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Points $L$ and $M$ lie on a circle $\omega_1$ centered at $O$.  The circle $\omega_2$ passing through points $O,$ $L,$ and $M$ is drawn.  If the measure of arc $LM$ in circle $\omega_1$ is $90^\circ,$ and the radius of $\omega_1$ is 1, then find the area of triangle $LOM$.

 Apr 30, 2024
 #1
avatar+178 
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The area of triangle LOM is just \((1/2) r^2 sin (90°)\)

 

\(sin(90) = 1\) amd the circle has radius 1, so we just have \((1/2)(1)^2 * (1) \)

 

This means our final answer is just \([ LOM ] = (1/2) r^2 sin (90°) = (1/2)(1)^2 * (1) = 1/2\)

 

1/2

 

Thanks! :)

 Apr 30, 2024

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