Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"
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~Wilhelm Z | ~Wilhelm Z | ||
| + | |||
| + | == Solution 2 == | ||
| + | The area is | ||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | \frac{1}{2} \left( 5 - 1 \right) 5 - \frac{1}{2} \left( 5 - 1 \right) 2 | ||
| + | & = 6 . | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(B) }6}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
==Discussion== | ==Discussion== | ||
Revision as of 21:15, 25 November 2021
Contents
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
Solution 2
The area is
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
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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