Difference between revisions of "2008 AMC 10A Problems/Problem 17"
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The region described contains three rectangles of dimensions <math>3 \times 6</math>, and three <math>120^{\circ}</math> degree arcs of circles of [[radius]] <math>3</math>. Thus the answer is <cmath>3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.</cmath> | The region described contains three rectangles of dimensions <math>3 \times 6</math>, and three <math>120^{\circ}</math> degree arcs of circles of [[radius]] <math>3</math>. Thus the answer is <cmath>3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.</cmath> | ||
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+ | ==Solution 2== | ||
+ | After a quick sketch of the problem, one can deduce that there are <math>3</math> rectangles in the figure, each with side length <math>6</math> and width <math>3</math>. Therefore, the combined areas of the rectangles is <math>54</math>. The other three regions are circle-shaped areas, probably expressed in some form of <math>\pi</math>. Answer choices <math>\mathrm{(A)}</math>, <math>\mathrm{(D)}</math>, <math>\mathrm{(E)}</math> are impossible because they either lack an integer or <math>\pi</math> in the answer, and <math>\mathrm{(C)}</math> is impossible since <math>18\sqrt{3}</math> clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is <math>\boxed{\mathrm{(B)}}</math>. | ||
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+ | Solution by Airplane50 | ||
==See also== | ==See also== |
Revision as of 19:08, 28 December 2019
Contents
Problem
An equilateral triangle has side length . What is the area of the region containing all points that are outside the triangle but not more than units from a point of the triangle?
Solution
The region described contains three rectangles of dimensions , and three degree arcs of circles of radius . Thus the answer is
Solution 2
After a quick sketch of the problem, one can deduce that there are rectangles in the figure, each with side length and width . Therefore, the combined areas of the rectangles is . The other three regions are circle-shaped areas, probably expressed in some form of . Answer choices , , are impossible because they either lack an integer or in the answer, and is impossible since clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is .
Solution by Airplane50
See also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
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