Difference between revisions of "2006 AIME I Problems/Problem 7"
m (→Solution) |
m (→Solution) |
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Line 11: | Line 11: | ||
:perpendicular to x-axis | :perpendicular to x-axis | ||
:& cross x-axis at 0, 1, 2... | :& cross x-axis at 0, 1, 2... | ||
− | *Base of | + | *Base of region <math>\mathcal{A}</math> be at <math>x = 1</math>; bigger base of region <math>\mathcal{D}</math> at <math>x = 7</math> |
*One side of the angle be x-axis. | *One side of the angle be x-axis. | ||
*The other side be <math>y = x - h</math> | *The other side be <math>y = x - h</math> |
Revision as of 20:26, 11 March 2007
Problem
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region to the area of shaded region
is 11/5. Find the ratio of shaded region
to the area of shaded region
Solution
Apex of the angle is not on the parallel lines.
Let...
- The set of parallel lines be
- perpendicular to x-axis
- & cross x-axis at 0, 1, 2...
- Base of region
be at
; bigger base of region
at
- One side of the angle be x-axis.
- The other side be
Then...
As area of triangle =.5 base x height...
By similar method, seems to be 408.