Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"
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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> | <math>\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12</math> | ||
| − | ==Solution 1 (Area Addition)== | + | == Solution 1 (Area Addition) == |
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> | ||
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~MRENTHUSIASM ~Wilhelm Z | ~MRENTHUSIASM ~Wilhelm Z | ||
| − | ==Solution 2 (Area Subtraction)== | + | == Solution 2 (Area Subtraction) == |
| − | + | 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> | |
| − | <cmath> | + | ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) |
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| − | \ | ||
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| − | </cmath> | ||
| − | + | == Solution 3 (Shoelace Theorem) == | |
| + | The consecutive vertices of the shaded figure are <math>(1,0),(3,5),(5,0),</math> and <math>(3,2).</math> By the [[Shoelace_Theorem|Shoelace Theorem]], we get the area <math>\boxed{\textbf{(B)} \: 6}.</math> | ||
| − | + | ~Taco12 ~I-AM-DA-KING | |
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| − | ~Taco12 | ||
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| − | ~I-AM-DA-KING | ||
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| + | == Solution 4 (Pick's Theorem)== | ||
| + | We have <math>4</math> interior points and <math>6</math> boundary points. 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 | ||
Revision as of 08:52, 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.
Contents
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]](http://latex.artofproblemsolving.com/9/5/0/950ff8f5847f0f8acb13a489295e0c08395cb448.png)
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 ![\[2\cdot\left(\frac12\cdot3\cdot2\right)=\boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/e/a/3/ea393c968cf9651354953c7f88bd794710e7ad03.png)
~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 
): ![\[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/f/4/5/f4549488e5575113e4bc3c0ded44f9db3c88a1f1.png)
~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 3 (Shoelace Theorem)
The consecutive vertices of the shaded figure are 
and 
By the Shoelace Theorem, we get the area 
~Taco12 ~I-AM-DA-KING
Solution 4 (Pick's Theorem)
We have interior points and 
boundary points. By Pick's Theorem, the area of the shaded figure is ![\[4+\frac{6}{2}-1 = 4+3-1 = \boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/8/1/4/81411734430f77e31c966a110c816c2ef4d4b3fc.png)
~danprathab
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. 
