Difference between revisions of "2005 AMC 10A Problems/Problem 25"

Revision as of 22:36, 9 June 2015

Problem

In $ABC$ we have $AB = 25$ , $BC = 39$ , and $AC=42$ . Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$ . What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$ ?

$\mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1$

Solution

The area of a triangle is $\frac{1}{2}bc\sin A$ .

Using this formula:

$[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A$

$[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A$

Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$ ,

$[BCED] = 525\sin A - 133\sin A = 392\sin A$ .

Therefore, the desired ratio is $\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}$

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math>
-==See Also==
+==See also==
-*[[2005 AMC 10A Problems]]
+{{AMC10 box|year=2005|ab=A|num-b=20|num-a=22}}
-*[[2005 AMC 10A Problems/Problem 24|Previous Problem]]
+[[Category:Introductory Number Theory Problems]]
-[[Category:Introductory Geometry Problems]]
-[[Category:Area Ratio Problems]]
 {{MAA Notice}}

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Difference between revisions of "2005 AMC 10A Problems/Problem 25"

Revision as of 22:36, 9 June 2015

Problem

Solution

See also