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| − | ==Problem==
| + | #REDIRECT [[2021_Fall_AMC_12B_Problems/Problem_2]] |
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| − | What is the area of the shaded figure shown below?
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| − | <asy>
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| − | size(200);
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| − | defaultpen(linewidth(0.4)+fontsize(12));
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| − | pen s = linewidth(0.8)+fontsize(8);
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| − | pair O,X,Y;
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| − | O = origin;
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| − | X = (6,0);
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| − | Y = (0,5);
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| − | fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2));
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| − | for (int i=1; i<7; ++i)
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| − | {
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| − | draw((i,0)--(i,5), gray+dashed);
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| − | label("${"+string(i)+"}$", (i,0), 2*S);
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| − | if (i<6)
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| − | {
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| − | draw((0,i)--(6,i), gray+dashed);
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| − | label("${"+string(i)+"}$", (0,i), 2*W);
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| − | }
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| − | }
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| − | label("$0$", O, 2*SW);
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| − | draw(O--X+(0.15,0), EndArrow);
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| − | draw(O--Y+(0,0.15), EndArrow);
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| − | draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5);
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| − | </asy>
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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>
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| − | ==Solution #1==
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| − | 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>
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| − | ~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=3\cdot2=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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| − | ==Solution #3 (Overkill)==
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| − | We start by finding the points. The outlined shape is made up of <math>(1,0),(3,5),(5,0),(3,2)</math>. By the Shoelace Theorem, we find the area to be <math>6</math>, or <math>\boxed{B}</math>.
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| − | ~Taco12
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}}
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| − | {{MAA Notice}}
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