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Difference between revisions of "2019 AMC 12B Problems/Problem 18"

(→‎Problem: Added problem statement (but not answer choices).)
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Square pyramid <math>ABCDE</math> has base <math>ABCD</math>, which measures <math>3</math> cm on a side, and altitude <math>AE</math> perpendicular to the base, which measures <math>6</math> cm. Point <math>P</math> lies on <math>BE</math>, one third of the way from <math>B</math> to <math>E</math>; point <math>Q</math> lies on <math>DE</math>, one third of the way from <math>D</math> to <math>E</math>; and point <math>R</math> lies on <math>CE</math>, two thirds of the way from <math>C</math> to <math>E</math>. What is the area, in square centimeters, of <math>\triangle{PQR}</math>?
 
Square pyramid <math>ABCDE</math> has base <math>ABCD</math>, which measures <math>3</math> cm on a side, and altitude <math>AE</math> perpendicular to the base, which measures <math>6</math> cm. Point <math>P</math> lies on <math>BE</math>, one third of the way from <math>B</math> to <math>E</math>; point <math>Q</math> lies on <math>DE</math>, one third of the way from <math>D</math> to <math>E</math>; and point <math>R</math> lies on <math>CE</math>, two thirds of the way from <math>C</math> to <math>E</math>. What is the area, in square centimeters, of <math>\triangle{PQR}</math>?
  
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==Solution==
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<math>\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</math>
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==Solution (Coordinate Bash)==
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Let <math>A(0, 0, 0), B(3, 0, 0), C(0, 3, 0), D(3, 3, 0),</math> and <math>E(0, 0, 6)</math>. We can figure out that <math>P(2, 0, 2), Q(0, 2, 2),</math> and <math>R(1, 1, 4)</math>.
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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>.
  
 
==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}}

Revision as of 17:47, 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 $\boxed{\textbf{(C) }2\sqrt{2}}$.

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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