Difference between revisions of "1999 AMC 8 Problems/Problem 23"
Mrdavid445 (talk | contribs) (Created page with "Since the squares have side length <math>3</math>, the area of the entire square is <math>9</math>. The segments divide the square into 3 equal parts, so the area of each part i...") |
Mrdavid445 (talk | contribs) |
||
| Line 7: | Line 7: | ||
<math>3-2=1</math>. | <math>3-2=1</math>. | ||
| − | <math>CM=\sqrt 3^2+2^2</math> = <math>\sqrt 13</math> | + | <math>CM=\sqrt {3^2+2^2</math>} = <math>\sqrt 13</math> |
Revision as of 15:28, 17 June 2011
Since the squares have side length 
, the area of the entire square is 
.
The segments divide the square into 3 equal parts, so the area of each part is .
The base of a triangle is , so the height must be 
.
.
$CM=\sqrt {3^2+2^2$ (Error compiling LaTeX. Unknown error_msg)} = 
