Difference between revisions of "1999 AMC 8 Problems/Problem 23"

Revision as of 14:28, 17 June 2011

Since the squares have side length $3$ , the area of the entire square is $9$ .

The segments divide the square into 3 equal parts, so the area of each part is $3$ .

The base of a triangle is $3$ , so the height must be $2$ .

$3-2=1$ .

$CM=\sqrt {3^2+2^2$ (Error compiling LaTeX. Unknown error_msg)} = $\sqrt 13$

-Mrdavid445

Revision as of 14:28, 17 June 2011 (view source) Mrdavid445 (talk \| contribs) ← Older edit		Revision as of 14:28, 17 June 2011 (view source) Mrdavid445 (talk \| contribs) Newer edit →
Line 8:		Line 8:

	<math>CM=\sqrt {3^2+2^2</math>} = <math>\sqrt 13</math>		<math>CM=\sqrt {3^2+2^2</math>} = <math>\sqrt 13</math>
		+
		+
		+	-Mrdavid445