Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"
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</asy> | </asy> | ||
| − | <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== | ||
| Line 35: | Line 35: | ||
By inspection | By inspection | ||
| − | <math>Total Area=2*Triangle Area=2*(\frac{1}{2}*3*2)=\boxed{ | + | <math>Total Area=2*Triangle Area=2*(\frac{1}{2}*3*2)=\boxed{\textbf{(B)} \: 6}</math>. |
~Wilhelm Z | ~Wilhelm Z | ||
| Line 48: | Line 48: | ||
</cmath> | </cmath> | ||
| − | Therefore, the answer is <math>\boxed{\textbf{(B) }6}</math>. | + | Therefore, the answer is <math>\boxed{\textbf{(B)} \: 6}</math>. |
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
| + | == Solution 3 == | ||
| + | We start by finding the points. The outlined shape is made up of <math>(1,0),(3,5),(5,0),(3,2)</math>. By the | ||
| + | Shoelace Theorem, we find the area to be <math>6</math>, or <math>\boxed{\textbf{(B)} \: 6}</math>. | ||
| + | |||
| + | 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 <math>4</math> interior points and <math>6</math> boundary points. By Pick's Theorem, we get <math>4+\frac{6}{2}-1 = 4+3-1 = 6.</math> Checking our answer choices, we find our answer to be <math>\boxed{\textbf{(B)} \: 6}</math>. | ||
| + | |||
| + | ~danprathab | ||
==See Also== | ==See Also== | ||
Revision as of 05:44, 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
By inspection
.
~Wilhelm Z
Solution 2
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
