Difference between revisions of "2020 AMC 10B Problems/Problem 10"
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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #10]] and [[2020 AMC 12B Problems|2020 AMC 12B #9]]}} | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #10]] and [[2020 AMC 12B Problems|2020 AMC 12B #9]]}} | ||
− | ==Problem | + | == Problem == |
+ | A three-quarter sector of a circle of radius <math>4</math> 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? | ||
− | |||
<asy> | <asy> | ||
− | |||
draw(Arc((0,0), 4, 0, 270)); | draw(Arc((0,0), 4, 0, 270)); | ||
draw((0,-4)--(0,0)--(4,0)); | draw((0,-4)--(0,0)--(4,0)); | ||
− | |||
label("$4$", (2,0), S); | label("$4$", (2,0), S); | ||
+ | </asy> | ||
− | |||
<math>\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math> | <math>\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math> | ||
− | ==Solution== | + | == Solutions == |
+ | === Solution 1 === | ||
+ | Notice that when the cone is created, the 2 shown radii when merged 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 <math>8\pi\cdot\frac{3}{4}=6\pi</math>. Since that is also equal to the circumference of the cone, the radius of the cone is <math>3</math>. We also have that the slant height of the cone is <math>4</math>. Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is <math>\sqrt{4^2-3^2}=\sqrt7</math>. The volume of the cone is <math>\frac{1}{3}\cdot\pi\cdot3^2\cdot\sqrt7=\boxed{\textbf{(C)}\ 3 \pi \sqrt7 }</math> | |
− | + | -PCChess | |
− | ==Solution 2 (Last Resort/Cheap)== | + | === 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 <math>\implies r=3</math>. You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height <math>h</math> is found to be <math>h^2 = 4^{2} - 3^{2} \implies h = \sqrt{7}</math>. The volume of a cone is <math>\frac{1}{3}\pi r^{2}h</math>. Plugging in we find <math>V = 3\pi \sqrt{7} \implies \boxed{\textbf{(C)}}</math> | 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 <math>\implies r=3</math>. You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height <math>h</math> is found to be <math>h^2 = 4^{2} - 3^{2} \implies h = \sqrt{7}</math>. The volume of a cone is <math>\frac{1}{3}\pi r^{2}h</math>. Plugging in we find <math>V = 3\pi \sqrt{7} \implies \boxed{\textbf{(C)}}</math> | ||
- DBlack2021 | - DBlack2021 | ||
− | ==Video Solution== | + | === Solution 3 === |
− | https://youtu.be/OHR_6U686Qg | + | The radius of the given <math>\frac{3}{4}</math> - 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 <math>4</math>. The volume of a cone is <math>\frac{1}{3}\pi r^{2}h</math>. Using this with the options, we can take out the <math>\pi</math> and multiply the coefficient of the radical by <math>3</math> to get <math>r^{2}h</math>. We can then use the <math>r</math> and <math>h</math> values to see that the only option that satisfies <math>r^2+h^2=4^2=16</math> is <math>\boxed{\textbf{(C)}}</math> |
− | https://youtu.be/6ujfjGLzVoE | + | |
+ | - ColtsFan10 | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/zQ9GHc6u6dM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution by TheBeautyOfMath == | ||
+ | https://youtu.be/OHR_6U686Qg | ||
+ | |||
+ | https://youtu.be/6ujfjGLzVoE | ||
− | + | == Video Solution by WhyMath == | |
+ | https://youtu.be/4OAhfceUJXc | ||
− | + | ~savannahsolver | |
+ | == See Also == | ||
{{AMC10 box|year=2020|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2020|ab=B|num-b=9|num-a=11}} | ||
{{AMC12 box|year=2020|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2020|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 08:42, 15 August 2024
- 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
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?
Solutions
Solution 1
Notice that when the cone is created, the 2 shown radii when merged 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 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution by TheBeautyOfMath
Video Solution by WhyMath
~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.