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

Revision as of 18:43, 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 $\Delta ABC$ ?

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

Solution 1

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 $1/2$ .

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)

Solution 2

~YTH (Working on it right now, please don't interfere. Thanks :) also if someone could help me create the solution 2 header that would be awesome)

~WIP (Header)

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

Revision as of 18:43, 8 November 2024

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 20: / Line 20: @@
 ~9897 (latex beginner here)
-==Solution 2== (WIP)
+==Solution 2==
 ~YTH (Working on it right now, please don't interfere. Thanks :) also if someone could help me create the solution 2 header that would be awesome)
+~WIP (Header)