Difference between revisions of "2021 Fall AMC 10B Problems/Problem 2"
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<math>(\textbf{A})\: 4\qquad(\textbf{B}) \: 6\qquad(\textbf{C}) \: 8\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 12</math> | <math>(\textbf{A})\: 4\qquad(\textbf{B}) \: 6\qquad(\textbf{C}) \: 8\qquad(\textbf{D}) \: 10\qquad(\textbf{E}) \: 12</math> | ||
| − | ==Solution== | + | ==Solution #1== |
We have <math>2</math> isosceles triangles. Thus, the area of the shaded region is <math>\frac{1}{2} \cdot 5 \cdot 4 - \left(\frac{1}{2} \cdot 4 \cdot 2\right) = 10 - 4 = 6.</math> Thus our answer is <math>\boxed{(\textbf{B}.)}.</math> | We have <math>2</math> isosceles triangles. Thus, the area of the shaded region is <math>\frac{1}{2} \cdot 5 \cdot 4 - \left(\frac{1}{2} \cdot 4 \cdot 2\right) = 10 - 4 = 6.</math> Thus our answer is <math>\boxed{(\textbf{B}.)}.</math> | ||
~NH14 | ~NH14 | ||
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| + | ==Solution #2== | ||
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| + | As we can see, the shape is symmetrical, so it will be equally valid to simply calculate one of the half's area and multiply by 2. One half's area is <math>\frac{bh}2</math>, so two halves would be <math>bh=2\cdot3=6</math>. Thus our answer is <math>\boxed{(\textbf{B}.)}.</math> | ||
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| + | ~Hefei417, or 陆畅 Sunny from China | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:38, 23 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
We have 
isosceles triangles. Thus, the area of the shaded region is 
Thus our answer is 
~NH14
Solution #2
As we can see, the shape is symmetrical, so it will be equally valid to simply calculate one of the half's area and multiply by 2. One half's area is 
, so two halves would be 
. Thus our answer is
~Hefei417, or 陆畅 Sunny from China
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
