+241 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 $LOM$.
+8 Points \( L \) and \( M \) lie on a circle \( \omega_1 \) centered at \( O \). The circle \( \omega_2 \) passes through points \( O \), \( L \), and \( M \). The measure of arc \( LM \) in circle \( \omega_1 \) is \( 90^\circ \), and the radius of \( \omega_1 \) is 1. We are tasked with finding the area of triangle \( LOM \).
The key observation here is that triangle \( LOM \) is a right triangle. Since the measure of arc \( LM \) in circle \( \omega_1 \) is \( 90^\circ \), the central angle \( \angle LOM \) subtended by arc \( LM \) is also \( 90^\circ \).
Therefore, triangle \( LOM \) is a right triangle with the right angle at \( O \). The lengths of the legs \( OL \) and \( OM \) are both equal to the radius of \( \omega_1 \), which is 1.
The area of a right triangle is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. \] In this case, the base and height are the lengths of \( OL \) and \( OM \), both of which are 1. Therefore, the area of triangle \( LOM \) is: \[ \text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}. \]
Thus, the area of triangle \( LOM \) is \( \boxed{\frac{1}{2}} \).
