Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"
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− | {{duplicate|[[2021 Fall AMC 10B Problems | + | {{duplicate|[[2021 Fall AMC 10B Problems/Problem 2|2021 Fall AMC 10B #2]] and [[2021 Fall AMC 12B Problems/Problem 2|2021 Fall AMC 12B #2]]}} |
==Problem== | ==Problem== | ||
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The line of symmetry divides the shaded figure into two congruent triangles, each with base <math>3</math> and height <math>2.</math> | The line of symmetry divides the shaded figure into two congruent triangles, each with base <math>3</math> and height <math>2.</math> | ||
− | Therefore, the area of the shaded figure is <cmath>2\cdot\left(\frac12\cdot3\cdot2\right)=\boxed{\textbf{(B)} \: 6}.</cmath> | + | Therefore, the area of the shaded figure is <cmath>2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.</cmath> |
~MRENTHUSIASM ~Wilhelm Z | ~MRENTHUSIASM ~Wilhelm Z | ||
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== Solution 3 (Shoelace Theorem) == | == Solution 3 (Shoelace Theorem) == | ||
− | The consecutive vertices of the shaded figure are <math>(1,0),(3, | + | The consecutive vertices of the shaded figure are <math>(1,0),(3,2),(5,0),</math> and <math>(3,5).</math> By the [[Shoelace_Theorem|Shoelace Theorem]], the area is <cmath>\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{\textbf{(B)} \: 6}.</cmath> |
− | |||
~Taco12 ~I-AM-DA-KING | ~Taco12 ~I-AM-DA-KING | ||
== Solution 4 (Pick's Theorem)== | == Solution 4 (Pick's Theorem)== | ||
− | We have <math>4</math> interior | + | We have <math>4</math> lattice points in the interior and <math>6</math> lattice points on the boundary. By [[Pick%27s_Theorem|Pick's Theorem]], the area of the shaded figure is <cmath>4+\frac{6}{2}-1 = 4+3-1 = \boxed{\textbf{(B)} \: 6}.</cmath> |
~danprathab | ~danprathab | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/p9_RH4s-kBA?t=110 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ZV-cQm5p7Pc | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/1sAevgxImQM | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512 | ||
+ | |||
+ | For AMC 12: https://youtu.be/yaE5aAmeesc?t=512 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:13, 12 November 2023
- The following problem is from both the 2021 Fall AMC 10B #2 and 2021 Fall AMC 12B #2, so both problems redirect to this page.
Contents
Problem
What is the area of the shaded figure shown below?
Solution 1 (Area Addition)
The line of symmetry divides the shaded figure into two congruent triangles, each with base and height
Therefore, the area of the shaded figure is ~MRENTHUSIASM ~Wilhelm Z
Solution 2 (Area Subtraction)
To find the area of the shaded figure, we subtract the area of the smaller triangle (base and height ) from the area of the larger triangle (base and height ): ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 3 (Shoelace Theorem)
The consecutive vertices of the shaded figure are and By the Shoelace Theorem, the area is ~Taco12 ~I-AM-DA-KING
Solution 4 (Pick's Theorem)
We have lattice points in the interior and lattice points on the boundary. By Pick's Theorem, the area of the shaded figure is ~danprathab
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=110
~Interstigation
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512
For AMC 12: https://youtu.be/yaE5aAmeesc?t=512
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 |
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.