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

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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>
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(Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.).
  
 
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==Solution 2 (no trig)==
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==See also==
 
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{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}}
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[[Category:Triangle Area Ratio Problems]]
 
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https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg
 
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Latest revision as of 13:31, 13 October 2024

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

(Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.).

[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]

$[BCED] = [ABC] - [ADE]$, 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 \[\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}\].

Video Solution

https://youtu.be/VXyOJWcpi00

Solution 2

[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, after some calculations, \[\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}\].

~ Nafer

~ LaTeX changes by tkfun


Solution 5

Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].

\[\text{Diagram:}\] [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]

Let the area of $\triangle ABC$ be $x$. $\triangle ABE$ and $\triangle BEC$ share a height, and the ratio of their bases are $1:2$, so the area of $\triangle ABE$ is $\frac{x}{3}$.

Similarly, $\triangle AED$ and $\triangle DEB$ share a height, and the ratio of their bases is $19:6$, so the ratio of $\frac{[AED]}{[AEB]}=\frac{19}{25}$. Therefore, \[[AED]=\frac{19}{25}\cdot\left[AEB\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot\left[ABC\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot x=\frac{19}{75}x\] \[[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x\] The ratio $\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}$ which is answer choice $\boxed{\textbf{(D) } \frac{19}{56}}$.


~JH. L

See also

2005 AMC 10A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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