Difference between revisions of "2012 AMC 10B Problems/Problem 17"
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We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.</math> Now we can find the GENERAL formulas for <math>r</math> and <math>h</math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us <math>C.</math> | We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.</math> Now we can find the GENERAL formulas for <math>r</math> and <math>h</math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us <math>C.</math> | ||
Revision as of 08:34, 17 August 2017
Problem
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?
Solution
Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is , so the circumference of the smaller cone would be . This means that the radius of the smaller cone is . Since the radius of the paper disk is , the slant height if the smaller cone would be . By the Pythagorean Theorem, the height of the cone is . Thus, the volume of the smaller cone is .
Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is and the slant height is . By the Pythagorean Theorem again, the height is . Thus, the volume of the larger cone is .
The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find after simplifying, or .
~A side note
We can first simplify the volume ratio: Now we can find the GENERAL formulas for and based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.