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Difference between revisions of "1981 AHSME Problems/Problem 2"

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Point <math>E</math> is on side <math>AB</math> of square <math>ABCD</math>. If <math>EB</math> has length one and <math>EC</math> has length two, then the area of the square is
 
Point <math>E</math> is on side <math>AB</math> of square <math>ABCD</math>. If <math>EB</math> has length one and <math>EC</math> has length two, then the area of the square is
 +
 +
<asy>
 +
unitsize(2cm);
 +
size(200);
 +
pair A=(0,0), B=(1.732,0), C=(1.732,1.732), D=(0,1.732), E=(0.732,0);
 +
draw(A--B--C--D--cycle,black);
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draw(C--E,black);
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,NE);
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label("$D$",D,NW);
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label("$E$",E,S);
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label("$2$", C--E, NW);
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label("$1$", B--E, S);
 +
</asy>
  
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5</math>
  
 
== Solution ==
 
== Solution ==
−
Note that <math>\triangle BCE</math> is a right triangle. Thus, we do Pythagorean theorem to find that side <math>BC=\sqrt{3}</math>. Since this is the side length of the square, the area of <math>ABCD</math> is <math>\boxed{\textbf{(C)}\ 3}</math>.  
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Note that <math>\triangle BCE</math> is a right triangle. By the Pythagorean theorem, <math>BC^2 = CE^2 - BE^2 = 2^2-1^2=3</math>, so the area of <math>ABCD</math> is <math>\boxed{\textbf{(C)}\ 3}</math>.
 +
 
 +
~superagh, edited by j314andrews.
  
−
~superagh
+
== See Also ==
 +
{{AHSME box|year=1981|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Latest revision as of 11:16, 29 June 2025

Problem

Point $E$ is on side $AB$ of square $ABCD$. If $EB$ has length one and $EC$ has length two, then the area of the square is

[asy] unitsize(2cm); size(200); pair A=(0,0), B=(1.732,0), C=(1.732,1.732), D=(0,1.732), E=(0.732,0); draw(A--B--C--D--cycle,black); draw(C--E,black); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,S); label("$2$", C--E, NW); label("$1$", B--E, S); [/asy]

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$

Solution

Note that $\triangle BCE$ is a right triangle. By the Pythagorean theorem, $BC^2 = CE^2 - BE^2 = 2^2-1^2=3$, so the area of $ABCD$ is $\boxed{\textbf{(C)}\ 3}$.

~superagh, edited by j314andrews.

See Also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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