Difference between revisions of "2024 AMC 10A Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
− | First, we should find the length of <math>AB</math>. In order to do this, as we see in the diagram, it can be split into 4 sections. Since diagram <math>K</math> shows us that it is made up of two <math>{30,60,90}</math> triangles, then the triangle outlined in red must be a <math>30,60,90</math> triangle. Also, since we know the length of the longest side is <math>{\sqrt3}</math>, then the side we are looking for, which is outlined in blue, must be <math>\frac{3}{2}</math> by the <math>{1,\sqrt3, 2}</math> relationship of <math>{30,60,90}</math> triangles. Therefore | + | First, we should find the length of <math>AB</math>. In order to do this, as we see in the diagram, it can be split into 4 sections. Since diagram <math>K</math> shows us that it is made up of two <math>{30,60,90}</math> triangles, then the triangle outlined in red must be a <math>30,60,90</math> triangle. Also, since we know the length of the longest side is <math>{\sqrt3}</math>, then the side we are looking for, which is outlined in blue, must be <math>\frac{3}{2}</math> by the <math>{1,\sqrt3, 2}</math> relationship of <math>{30,60,90}</math> triangles. Therefore <math>AB</math>, which is the base of the triangle we are looking, for must be {6}. |
Revision as of 16:13, 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 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 , which 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)
Solution 3
Let's start by looking at kite . We can quickly deduce based off of the side lengths that the kite can be split into two triangles. Going back to the triangle , focus on side . There are kites, they are all either reflected over the line or a line perpendicular to , meaning the length of can be split up into 4 equal parts.
Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and share a degree angle. (this was deduced from the triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a angle. Because that is also a triangle with a hypotenuse of , so we find the length of AB to be , which is .
Then, we can drop an altitude from to . We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and . (Look at the line formed on the left of that drops down to if you are confused) We already have those values from the triangles, so we can just plug it into the triangle area formula, . We get
~YTH (Need help with Latex and formatting)
~WIP (Header)
~Tacos_are_yummy_1 ( & Formatting)
Solution 4 (On Paper)
(latexing a WIP) ~mathboy282 ~Thesmartgreekmathdude
Solution 5
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.