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); | ||
| + | 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> | ||
<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> | ||
Revision as of 11:12, 29 June 2025
Problem
Point 
is on side 
of square 
. If 
has length one and 
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]](http://latex.artofproblemsolving.com/3/9/b/39bd080977330af9a59fc9b1c37d7b0975d1dc58.png)
Solution
Note that 
is a right triangle. Thus, we do Pythagorean theorem to find that side 
. Since this is the side length of the square, the area of is 
.
~superagh
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. 
