Difference between revisions of "2024 AMC 10A Problems/Problem 22"

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Due to the many similarities present, we can find that <math>AB</math>
  
 
==See also==
 
==See also==
 
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{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
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{{MAA Notice}}

Revision as of 17:39, 8 November 2024

Problem

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 $\Delta ABC$?

Asset-ddfea426a1acee64ea44467d8aa8797a.png

$\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3$

Solution

Let $\mathcal K$ be quadrilateral MNOP. Drawing line MO splits the triangle into $\Delta MNO$. Drawing the altitude from N to point Q on line MO, we know NQ is $\sqrt3/2$, MQ is $3/2$, and QO is $1/2$.

Screenshot 2024-11-08 2.33.52 PM.png

Due to the many similarities present, we can find that $AB$

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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