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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==Solution 3== | ==Solution 3== |
Revision as of 16:10, 9 November 2024
Contents
[hide]Problem
Let be the kite formed by joining two right triangles with legs and along a common hypotenuse. Eight copies of are used to form the polygon shown below. What is the area of triangle ?
Solution 3
(latexing a WIP) ~mathboy282 ~Thesmartgreekmathdude
Solution 4
Let the point of intersection of and the kite with as vertex be .
Let the left kite with as a vertex touch the kite with as vertex at point .
is a so and .
So, and , and the area is
~Mintylemon66
Solution 1
First, we should find the length of . In order to do this, as we see in the diagram, it can be split into 4 sections. Since diagram shows us that it is made up of two triangles, then the triangle outlined in red must be a triangle. Also, since we know the length of the longest side is , then the side we are looking for, which is outlined in blue, must be by the relationship of 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 , as shows us. Also, the light green section must be equal to , as in the previous paragraph, the triangle outlined in red is . Then, the green section, which is the height, must be , which is just .
Then the area of the triangle must be , which is just
~Solution by HappySharks
Solution 2
Let be quadrilateral . Drawing line splits the triangle into . Drawing the altitude from to point on line , we know is , is , and is .
Due to the many similarities present, we can find that is , and the height of is
is and the height of is .
Solving for the area of gives which is
~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
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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