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Difference between revisions of "1996 AHSME Problems/Problem 17"
 
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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>.
 
Then <math>x-2 = 6\sqrt{3} - 2</math>, and <math>y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}</math>.
  
−
The area of the square is thus <math>xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36</math>.
+
The area of the rectangle is thus <math>xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36</math>.
  
 
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{\text{E}}</math>.  (The actual area is actually greater, since <math>\sqrt{3} > 1.7</math>).
 
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{\text{E}}</math>.  (The actual area is actually greater, since <math>\sqrt{3} > 1.7</math>).
 +
 +
== Solution 1.1 (Better) ==
 +
Use the process above, but use <math>\sqrt{3} \approx 1.73</math>. You should get <math>[ABCD]=150.84</math>, which then you select <math>\boxed{\text{E}}</math>. Notice that the actual area, when plugged into a calculator, yields about <math>151.0614872</math>.
 +
 +
~hastapasta
  
 
==See also==
 
==See also==

Latest revision as of 15:24, 12 May 2022

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), plain.E, fontsize(10)); label("$x$", (9,3.5), E, fontsize(10)); label("$x-2$", (0,5), plain.E, fontsize(10)); label("$y$", (5,7), N, fontsize(10)); label("$6$", (7.5,0), S, fontsize(10));[/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 rectangle 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 $\boxed{\text{E}}$. (The actual area is actually greater, since $\sqrt{3} > 1.7$).

Solution 1.1 (Better)

Use the process above, but use $\sqrt{3} \approx 1.73$. You should get $[ABCD]=150.84$, which then you select $\boxed{\text{E}}$. Notice that the actual area, when plugged into a calculator, yields about $151.0614872$.

~hastapasta

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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