Difference between revisions of "2015 AMC 8 Problems/Problem 2"
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− | The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>. | + | The octagon has been divided up into <math>16</math> identical triangles (and thus they each have equal area). Since the shaded region occupies <math>7</math> out of the <math>16</math> total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 11:46, 26 November 2015
Point is the center of the regular octagon , and is the midpoint of the side What fraction of the area of the octagon is shaded?
Contents
[hide]Solution 1
Since octagon is a regular octagon, it is split into 8 equal parts, such as triangles , etc. These parts, since they are all equal, are of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is
Solution 2
The octagon has been divided up into identical triangles (and thus they each have equal area). Since the shaded region occupies out of the total triangles, the answer is .
Solution 3
For starters what I find helpful is to divide the whole octagon up into triangles as shown here:
Now it is just a matter of counting the larger triangles remember that and are not full triangles and are only half for these purposes. We count it up and we get a total of of the shape shaded. We then simplify it to get our answer: or .
See Also
2015 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
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