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Difference between revisions of "2008 AMC 12B Problems/Problem 18"
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==Solution==
 
==Solution==
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Let <math>h</math> be the height of the pyramid and <math>a</math> be the distance from <math>h</math> to <math>CD</math>. The side length of the base is 14. The side lengths of <math>\triangle ABE</math> and <math>\triangle CDE</math> are <math>2\cdot105\div14=15</math> and <math>2\cdot91\div14=13</math>, respectively. We have a systems of equations through the Pythagorean Theorem:
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Let <math>h</math> be the height of the pyramid and <math>a</math> be the distance from <math>h</math> to <math>CD</math>. The side length of the base is <math>14</math>. The heights of <math>\triangle ABE</math> and <math>\triangle CDE</math> are <math>2\cdot105\div14=15</math> and <math>2\cdot91\div14=13</math>, respectively. Consider a side view of the pyramid from <math>\triangle BCE</math>.  We have a systems of equations through the Pythagorean Theorem:
  
 
<math>13^2-(14-a)^2=h^2 \\
 
<math>13^2-(14-a)^2=h^2 \\
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Therefore, <math>h=\sqrt{15^2-9^2}=12</math>, and the volume of the pyramid is <math>\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}</math>.
 
Therefore, <math>h=\sqrt{15^2-9^2}=12</math>, and the volume of the pyramid is <math>\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}</math>.
 +
 
==See also==
 
==See also==
 
{{AMC12 box|year=2008|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2008|ab=B|num-b=17|num-a=19}}

Latest revision as of 11:01, 2 February 2015

Problem

A pyramid has a square base $ABCD$ and vertex $E$. The area of square $ABCD$ is $196$, and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$, respectively. What is the volume of the pyramid?

$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$

Solution

Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$. The side length of the base is $14$. The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$, respectively. Consider a side view of the pyramid from $\triangle BCE$. We have a systems of equations through the Pythagorean Theorem:

$13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$

Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$.

Therefore, $h=\sqrt{15^2-9^2}=12$, and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$.

See also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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