2021 Fall AMC 12B Problems/Problem 2: Difference between revisions
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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== | ==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> | |||
Therefore, the area of the shaded figure is <cmath>2\cdot\left(\frac12\cdot3\cdot2\right)=\boxed{\textbf{(B)} \: 6}.</cmath> | |||
~MRENTHUSIASM ~Wilhelm Z | |||
==Solution 2 (Area Subtraction)== | |||
The area is | The area is | ||
<cmath> | <cmath> | ||
Revision as of 07:16, 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]](http://latex.artofproblemsolving.com/1/7/1/1714dc248f12bd0640e9ec37023a14eddda46d9c.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)
The area is
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Solution 3
We start by finding the points. The outlined shape is made up of 
. By the
Shoelace Theorem, we find the area to be 
, or .
https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem
~Taco12
~I-AM-DA-KING for the link
Solution 4
We can use Pick's Theorem. We have 
interior points and boundary points. By Pick's Theorem, we get 
Checking our answer choices, we find our answer to be .
~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 | |
These problems are copyrighted © by the Mathematical Association of America. 
