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

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== Problem ==
 
== Problem ==
−
How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
+
In quadrilateral <math>ABCD</math>, <math>m\angle B = m \angle C = 120^{\circ}, AB=3, BC=4,</math> and <math>CD=5.</math> Find the area of <math>ABCD.</math>
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
+
<math>
 +
\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}
 +
</math>
  
 
== Solution ==
 
== Solution ==
−
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>.
+
Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG \perp AE</math>, where <math>F</math> and <math>G</math> are on <math>AE</math>.
 +
 
 +
It is clear that triangles <math>AFB</math> and <math>EGC</math> are congruent 30-60-90 triangles. Therefore, <math>AF = EG = \frac{3}{2}</math> and <math>BF = CG = \frac{3\sqrt{3}}{2}</math>.
 +
 
 +
Therefore, <math>AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7</math> and the area of trapezoid <math>ABCE</math> is <math>(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}</math>.
 +
 
 +
It remains to find the area of triangle <math>AED</math>, which is <math>(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}</math>.
 +
 
 +
Therefore, the total area of quadrilateral <math>ABCD</math> is <math>\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\frac{47\sqrt{3}}{4} \text{(D) }}</math>.
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2000|num-b=6|num-a=8}}
+
{{AMC12 box|year=2002|ab=P|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:22, 11 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 $CG \perp AE$, where $F$ and $G$ are on $AE$.

It is clear that triangles $AFB$ and $EGC$ are congruent 30-60-90 triangles. Therefore, $AF = EG = \frac{3}{2}$ and $BF = CG = \frac{3\sqrt{3}}{2}$.

Therefore, $AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7$ and the area of trapezoid $ABCE$ is $(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}$.

It remains to find the area of triangle $AED$, which is $(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}$.

Therefore, the total area of quadrilateral $ABCD$ is $\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\frac{47\sqrt{3}}{4} \text{(D) }}$.

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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