Difference between revisions of "2008 AMC 10A Problems/Problem 20"

Revision as of 19:34, 17 February 2008 (view source) Daman71692 (talk \| contribs) (New page: Draw isoceles trapezoid ABCD. Draw diagonals intersecting at K. Obviously, AKD is congruent to BKC. Likelywise, AKB is similar to CKD. Call the altitude to base AB of triangle AKB 3x and t...)		Revision as of 19:34, 17 February 2008 (view source) Daman71692 (talk \| contribs) (Removing all content from page) Newer edit →
Line 1:		Line 1:
−	Draw isoceles trapezoid ABCD. Draw diagonals intersecting at K. Obviously, AKD is congruent to BKC. Likelywise, AKB is similar to CKD. Call the altitude to base AB of triangle AKB 3x and the altitude to base CD of triangle CKD 4x. Now the area of the trapezoid is the sum of areas AKB, CKD, AKD, and BKC which equals . Now take isosceles trapezoid ABCD again. Drop altitudes to the base CD from point A and point B to get a rectangle and two triangles. The area of the rectangle is and the area of the two triangles is as the lengths of the legs are 1.5 and 7x. So now you have and simple arithmetic gives you . .	+

Page