Difference between revisions of "2015 AMC 8 Problems/Problem 2"
(→Solution 2) |
(Adding solution from http://artofproblemsolving.com/community/c183004h1167227_daily_problem_5_amc8_20152) |
||
Line 82: | Line 82: | ||
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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | For starters what I find helpful is to divide the whole octagon up into triangles as shown here: | ||
+ | <asy> | ||
+ | pair A,B,C,D,E,F,G,H,O,X; | ||
+ | A=dir(45); | ||
+ | B=dir(90); | ||
+ | C=dir(135); | ||
+ | D=dir(180); | ||
+ | E=dir(-135); | ||
+ | F=dir(-90); | ||
+ | G=dir(-45); | ||
+ | H=dir(0); | ||
+ | O=(0,0); | ||
+ | X=midpoint(A--B); | ||
+ | |||
+ | fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); | ||
+ | draw(A--B--C--D--E--F--G--H--cycle); | ||
+ | |||
+ | dot("$A$",A,dir(45)); | ||
+ | dot("$B$",B,dir(90)); | ||
+ | dot("$C$",C,dir(135)); | ||
+ | dot("$D$",D,dir(180)); | ||
+ | dot("$E$",E,dir(-135)); | ||
+ | dot("$F$",F,dir(-90)); | ||
+ | dot("$G$",G,dir(-45)); | ||
+ | dot("$H$",H,dir(0)); | ||
+ | dot("$X$",X,dir(135/2)); | ||
+ | dot("$O$",O,dir(0)); | ||
+ | draw(E--O--X); | ||
+ | draw(C--O--B); | ||
+ | draw(B--O--A); | ||
+ | draw(A--O--H); | ||
+ | draw(H--O--G); | ||
+ | draw(G--O--F); | ||
+ | draw(F--O--E); | ||
+ | draw(E--O--D); | ||
+ | draw(D--O--C); | ||
+ | </asy> | ||
+ | |||
+ | Now it is just a matter of counting the larger triangles remember that <math>\triangle BOX</math> and <math>\triangle XOA</math> are not full triangles and are only half for these purposes. We count it up and we get a total of <math>\frac{3.5}{8}</math> of the shape shaded. We then simplify it to get our answer: <math>\frac{7}{6}</math> or <math>\textbf{(D)}</math>. | ||
==See Also== | ==See Also== |
Revision as of 07:20, 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
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.