1988 AIME Problems/Problem 7: Difference between revisions
Mathgeek2006 (talk | contribs) |
Burunduchok (talk | contribs) |
||
| Line 18: | Line 18: | ||
</cmath> | </cmath> | ||
The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = 110</math>. | The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}</math>. | ||
== See also == | == See also == | ||
Revision as of 17:40, 23 April 2017
Problem
In triangle 
, 
, and the altitude from 
divides 
into segments of length 3 and 17. What is the area of triangle ?
Solution
Let 
be the intersection of the altitude with 
, and 
be the length of the altitude. Without loss of generality, let 
and 
. Then 
and 
. Using the tangent sum formula,
The postive value of 
, so the area is 
.
See also
| 1988 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
