Difference between revisions of "1981 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | 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> <math>\boxed{\textbf{(C)}\ 3}</math>. | + | 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, edited by j314andrews. | ||
Latest revision as of 11:16, 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. By the Pythagorean theorem, 
, so the area of is 
.
~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. 
