2021 Fall AMC 12B Problems/Problem 2

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

$Area=2*Triangle=2*(\frac{1}{2}*3*2)=\boxed{(\textbf{B})\ 6}$ .

~Wilhelm Z

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 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

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2021 Fall AMC 12B Problems/Problem 2

Problem

Solution 1

Discussion