Difference between revisions of "1981 AHSME Problems/Problem 2"
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| + | == Problem == | ||
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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 | ||
<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> | + | 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>. |
~superagh | ~superagh | ||
Latest revision as of 13:33, 16 July 2024
Problem
Point 
is on side 
of square 
. If 
has length one and 
has length two, then the area of the square is
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
