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? $[asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("{"+string(i)+"}", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("{"+string(i)+"}", (0,i), 2*W); } } label("0", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]$

$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$

The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$

Therefore, the area of the shaded figure is $$2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.$$ ~MRENTHUSIASM ~Wilhelm Z

Solution 2 (Area Subtraction)

To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$) from the area of the larger triangle (base $4$ and height $5$): $$\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{\textbf{(B)} \: 6}.$$ ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)

Solution 3 (Shoelace Theorem)

The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem, the area is $$\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{\textbf{(B)} \: 6}.$$ ~Taco12 ~I-AM-DA-KING

Solution 4 (Pick's Theorem)

We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem, the area of the shaded figure is $$4+\frac{6}{2}-1 = 4+3-1 = \boxed{\textbf{(B)} \: 6}.$$ ~danprathab

~Interstigation

Video Solution

~Education, the Study of Everything

~savannahsolver