2024 AMC 10A Problems/Problem 22
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[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
Let be quadrilateral MNOP. Drawing line MO splits the triangle into . Drawing the altitude from N to point Q on line MO, we know NQ is , MQ is , and QO is .
Due to the many similarities present, we can find that AB is , and the height of is
AB is and the height of is .
Solving for the area of gives which is
~9897 (latex beginner here)
Solution 2
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 3
~mathboy282
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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