2024 AMC 10A Problem 22Let KKK be the kite formed by joining two right triangles with legs 111 and 3\sqrt{3}3 along a common hypotenuse. Eight copies of KKK are used to form the polygon shown below. What is the area of triangle △ABC\triangle ABC△ABC?(A) 2+33\text{(A)}\;2+3\sqrt{3}(A)2+33(B) 932\text{(B)}\;\frac{9\sqrt{3}}{2}(B)293(C) 10+833\text{(C)}\;\frac{10+8\sqrt{3}}{3}(C)310+83(D) 8\text{(D)}\;8(D)8(E) 53\text{(E)}\;5\sqrt{3}(E)53Related TopicsCoreSpecial Right TrianglesCheck AnswerYour answer:ABCDECheckHints (5)Hint 1Hint 2Hint 3ADFEADFEADFE is an isosceles trapezoid.Hint 4Hint 5Height=HD=HG+GD=3+32=332\text{Height}=HD=HG+GD=\sqrt{3}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}Height=HD=HG+GD=3+23=233Base=AB=AD+DE+EF+FB=4×32=6\text{Base}=AB=AD+DE+EF+FB=4\times\frac{3}{2}=6Base=AB=AD+DE+EF+FB=4×23=6Area=332×62=932\text{Area}=\frac{\frac{3\sqrt{3}}{2}\times6}{2}=\frac{9\sqrt{3}}{2}Area=2233×6=293Related Problems (3)AMC 10B 2025 (Problem 12)2024 AMC 10A Problem 14AIME II 2026 (Problem 3)Final Answer(B) 932\frac{9\sqrt{3}}{2}293
