AMC10 2024 A

Error creating thumbnail: Unable to save thumbnail to destination First, we should find the length of $AB$. In order to do this, as we see in the diagram, it can be split into 4 equal sections. Since diagram $K$ shows us that it is made up of two ${30,60,90}$ triangles, then the triangle outlined in red must be a $30,60,90$ triangle, as ${30+30=60}$, and the two lines are perpendicular (it is proveable, but during competition, it is best to assume this is true, as the diagram is drawn pretty well to scale). Also, since we know the length of the longest side of the red triangle is ${\sqrt3}$, then the side we are looking for, which is outlined in blue, must be $\frac{3}{2}$ by the ${1,\sqrt3, 2}$ relationship of ${30,60,90}$ triangles. Therefore $AB$, which is the base of the triangle we are looking, for must be $6.$ Now all we have to do is find the height. We can split the height into 2 sections, the green and the light green. The green section must be ${\sqrt3}$, as $K$ shows us. Also, the light green section must be equal to ${\frac{\sqrt3}{2}}$, as in the previous paragraph, the triangle outlined in red is $30,60,90$. Then, the green section, which is the height, must be ${\sqrt3}+{\frac{\sqrt3}{2}}$, which is just ${\frac{3\sqrt3}{2}}$. Then the area of the triangle must be ${\frac{1}{2}}\cdot {b} \cdot {h}$, which is just $\boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.$ Simple Proof of the red triangle having $30^\circ$, $60^\circ$, and $90^\circ$ angles: In the small kite illustrated in the problem, we can see that the two angles on the left side add up to $60^\circ$ (this is easily noticeable if you look hard enough.) So, the angle that the arrow is pointing to is $60^\circ$. Since another angle in the red triangle is obviously $90^\circ$, the portion of the angle that is in the red triangle and at point A is $180^\circ - 90^\circ - 60^\circ = 30^\circ$. Therefore, the red triangle is a $30^\circ$ $60^\circ$ $90^\circ$ triangle.