Finding the area of a circle...with a twist!

\(\angle AOC = 90^\circ \)

The angle AOC is twice the angle ABC. Now use Pythagoras theorem.

Cool problem. :) Basically, we want to find the radius. If we look at it a bit, we can see two radii as the sides of triangle AOC, so that triangle must be isosceles. If we use the "Central Angle - Inscribed Angle" idea that the central angle is twice the inscribed angle, we can find the angle that triangle AOC opens at the vertex. Then, you have to use a bit of trigonometry to find the radius from that isosceles triangle, and then area formula should work out from there.

You don't need trig, because the central angle is a special angle :)

\(2r^2=9 \implies r = 1.22474 \) Area \(= 14.1372 \; ft^2 \)

Awesome. thanks guys!

:)

You're welcome! I usually just refer to general triangle magics as Trigonometry. :P

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