Question

Let $\mathcal{K}$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal{K}$ are used to form the polygon shown below. What is the area of $\triangle ABC?$

$2 + 3\sqrt3$

$\dfrac{9}{2}\sqrt3$

$\dfrac{10 + 8\sqrt3}{3}$

$8$

$5\sqrt3$

Answer: B

Concepts:kite special right triangle coordinate geometry triangle area

Difficulty rating: 2120

Solution:

Each half of a kite is a $30$ - $60$ - $90$ triangle, so its edges have the shown lengths and directions. Take $A=(0,0)$ and $AB$ horizontal. The horizontal span in the figure is six unit lengths, so $B=(6,0).$ Along the outer boundary from $A$ to $C,$ the three edges have vectors $\sqrt3(\cos30^\circ,\sin30^\circ),$ $\sqrt3(\cos90^\circ,\sin90^\circ),$ and $(1,0).$ Their sum is $(\tfrac52,\tfrac{3\sqrt3}{2}),$ so $C=(\tfrac52,\tfrac{3\sqrt3}{2}).$ Thus $AB=6$ and the altitude from $C$ is $\tfrac{3\sqrt3}{2},$ giving area $\tfrac12\cdot6\cdot\tfrac{3\sqrt3}{2}=\tfrac{9\sqrt3}{2}.$ Therefore, the answer is B.

2024 AMC 10A Problem 22

Answer: B

Difficulty rating: 2120

Solution:

Problem 22 in Other Years