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Difference between revisions of "2021 Fall AMC 12B Problems/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==
 
==Problem==
 
What is the area of the shaded figure shown below?
 
What is the area of the shaded figure shown below?
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Therefore, the answer is <math>\boxed{\textbf{(B) }6}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(B) }6}</math>.
  
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~Steven Chen (www.professorchenedu.com)
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~Steven Chen (www.professorchenedu.com)  
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==Discussion==
 
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To find the area of the figure, it can be divided along the line <math>x=3</math> into two congruent triangles, or the area of the triangle with vertices <math>(1,0)</math>, <math>(3,2)</math>, and <math>(5,0)</math> can be subtracted from the area of the triangle with vertices <math>(1,0)</math>, <math>(3,5)</math>, and <math>(5,0)</math>. Alternatively, [[Pick's Theorem]] or the [[Shoelace Theorem]] can be used.
 
  
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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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:39, 3 January 2022

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+(0.15,0), EndArrow); draw(O--Y+(0,0.15), EndArrow); 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

By inspection

$Total Area=2*Triangle Area=2*(\frac{1}{2}*3*2)=\boxed{(\textbf{B})\ 6}$.

~Wilhelm Z

Solution 2

The area is \begin{align*} \frac{1}{2} \left( 5 - 1 \right) 5 - \frac{1}{2} \left( 5 - 1 \right) 2 & = 6 . \end{align*}

Therefore, the answer is $\boxed{\textbf{(B) }6}$.

~Steven Chen (www.professorchenedu.com)

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

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