Difference between revisions of "2015 AMC 8 Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | + | <asy> | |
pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; | pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g; | ||
A=dir(45); | A=dir(45); | ||
Line 60: | Line 60: | ||
draw(A--B--C--D--E--F--G--H--cycle); | draw(A--B--C--D--E--F--G--H--cycle); | ||
− | dot(" | + | dot("$A$",A,dir(45)); |
− | dot(" | + | dot("$B$",B,dir(90)); |
− | dot(" | + | dot("$C$",C,dir(135)); |
− | dot(" | + | dot("$D$",D,dir(180)); |
− | dot(" | + | dot("$E$",E,dir(-135)); |
− | dot(" | + | dot("$F$",F,dir(-90)); |
− | dot(" | + | dot("$G$",G,dir(-45)); |
− | dot(" | + | dot("$H$",H,dir(0)); |
− | dot(" | + | dot("$X$",X,dir(135/2)); |
− | dot(" | + | dot("$O$",O,dir(0)); |
draw(E--O--X); | draw(E--O--X); | ||
draw(B--F); | draw(B--F); | ||
Line 79: | Line 79: | ||
draw(c--g); | draw(c--g); | ||
draw(d--O); | draw(d--O); | ||
− | + | </asy> | |
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 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>. |
Revision as of 18:51, 25 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?
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 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.