Difference between revisions of "1989 AHSME Problems/Problem 4"
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<math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>. | <math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>. | ||
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+ | == See also == | ||
+ | {{AHSME box|year=1989|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:42, 22 October 2014
Problem
In the figure, is an isosceles trapezoid with side lengths
,
, and
. The point
is on
and
is the midpoint of hypotenuse
in right triangle
. Then
Solution
Drop perpendiculars from and
; then the triangle
is similar to
but with corresponding sides of half the length.
and
, hence
.
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.