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

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Solving for the area of <math>\Delta ABC</math> gives <math>6*3\sqrt3/2*1/2</math> which is <math>\textbf{(B) }\dfrac92\sqrt3\qquad</math>
 
Solving for the area of <math>\Delta ABC</math> gives <math>6*3\sqrt3/2*1/2</math> which is <math>\textbf{(B) }\dfrac92\sqrt3\qquad</math>
  
−
latex beginner here
+
~9897 (latex beginner here)
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:56, 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 is $4(MQ)$, and the height of $\Delta ABC$ is $NQ+MN$

AB is $4(3/2)=6$ and the height of $\Delta ABC$ is $\sqrt3+\sqrt3/2=3\sqrt3/2$.

Solving for the area of $\Delta ABC$ gives $6*3\sqrt3/2*1/2$ which is $\textbf{(B) }\dfrac92\sqrt3\qquad$

~9897 (latex beginner here)

See also

2024 AMC 10A (Problems • Answer Key • Resources)
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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