Difference between revisions of "2005 CEMC Gauss (Grade 7) Problems/Problem 13"
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Latest revision as of 01:14, 23 October 2014
Problem
In the diagram, the length of 
is twice the length of 
. What is the area of the triangle 
?
![[asy] draw((0,0)--(-3,0)--(0,4)--cycle); draw((0,0)--(6,0)--(0,4)--cycle); label("3",(-1.5,0),N); label("4",(0,2),E); label("$A$",(0,4),N); label("$B$",(-3,0),S); label("$C$",(6,0),S); label("$D$",(0,0),S); draw((0,0.4)--(0.4,0.4)--(0.4,0)); [/asy]](http://latex.artofproblemsolving.com/4/3/3/43330d5778a7805f72e4c2b774d2b4f1316a93db.png)
Solution
Since 
and is twice the length of , then 
. Therefore, triangle has a base of length 
and a height of length 
. Therefore, the area of triangle is 
. Therefore, the correct answer is 
See Also
| 2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| CEMC Gauss (Grade 7) | ||
