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Difference between revisions of "2019 AMC 12B Problems/Problem 18"
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Using the distance formula, <math>PQ = 2\sqrt{2}</math>, <math>PR = \sqrt{6}</math>, and <math>QR = \sqrt{6}</math>. Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of <math>\triangle{PQR}</math> is <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>.
 
Using the distance formula, <math>PQ = 2\sqrt{2}</math>, <math>PR = \sqrt{6}</math>, and <math>QR = \sqrt{6}</math>. Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of <math>\triangle{PQR}</math> is <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>.
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==Alternative Finish (Vectors)==
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 +
Upon solving for <math>P,Q,</math> and <math>R</math>, we can find vectors <math>\overrightarrow{PQ}=</math><<math>-2,2,0</math>> and <math>\overrightarrow{PR}=</math><<math>-1,1,2</math>>, take the cross product's magnitude and divide by 2. Then the cross product equals <<math>4,4,0</math>> with magnitude <math>4\sqrt{2}</math>, yielding <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2019|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:35, 14 February 2019

Problem

Square pyramid $ABCDE$ has base $ABCD$, which measures $3$ cm on a side, and altitude $AE$ perpendicular to the base, which measures $6$ cm. Point $P$ lies on $BE$, one third of the way from $B$ to $E$; point $Q$ lies on $DE$, one third of the way from $D$ to $E$; and point $R$ lies on $CE$, two thirds of the way from $C$ to $E$. What is the area, in square centimeters, of $\triangle{PQR}$?

$\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$

Solution (Coordinate Bash)

Let $A(0, 0, 0), B(3, 0, 0), C(0, 3, 0), D(3, 3, 0),$ and $E(0, 0, 6)$. We can figure out that $P(2, 0, 2), Q(0, 2, 2),$ and $R(1, 1, 4)$.

Using the distance formula, $PQ = 2\sqrt{2}$, $PR = \sqrt{6}$, and $QR = \sqrt{6}$. Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of $\triangle{PQR}$ is $\boxed{\textbf{(C) }2\sqrt{2}}$.

Alternative Finish (Vectors)

Upon solving for $P,Q,$ and $R$, we can find vectors $\overrightarrow{PQ}=$<$-2,2,0$> and $\overrightarrow{PR}=$<$-1,1,2$>, take the cross product's magnitude and divide by 2. Then the cross product equals <$4,4,0$> with magnitude $4\sqrt{2}$, yielding $\boxed{\textbf{(C) }2\sqrt{2}}$.

See Also

2019 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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