Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | ||
− | <math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{ | + | <math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{1}{3} \Rightarrow C</math> |
==Solution 2== | ==Solution 2== |
Revision as of 00:24, 3 June 2015
Contents
Problem
Let be a diameter of a circle and let
be a point on
with
. Let
and
be points on the circle such that
and
is a second diameter. What is the ratio of the area of
to the area of
?
Solution 1
Let us assume that the diameter is of length .
is
of diameter and
is
.
is the radius of the circle, so using the Pythagorean theorem height
of
is
. This is also the height of the
.
Area of the is
=
.
The height of can be found using the area of
and
as base.
Hence the height of is
=
.
The diameter is the base for both the triangles and
.
Hence, the ratio of the area of to the area of
is
=
Solution 2
Since and
share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from
to
.
.
Since , then
. So the ratio of the two altitudes is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.