Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 5"
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==Solution== | ==Solution== | ||
| + | In general, if <math>G</math> and <math>H</math> are similar triangles with sides <math>s_1</math> and <math>s_2</math> , then <math>\frac{Area G}{Area H}=\frac{(s_1)^2}{(s_2)^2}</math>. | ||
| + | Here, <math>\triangle{BCD}</math> is <math>\frac{50}{128}=\frac{25}{64}</math> of <math>\triangle{ACE}</math>, so BC is <math>\frac{5}{8}</math> of AC or AB is <math>\frac{3}{8}</math> of AC | ||
| + | then <math>\triangle{ABF}</math> is <math>\frac{9}{64}</math> of <math>\triangle{ACE}</math> so | ||
| + | <math>\triangle ABF = \boxed{18}</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 02:20, 13 January 2019
Problem
In the figure 
is parallel to 
and also 
is parallel to 
. The area of the larger triangle 
is 
.
The area of the trapezoid 
is 
. Determine the area of triangle 
.
![[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW); MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S); [/asy]](http://latex.artofproblemsolving.com/8/b/c/8bc13e1d9d8740bc4f6f48868cdc693e5c1b390c.png)
Solution
In general, if 
and 
are similar triangles with sides 
and 
, then 
.
Here, 
is 
of 
, so BC is 
of AC or AB is 
of AC
then 
is 
of so
See Also
| 2006 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
