Difference between revisions of "1985 AIME Problems/Problem 2"
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When a [[right triangle]] is rotated about one leg, the [[volume]] of the [[cone]] produced is <math>800\pi \;\textrm{ cm}^3</math>. When the [[triangle]] is rotated about the other leg, the volume of the cone produced is <math>1920\pi \;\textrm{ cm}^3</math>. What is the length (in cm) of the [[hypotenuse]] of the triangle? | When a [[right triangle]] is rotated about one leg, the [[volume]] of the [[cone]] produced is <math>800\pi \;\textrm{ cm}^3</math>. When the [[triangle]] is rotated about the other leg, the volume of the cone produced is <math>1920\pi \;\textrm{ cm}^3</math>. What is the length (in cm) of the [[hypotenuse]] of the triangle? | ||
==Solution== | ==Solution== | ||
| − | Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>. When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>. Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>. If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>. Then <math>\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>. Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = 026</math>. | + | Let one leg of the triangle have length <math>a</math> and let the other leg have length <math>b</math>. When we rotate around the leg of length <math>a</math>, the result is a cone of height <math>a</math> and [[radius]] <math>b</math>, and so of volume <math>\frac 13 \pi ab^2 = 800\pi</math>. Likewise, when we rotate around the leg of length <math>b</math> we get a cone of height <math>b</math> and radius <math>a</math> and so of volume <math>\frac13 \pi b a^2 = 1920 \pi</math>. If we divide this equation by the previous one, we get <math>\frac ab = \frac{\frac13 \pi b a^2}{\frac 13 \pi ab^2} = \frac{1920}{800} = \frac{12}{5}</math>, so <math>a = \frac{12}{5}b</math>. Then <math>\frac{1}{3} \pi (\frac{12}{5}b)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>. Then by the [[Pythagorean Theorem]], the hypotenuse has length <math>\sqrt{a^2 + b^2} = \boxed{026}</math>. |
== See also == | == See also == | ||
Revision as of 22:32, 12 March 2009
Problem
When a right triangle is rotated about one leg, the volume of the cone produced is 
. When the triangle is rotated about the other leg, the volume of the cone produced is 
. What is the length (in cm) of the hypotenuse of the triangle?
Solution
Let one leg of the triangle have length 
and let the other leg have length 
. When we rotate around the leg of length , the result is a cone of height and radius , and so of volume 
. Likewise, when we rotate around the leg of length we get a cone of height and radius and so of volume 
. If we divide this equation by the previous one, we get 
, so 
. Then 
so 
and 
so 
. Then by the Pythagorean Theorem, the hypotenuse has length 
.
See also
| 1985 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
