Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | 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, < | + | 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, <cmath>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}.</cmath> Also, <math>[AED]=2\cdot2\cdot\frac{1}{2}\cdot\sin{120^o}=\sqrt{3}</math> and so <cmath>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}.</cmath> |
==Solution 2== | ==Solution 2== |
Revision as of 17:58, 3 July 2021
- The following problem is from both the 2021 AMC 10B #20 and 2021 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
The figure is constructed from line segments, each of which has length
. The area of pentagon
can be written as
, 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
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=1226
~IceMatrix
Video Solution by Interstigation (Ignore Useless Segments)
~Interstigation
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.