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

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Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math>
 
Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math>
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Note: <math>BC=39</math> was not used in this problem
  
 
==See also==
 
==See also==

Revision as of 12:07, 8 November 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 $\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}$


Note: $BC=39$ was not used in this problem

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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

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