Difference between revisions of "2003 AMC 10A Problems/Problem 22"
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<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath> | <cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath> | ||
− | We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF is twice of 10, or < | + | We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>20\Rightarrow \mathrm{(B)}</math> |
== See Also == | == See Also == |
Revision as of 09:59, 6 March 2008
Problem
In rectangle , we have
,
,
is on
with
,
is on
with
, line
intersects line
at
, and
is on line
with
. Find the length of
.
Solution
Solution 1
Since is a rectangle,
.
Since is a rectangle and
,
.
Since is a rectangle,
.
So, is a transversal, and
.
This is sufficient to prove that and
.
Using ratios:
Since can't have 2 different lengths, both expressions for
must be equal.
Solution 2
Since is a rectangle,
,
, and
. From the Pythagorean Theorem,
.
Lemma
Statement:
Proof: , obviously.
$
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.
Let .
Also, , therefore
We can multiply both sides by to get that
is twice of 10, or
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |