2020 AMC 10B Problems/Problem 10
- The following problem is from both the 2020 AMC 10B #10 and 2020 AMC 12B #9, so both problems redirect to this page.
Contents
Problem 10
A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?
Solution
Notice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.
We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is . The volume of the cone is -PCChess
Solution 2 (Last Resort/Cheap)
Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6 cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found to be . The volume of a cone is . Plugging in we find
- DBlack2021
Solution 3
The radius of the given - circle will end up being the slant height of the cone. Thus, the radius and height of the cone are legs of a right triangle with hypotenuse . The volume of a cone is . Using this with the options, we can take out the and multiply the coefficient of the radical by to get . We can then use the and values to see that the only option that satisfies is
- ColtsFan10
Video Solution
https://youtu.be/OHR_6U686Qg (for AMC 10) https://youtu.be/6ujfjGLzVoE (for AMC 12)
~IceMatrix
~savannahsolver
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.