Difference between revisions of "2016 AMC 10A Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | The big cookie has radius <math>3</math>, since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is <math>2 \pi</math>. The scrap cookie has this area, so its radius must be <math>\boxed{\textbf{(A) }\sqrt 2}</math>. | + | The big cookie has radius <math>3</math>, since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is <math>3^2\pi-7\pi=9\pi-7\pi=2 \pi</math>. The scrap cookie has this area, so its radius must be <math>\boxed{\textbf{(A) }\sqrt 2}</math>. |
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+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/1ocp50QWOzU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
==Video Solution 1 == | ==Video Solution 1 == | ||
https://youtu.be/dHY8gjoYFXU?t=1290 | https://youtu.be/dHY8gjoYFXU?t=1290 | ||
− | + | == Video Solution == | |
− | |||
− | == Video Solution | ||
https://youtu.be/NsQbhYfGh1Q?t=899 | https://youtu.be/NsQbhYfGh1Q?t=899 | ||
Latest revision as of 14:53, 25 June 2023
Contents
Problem
Seven cookies of radius inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
Solution
The big cookie has radius , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is
. The scrap cookie has this area, so its radius must be
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution 1
https://youtu.be/dHY8gjoYFXU?t=1290
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=899
~ pi_is_3.14
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.