2021 Fall AMC 12B Problems/Problem 2: Difference between revisions
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By inspection | By inspection | ||
<math>Area=3*2=\boxed{(\textbf{B})\ 6}</math>. | <math>Area=2*Triangle=2*(\frac{1}{2}*3*2)=\boxed{(\textbf{B})\ 6}</math>. | ||
~Wilhelm Z | ~Wilhelm Z | ||
Revision as of 19:50, 24 November 2021
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]](http://latex.artofproblemsolving.com/1/7/1/1714dc248f12bd0640e9ec37023a14eddda46d9c.png)
Solution 1
By inspection
.
~Wilhelm Z
Discussion
To find the area of the figure, it can be divided along the line 
into two congruent triangles, or the area of the triangle with vertices 
, 
, and 
can be subtracted from the area of the triangle with vertices , 
, and . Alternatively, Pick's Theorem or the Shoelace Theorem can be used.
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
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Followed by Problem 3 |
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