2021 Fall AMC 12B Problems/Problem 2

Revision as of 21:15, 25 November 2021 by Stevenyiweichen (talk | contribs)

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)

Discussion

To find the area of the figure, it can be divided along the line $x=3$ into two congruent triangles, or the area of the triangle with vertices $(1,0)$, $(3,2)$, and $(5,0)$ can be subtracted from the area of the triangle with vertices $(1,0)$, $(3,5)$, and $(5,0)$. Alternatively, Pick's Theorem or the Shoelace Theorem can be used.

2021 Fall AMC 12B (ProblemsAnswer KeyResources)
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. AMC logo.png