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

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<math>\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</math>
 
<math>\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</math>
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[https://wiki-images.artofproblemsolving.com//2/2d/Drawing.sketchpad.png]
 
  
 
==Solution==
 
==Solution==
 
Let <math>\mathcal K</math> be quadrilateral MNOP. Drawing line MO splits the triangle into <math>\Delta MNO</math>.
 
Let <math>\mathcal K</math> be quadrilateral MNOP. Drawing line MO splits the triangle into <math>\Delta MNO</math>.
 
Drawing the altitude from N to point Q on line MO, we know NQ is <math>\sqrt3/2</math>, MQ is <math>3/2</math>, and QO is <math>1/2</math>.
 
Drawing the altitude from N to point Q on line MO, we know NQ is <math>\sqrt3/2</math>, MQ is <math>3/2</math>, and QO is <math>1/2</math>.

Revision as of 17:30, 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 $\Delta ABC$?

[1]

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

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 $1/2$.

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