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> | |
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==See Also== | ==See Also== |
Latest revision as of 03:20, 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 .
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 |