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>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== | ||
+ | https://youtu.be/dHY8gjoYFXU?t=1290 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2016|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:42, 19 May 2020
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
https://youtu.be/dHY8gjoYFXU?t=1290
~IceMatrix
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 |
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