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Difference between revisions of "1996 AHSME Problems/Problem 17"

(Created page with "==See also== {{AHSME box|year=1996|num-b=16|num-a=18}}")
 
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==Problem==
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In rectangle <math>ABCD</math>, angle <math>C</math> is trisected by <math>\overline{CF}</math> and <math>\overline{CE}</math>, where <math>E</math> is on <math>\overline{AB}</math>, <math>F</math> is on <math>\overline{AD}</math>, <math>BE=6</math> and <math>AF=2</math>. Which of the following is closest to the area of the rectangle <math>ABCD</math>?
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<asy>
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pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2);
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draw(A--B--C--D--cycle, linewidth(0.8));
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draw(E--C--F);
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dot(A^^B^^C^^D^^E^^F);
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label("$A$", A, dir((5, 3.5)--A));
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label("$B$", B, dir((5, 3.5)--B));
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label("$C$", C, dir((5, 3.5)--C));
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label("$D$", D, dir((5, 3.5)--D));
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label("$E$", E, dir((5, 3.5)--E));
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label("$F$", F, dir((5, 3.5)--F));
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label("$2$", (0,1), dir(0));
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label("$6$", (7.5,0), N);</asy>
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<math> \text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150 </math>
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==Solution==
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Since <math>\angle C = 90^\circ</math>, each of the three smaller angles is <math>30^\circ</math>, and <math>\triangle BEC</math> and <math>\triangle CDF</math> are both <math>30-60-90</math> triangles.
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<asy>
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pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2);
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draw(A--B--C--D--cycle, linewidth(0.8));
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draw(E--C--F);
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dot(A^^B^^C^^D^^E^^F);
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label("$A$", A, dir((5, 3.5)--A));
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label("$B$", B, dir((5, 3.5)--B));
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label("$C$", C, dir((5, 3.5)--C));
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label("$D$", D, dir((5, 3.5)--D));
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label("$E$", E, dir((5, 3.5)--E));
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label("$F$", F, dir((5, 3.5)--F));
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label("$2$", (0,1), W);
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label("$x$", (9,3.5), E);
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label("$x-2$", (.2,5), W);
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label("$y$", (5,7), N);
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label("$6$", (7.5,0), S);</asy>
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Defining the variables as illustrated above, we have <math>x = 6\sqrt{3}</math> from <math>\triangle BEC</math>
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Then <math>x-2 = 6\sqrt{3} - 2</math>, and <math>y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}</math>.
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The area of the square is thus <math>xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36</math>.
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Using the approximation <math>\sqrt{3} \approx 1.7</math>, we get an area of just under <math>147.6</math>, which is closest to answer <math>\boxed{D}</math>.  (The actual area is actually greater, since <math>\sqrt{3} > 1.7</math>).
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=16|num-a=18}}
 
{{AHSME box|year=1996|num-b=16|num-a=18}}

Revision as of 15:11, 19 August 2011

Problem

In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$? [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), dir(0)); label("$6$", (7.5,0), N);[/asy] $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$

Solution

Since $\angle C = 90^\circ$, each of the three smaller angles is $30^\circ$, and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.

[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), W); label("$x$", (9,3.5), E); label("$x-2$", (.2,5), W); label("$y$", (5,7), N); label("$6$", (7.5,0), S);[/asy]

Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$

Then $x-2 = 6\sqrt{3} - 2$, and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$.

The area of the square is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$.

Using the approximation $\sqrt{3} \approx 1.7$, we get an area of just under $147.6$, which is closest to answer $\boxed{D}$. (The actual area is actually greater, since $\sqrt{3} > 1.7$).

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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