), I'm not sure how to start it.

For (ii), I know I need to find the area of the sector, which I found:

. . $\displaystyle \tfrac{1}{2}r^2\theta \:=\:\tfrac{1}{2}(12^2)(\tfrac{\pi}{3}) \:=\: 24\pi$ . Good!

$\displaystyle \text{Draw }OX.$

$\displaystyle \text{Then }\Delta X\!AO\text{ is a 30-60 right triangle.}$

$\displaystyle \text{Hence: }\:OA \,=\, 12,\;\;O\!X \,=\, 8\sqrt{3},\;\;\boxed{AX \,=\, 4\sqrt{3}}$


$\displaystyle \text{The area of }\Delta X\!AO \:=\:\tfrac{1}{2}(12)(4\sqrt{3}) \:=\:24\sqrt{3}$

$\displaystyle \text{The area of kite }OAXB \:=\:2\times 24\sqrt{3} \:=\:48\sqrt{3}$


$\displaystyle \text{Therefore: }\;\text{(Shaded area)} \;=\;\text{(Kite)} - \text{(Sector)}$

. . . . . . . . . . . . . . . . . . $\displaystyle \;=\;48\sqrt{3} - 24\pi$

. . . . . . . . . . . . . . . . . . $\displaystyle =\;24(2\sqrt{3} - \pi)$