2005 AMC 10A Problems/Problem 25

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 $BCED$ ?

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

Solution 1

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

$[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$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 $\begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{\frac{ABC}{ADE} - 1} \\ &= \frac{1}{\frac{75}{19} - 1} \\ &= \boxed{\textbf{(D) }\frac{19}{56}}. \end{align*}$

Note: If it is hard to understand why $\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}$ , you can use the fact that the area of a triangle equals $\frac{1}{2} \cdot ab \cdot \sin(C)$ . If angle $DAE = Z$ , we have that $\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}$ . - SuperJJ

Video Solution

CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00

Solution 2 (no trig)

$[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$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]$

We can let $[ADE]=x$ . Since $EC=2 \cdot EA$ , $[DEC]=2x$ . So, $[ADC]=3x$ . This means that $[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}$ . Thus, $\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.$

-Conantwiz2023

Solution 3 (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 $\frac{133\sin A}{392\sin A}=\boxed{\textbf{(D) }\frac{19}{56}}$

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

Solution 4

Let $F$ be on $AC$ such that $DE\parallel BF$ then we have $\frac{[ADE]}{[ABF]}=\left(\frac{AD}{AB}\right)^2=\left(\frac{19}{25}\right)^2=\frac{361}{625}$ $\frac{[ADE]}{[DEFB]}=\frac{361}{625-361}=\frac{361}{364}$ Since $\bigtriangleup ADE\sim\bigtriangleup ABF$ we have $\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}$ Thus $FC=EC-EF=\frac{448}{19}$ and $\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}$ $\frac{[ADE]}{[BFC]}=\left(\frac{[ADE]}{[ABF]}\right)\left(\frac{[ABF]}{[BFC]}\right)=\left(\frac{361}{625}\right)\left(\frac{350}{448}\right)=\frac{126350}{280000}$ Finally, $\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}$ after some calculations.

~ Nafer

~ LaTeX changes by tkfun

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

Page

Toolbox

Search