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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"

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==Problem 2==
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{{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]]}}
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==Problem==
 
What is the area of the shaded figure shown below?
 
What is the area of the shaded figure shown below?
 
<asy>
 
<asy>
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}
 
}
 
label("$0$", O, 2*SW);
 
label("$0$", O, 2*SW);
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draw(O--X+(0.15,0), EndArrow);
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draw(O--X+(0.35,0), black+1.5, EndArrow(10));
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draw(O--Y+(0,0.15), EndArrow);
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draw(O--Y+(0,0.35), black+1.5, EndArrow(10));
 
draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5);
 
draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5);
 
</asy>
 
</asy>
  
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<math>(\textbf{A})\: 4\qquad(\textbf{B}) \: 6\qquad(\textbf{C}) \: 8\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 12</math>
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<math>\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12</math>
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== Solution 1 (Area Addition) ==
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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>
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Therefore, the area of the shaded figure is <cmath>2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.</cmath>
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~MRENTHUSIASM ~Wilhelm Z
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== Solution 2 (Area Subtraction) ==
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To find the area of the shaded figure, we subtract the area of the smaller triangle (base <math>4</math> and height <math>2</math>) from the area of the larger triangle (base <math>4</math> and height <math>5</math>): <cmath>\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{\textbf{(B)} \: 6}.</cmath>
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~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
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== Solution 3 (Shoelace Theorem) ==
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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>
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~Taco12 ~I-AM-DA-KING
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== Solution 4 (Pick's Theorem)==
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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>
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~danprathab
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==Video Solution by Interstigation==
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https://youtu.be/p9_RH4s-kBA?t=110
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~Interstigation
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==Video Solution==
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https://youtu.be/ZV-cQm5p7Pc
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~Education, the Study of Everything
  
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==Solution 1==
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==Video Solution by WhyMath==
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https://youtu.be/1sAevgxImQM
  
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By inspection
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~savannahsolver
  
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<math>Area=3*2=6 \Rightarrow \boxed{(\textbf{B})\ 6}</math>.
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512
  
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~Wilhelm Z
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For AMC 12: https://youtu.be/yaE5aAmeesc?t=512
  
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~IceMatrix
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==See Also==
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{{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}}
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 +
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:13, 11 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.

Problem

What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]

$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$

Solution 1 (Area Addition)

The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$

Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.\] ~MRENTHUSIASM ~Wilhelm Z

Solution 2 (Area Subtraction)

To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$) from the area of the larger triangle (base $4$ and height $5$): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{\textbf{(B)} \: 6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution 3 (Shoelace Theorem)

The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem, the area is \[\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{\textbf{(B)} \: 6}.\] ~Taco12 ~I-AM-DA-KING

Solution 4 (Pick's Theorem)

We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem, the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{\textbf{(B)} \: 6}.\] ~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

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. AMC logo.png

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