Difference between revisions of "2002 AIME I Problems/Problem 10"
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Since the area of a triangle is <math>\frac{ab\sin{C}}2</math>, the area of <math>AEF</math> is <math>525/37</math> and the area of <math>AGF</math> is <math>5250/481</math>. | Since the area of a triangle is <math>\frac{ab\sin{C}}2</math>, the area of <math>AEF</math> is <math>525/37</math> and the area of <math>AGF</math> is <math>5250/481</math>. | ||
− | The area of triangle <math>ABD</math> is <math>360/7</math>, and the area of the entire triangle <math>ABC</math> is <math>210</math>. Subtracting the areas of <math>ABD</math> and <math>AGF</math> from <math>210</math> and finding the closest integer gives <math>148</math> as the answer. | + | The area of triangle <math>ABD</math> is <math>360/7</math>, and the area of the entire triangle <math>ABC</math> is <math>210</math>. Subtracting the areas of <math>ABD</math> and <math>AGF</math> from <math>210</math> and finding the closest integer gives <math>\boxed{148}</math> as the answer. |
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=9|num-a=11}} | {{AIME box|year=2002|n=I|num-b=9|num-a=11}} |
Revision as of 16:00, 22 October 2011
Problem
In the diagram below, angle is a right angle. Point
is on
, and
bisects angle
. Points
and
are on
and
, respectively, so that
and
. Given that
and
, find the integer closest to the area of quadrilateral
.
![AIME 2002I Problem 10.png](https://wiki-images.artofproblemsolving.com//8/8d/AIME_2002I_Problem_10.png)
Solution
By the Pythagorean Theorem, . Letting
we can use the angle bisector theorem on triangle
to get
, and solving gives
and
.
The area of triangle is
that of triangle
, since they share a common side and angle, so the area of triangle
is
the area of triangle
.
Since the area of a triangle is , the area of
is
and the area of
is
.
The area of triangle is
, and the area of the entire triangle
is
. Subtracting the areas of
and
from
and finding the closest integer gives
as the answer.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |