Difference between revisions of "2021 AMC 12B Problems/Problem 15"
m (→Solution 2: Added a diagram) |
(→Solution 1) |
||
Line 27: | Line 27: | ||
Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}</math>. | Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}</math>. | ||
− | |||
− | |||
==Solution 2== | ==Solution 2== |
Revision as of 23:25, 12 February 2021
Contents
Problem
The figure is constructed from line segments, each of which has length . The area of pentagon can be written is , where and are positive integers. What is
Solution 1
Let be the midpoint of . Noting that and are triangles because of the equilateral triangles, . Also, and so .
Solution 2
Draw diagonals and to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles and , they each have area . For triangle , we can see that and . Using Pythagorean Theorem, the altitude for this triangle is , so the area is . Adding each part up, we get .
Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.