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>.
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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)==
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https://youtu.be/1ocp50QWOzU
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~Education, the Study of Everything
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==Video Solution 1 ==
 
==Video Solution 1 ==
 
https://youtu.be/dHY8gjoYFXU?t=1290
 
https://youtu.be/dHY8gjoYFXU?t=1290
  
~IceMatrix
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== Video Solution ==
 
 
== Video Solution 2==
 
 
https://youtu.be/NsQbhYfGh1Q?t=899
 
https://youtu.be/NsQbhYfGh1Q?t=899
  

Latest revision as of 13:53, 25 June 2023

Problem

Seven cookies of radius $1$ 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?

[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1));  draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1));  draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]

$\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi$

Solution

The big cookie has radius $3$, 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 $3^2\pi-7\pi=9\pi-7\pi=2 \pi$. The scrap cookie has this area, so its radius must be $\boxed{\textbf{(A) }\sqrt 2}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/1ocp50QWOzU

~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 (ProblemsAnswer KeyResources)
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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