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Difference between revisions of "2002 AMC 12P Problems/Problem 19"

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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG</math> perpendicular to <math>AE</math>, where <math>F</math> and <math>G</math> are on <math>AE</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=18|num-a=20}}
 
{{AMC12 box|year=2002|ab=P|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:26, 10 March 2024

Problem

In quadrilateral $ABCD$, $m\angle B = m \angle C = 120^{\circ}, AB=3, BC=4,$ and $CD=5.$ Find the area of $ABCD.$

$\text{(A) }15 \qquad \text{(B) }9 \sqrt{3} \qquad \text{(C) }\frac{45 \sqrt{3}}{4} \qquad \text{(D) }\frac{47 \sqrt{3}}{4} \qquad \text{(E) }15 \sqrt{3}$

Solution

Draw $AE$ parallel to $BC$ and draw $BF$ and $CG$ perpendicular to $AE$, where $F$ and $G$ are on $AE$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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