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Difference between revisions of "2020 AMC 10B Problems/Problem 10"

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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
 
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
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==Solution 2 (Last Resort/Cheap)==
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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>
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- DBlack2021
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:20, 9 February 2020

Problem 10

A three-quarter sector of a circle of radius $4$ 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] draw(Arc((0,0), 4, 0, 270)); draw((0,-4)--(0,0)--(4,0)); label("$4$", (2,0), S); [/asy] $\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7$

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 $8\pi\cdot\frac{3}{4}=6\pi$. Since that is also equal to the circumference of the cone, the radius of the cone is $3$. We also have that the slant height of the cone is $4$. Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is $\sqrt{4^2-3^2}=\sqrt7$. The volume of the cone is $\frac{1}{3}\cdot\pi\cdot3^2\cdot\sqrt7=\boxed{\textbf{(C)}\ 3 \pi \sqrt7 }$ -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 $\implies r=3$. You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height $h$ is found to be $h^2 = 4^{2} - 3^{2} \implies h = \sqrt{7}$. The volume of a cone is $\frac{1}{3}\pi r^{2}h$. Plugging in we find $V = 3\pi \sqrt{7} \implies \boxed{\textbf{(C)}}$

- DBlack2021

Video Solution

https://youtu.be/OHR_6U686Qg (for AMC 10) https://youtu.be/6ujfjGLzVoE (for AMC 12)

~IceMatrix

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

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