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

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<math> \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 </math>
 
<math> \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 </math>
  
−
==Solution==
+
==Solution (no trig)==
 +
We have that
 +
<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);
 +
 
 +
pair A, B, C, D, E;
 +
 
 +
A = (191/39,28*sqrt(1166)/39);
 +
B = (0,0);
 +
C = (39,0);
 +
D = (6*A + 19*B)/25;
 +
E = (28*A + 14*C)/42;
 +
 
 +
draw(A--B--C--cycle);
 +
draw(D--E);
 +
 
 +
label("<math>A</math>", A, N);
 +
label("<math>B</math>", B, SW);
 +
label("<math>C</math>", C, SE);
 +
label("<math>D</math>", D, W);
 +
label("<math>E</math>", E, NE);
 +
label("<math>19</math>", (A + D)/2, W);
 +
label("<math>6</math>", (B + D)/2, W);
 +
label("<math>14</math>", (A + E)/2, NE);
 +
label("<math>28</math>", (C + E)/2, NE);
 +
[/asy]
 +
 
 +
But <math>[BCED] = [ABC] - [ADE]</math>, so
 +
\begin{align*}
 +
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
 +
&= \frac{1}{[ABC]/[ADE] - 1} \\
 +
&= \frac{1}{75/19 - 1} \\
 +
&= \boxed{\frac{19}{56}}.
 +
\end{align*}
 +
The answer is therefore <math>D</math>.
 +
 
 +
==Solution (trig)==
 
The [[area]] of a [[triangle]] is <math>\frac{1}{2}bc\sin A</math>.  
 
The [[area]] of a [[triangle]] is <math>\frac{1}{2}bc\sin A</math>.  
  

Revision as of 19:52, 3 November 2016

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 (no trig)

We have that \[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.\]

[asy] unitsize(0.15 cm);

pair A, B, C, D, E;

A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42;

draw(A--B--C--cycle); draw(D--E);

label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, NE); label("$19$", (A + D)/2, W); label("$6$", (B + D)/2, W); label("$14$", (A + E)/2, NE); label("$28$", (C + E)/2, NE); [/asy]

But $[BCED] = [ABC] - [ADE]$, so \begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{[ABC]/[ADE] - 1} \\ &= \frac{1}{75/19 - 1} \\ &= \boxed{\frac{19}{56}}. \end{align*} The answer is therefore $D$.

Solution (trig)

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 (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

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