Difference between revisions of "2019 AIME I Problems/Problem 3"
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Knowing that <math>\triangle{PQR}</math> has area <math>150</math> and is a <math>3</math>-<math>4</math>-<math>5</math> triangle, we can find the area of the smaller triangles <math>\triangle{DRE}</math>, <math>\triangle{APF}</math>, and <math>\triangle{CQB}</math> and subtract them from <math>\triangle{PQR}</math> to obtain our answer. First off, we know <math>\triangle{DRE}</math> has area <math>12.5</math> since it is a right triangle. To the find the areas of <math>\triangle{APF}</math> and <math>\triangle{CQB}</math> , we can use Law of Cosines (<math>c^2 = a^2 + b^2 - 2ab\cos C</math>) to find the lengths of <math>AF</math> and <math>CB</math>, respectively. Computing gives <math>AF = \sqrt{20}</math> and <math>CB = \sqrt{10}</math>. Now, using Heron's Formula, we find <math>\triangle{APF} = 10</math> and <math>\triangle{CQB} = 7.5</math>. Adding these and subtracting from <math>\triangle{PQR}</math>, we get <math>150 - (10 + 7.5 + 12.5) = \boxed{120}</math> -Starsher | Knowing that <math>\triangle{PQR}</math> has area <math>150</math> and is a <math>3</math>-<math>4</math>-<math>5</math> triangle, we can find the area of the smaller triangles <math>\triangle{DRE}</math>, <math>\triangle{APF}</math>, and <math>\triangle{CQB}</math> and subtract them from <math>\triangle{PQR}</math> to obtain our answer. First off, we know <math>\triangle{DRE}</math> has area <math>12.5</math> since it is a right triangle. To the find the areas of <math>\triangle{APF}</math> and <math>\triangle{CQB}</math> , we can use Law of Cosines (<math>c^2 = a^2 + b^2 - 2ab\cos C</math>) to find the lengths of <math>AF</math> and <math>CB</math>, respectively. Computing gives <math>AF = \sqrt{20}</math> and <math>CB = \sqrt{10}</math>. Now, using Heron's Formula, we find <math>\triangle{APF} = 10</math> and <math>\triangle{CQB} = 7.5</math>. Adding these and subtracting from <math>\triangle{PQR}</math>, we get <math>150 - (10 + 7.5 + 12.5) = \boxed{120}</math> -Starsher | ||
− | ==Video Solution #1( | + | ==Video Solution #1(Complementary Area Counting?)== |
https://youtu.be/JQdad7APQG8?t=417 | https://youtu.be/JQdad7APQG8?t=417 | ||
Revision as of 22:11, 18 February 2021
Contents
[hide]Problem
In , , , and . Points and lie on , points and lie on , and points and lie on , with . Find the area of hexagon .
Diagram
Solution 1
We know the area of the hexagon to be . Since , we know that is a right triangle. Thus the area of is . Another way to compute the area is Then the area of . Preceding in a similar fashion for , the area of is . Since , the area of . Thus our desired answer is
Solution 2
Let be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that , and . Using the shoelace theorem, the area is . Shoelace theorem:Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
.
Solution 3 (Easiest, uses only basic geometry too)
Note that has area and is a -- right triangle. Then, by similar triangles, the altitude from to has length and the altitude from to has length , so , meaning that . -Stormersyle
Solution 4
Knowing that has area and is a -- triangle, we can find the area of the smaller triangles , , and and subtract them from to obtain our answer. First off, we know has area since it is a right triangle. To the find the areas of and , we can use Law of Cosines () to find the lengths of and , respectively. Computing gives and . Now, using Heron's Formula, we find and . Adding these and subtracting from , we get -Starsher
Video Solution #1(Complementary Area Counting?)
https://youtu.be/JQdad7APQG8?t=417
Video Solution
https://www.youtube.com/watch?v=4jOfXNiQ6WM
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=941
~IceMatrix
Video Solution 3
~Shreyas S
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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