Difference between revisions of "1981 AHSME Problems/Problem 2"
(Created page with "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...") |
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==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>3, \boxed{C}</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>3, \boxed{\qquad\textbf{(C)}\ 3\qquad\textbf}</math>. |
~superagh | ~superagh | ||
Revision as of 21:15, 7 August 2020
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 $3, \boxed{\qquad\textbf{(C)}\ 3\qquad\textbf}$ (Error compiling LaTeX. Unknown error_msg).
~superagh
