2024 AMC 10A Problems/Problem 22

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 3

(latexing a WIP) ~mathboy282 ~Thesmartgreekmathdude

Solution 4

Let the point of intersection of $AB$ and the kite with $A$ as vertex be $D$ .

Let the left kite with $C$ as a vertex touch the kite with $A$ as vertex at point $E$ .

$\triangle ADE$ is a $30-60-90$ so $AD = \frac{3}{2}$ and $DE = \frac{\sqrt3}{2}$ .

So, $AB = 4\cdot AD = 6$ and $CD=CE+DE= \frac{3\sqrt3}{2}$ , and the area is $\frac12\cdot AB \cdot CD = \boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.$

~Mintylemon66

Solution 1

First, we should find the length of $AB$ . In order to do this, as we see in the diagram, it can be split into 4 sections. Since diagram $K$ shows us that it is made up of two ${30,60,90}$ triangles, then the triangle outlined in red must be a $30,60,90$ triangle. Also, since we know the length of the longest side is ${\sqrt3}$ , then the side we are looking for, which is outlined in blue, must be $\frac{3}{2}$ by the ${1,\sqrt3, 2}$ relationship of ${30,60,90}$ triangles. Therefore the long side that is the base of the triangle we are looking for must be {6}.

Now all we have to do is find the height. We can split the height into 2 sections, the green and the light green. The green section must be ${\sqrt3}$ , as $K$ shows us. Also, the light green section must be equal to ${\frac{\sqrt3}{2}}$ , as in the previous paragraph, the triangle outlined in red is $30,60,90$ . Then, the green section, which is the height, must be ${\sqrt3}+{\frac{\sqrt3}{2}}$ , which is just ${\frac{3\sqrt3}{2}}$ .

Then the area of the triangle must be ${\frac{1}{2}}*base*height$ , which is just $\boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.$

~Solution by HappySharks

Solution 2

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 $\frac{\sqrt{3}}{2}$ , $MQ$ is $\frac{3}{2}$ , and $QO$ is $\frac{1}{2}$ .

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 $4(\frac{3}{2})=6$ and the height of $\Delta ABC$ is $\sqrt3+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$ .

Solving for the area of $\Delta ABC$ gives $6\cdot\frac{3\sqrt{3}}{2}\cdot\frac{1}{2}$ which is $\boxed{\textbf{(B) }\dfrac92\sqrt3}$

~9897 (latex beginner here)

~i_am_suk_at_math(very minor latex edits)

Video Solution by Innovative Minds

https://www.youtube.com/watch?v=bhC58BB3kJA

~i_am_suk_at_math_2

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
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