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 $ABC?$

$2 + 3\sqrt3$

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

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

$8$

$5\sqrt3$

Solution:

Each kite is two $30$ - $60$ - $90$ triangles with legs $1$ and $\sqrt3$ and hypotenuse $2.$ Trace the eight-kite figure in coordinates and the outer vertices come out to $A = (0, 0),$ $B = (6, 0),$ and $C = \left(\tfrac52, \tfrac{3\sqrt3}{2}\right).$ So triangle $ABC$ has base $AB = 6$ and height $\tfrac{3\sqrt3}{2},$ and its area is $\tfrac12 \cdot 6 \cdot \tfrac{3\sqrt3}{2} = \tfrac{9\sqrt3}{2}.$ Therefore, the answer is B.

2024 AMC 10A Problem 22

Solution:

Problem 22 in Other Years