Difference between revisions of "2005 AMC 10A Problems/Problem 25"
Mathwiz0803 (talk | contribs) (→Solution) |
Mathwiz0803 (talk | contribs) (→Solution (no trig)) |
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<cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | ||
− | + | <asy> | |
unitsize(0.15 cm); | unitsize(0.15 cm); | ||
Line 22: | Line 22: | ||
draw(D--E); | draw(D--E); | ||
− | label(" | + | label("$A$", A, N); |
− | label(" | + | label("$B$", B, SW); |
− | label(" | + | label("$C$", C, SE); |
− | label(" | + | label("$D$", D, W); |
− | label(" | + | label("$E$", E, NE); |
− | label(" | + | label("$19$", (A + D)/2, W); |
− | label(" | + | label("$6$", (B + D)/2, W); |
− | label(" | + | label("$14$", (A + E)/2, NE); |
− | label(" | + | label("$28$", (C + E)/2, NE); |
− | + | </asy> | |
But <math>[BCED] = [ABC] - [ADE]</math>, so | But <math>[BCED] = [ABC] - [ADE]</math>, so |
Revision as of 19:53, 3 November 2016
Contents
[hide]Problem
In we have
,
, and
. Points
and
are on
and
respectively, with
and
. What is the ratio of the area of triangle
to the area of the quadrilateral
?
Solution (no trig)
We have that
But , so
.
Solution (trig)
Using this formula:
Since the area of is equal to the area of
minus the area of
,
.
Therefore, the desired ratio is
Note: was not used in this problem
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.