2021 Fall AMC 12B Problems/Problem 2: Difference between revisions
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<math>Area=3*2=6 \Rightarrow \boxed{(\textbf{B})6}</math>. | <math>Area=3*2=6 \Rightarrow \boxed{(\textbf{B})6}</math>. | ||
~Wilhelm | ~Wilhelm Z | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:56, 24 November 2021
Problem 2
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
| 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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