2021 Fall AMC 12B Problems/Problem 2

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)

Solution 3

We start by finding the points. The outlined shape is made up of $(1,0),(3,5),(5,0),(3,2)$ . By the Shoelace Theorem, we find the area to be $6$ , or $\boxed{\textbf{(B)} \: 6}$ .

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 $4$ interior points and $6$ boundary points. By Pick's Theorem, we get $4+\frac{6}{2}-1 = 4+3-1 = 6.$ Checking our answer choices, we find our answer to be $\boxed{\textbf{(B)} \: 6}$ .

~danprathab

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.

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