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

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(24 intermediate revisions by 9 users not shown)
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==Problem==
 
==Problem==
−
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the [[ratio]] of the area of triangle <math>ADE</math> to the area of the [[quadrilateral]] <math>BCED</math>?
+
In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39</math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the [[ratio]] of the area of triangle <math>ADE</math> to the area of the [[quadrilateral]] <math>BCED</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>
+
<math> \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 </math>
  
 
==Solution 1==
 
==Solution 1==
−
We have that
+
We have
 
<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>
 +
 +
(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>
 
<asy>
Line 33: Line 35:
 
</asy>
 
</asy>
  
−
But <math>[BCED] = [ABC] - [ADE]</math>, so
+
<math>[BCED] = [ABC] - [ADE]</math>, so
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
 
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
 
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
−
&= \frac{1}{[ABC]/[ADE] - 1} \\
+
&= \frac{1}{\frac{ABC}{ADE} - 1} \\
−
&= \frac{1}{75/19 - 1} \\
+
&= \frac{1}{\frac{75}{19} - 1} \\
−
&= \boxed{\frac{19}{56}\Longrightarrow D}.
+
&= \boxed{\textbf{(D) }\frac{19}{56}}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
−
==Solution 2(no trig)==
+
 
 +
 
 +
Note: If it is hard to understand why <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}</cmath>, you can use the fact that the area of a triangle equals <math>\frac{1}{2} \cdot ab \cdot \sin(C)</math>. If angle <math>DAE = Z</math>, we have that <cmath>\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}</cmath>.
 +
 
 +
==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 <math>[ADE]=x</math>.  
 
We can let <math>[ADE]=x</math>.  
−
Since <math>EC=2*EA</math>, <math>[DEC]=2x</math>.  
+
Since <math>EC=2 \cdot EA</math>, <math>[DEC]=2x</math>.  
 
So, <math>[ADC]=3x</math>.  
 
So, <math>[ADC]=3x</math>.  
−
This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>.  
+
This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>.  
−
Thus, <cmath>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\frac{19}{56}\Longrightarrow D}.</cmath>
+
Thus, <math>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.</math>
  
 
-Conantwiz2023
 
-Conantwiz2023
  
−
==Solution 3(trig)==
+
==Solution 3 (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>.  
  
 
Using this formula:
 
Using this formula:
Line 66: Line 102:
 
<math>[BCED] = 525\sin A - 133\sin A = 392\sin A</math>.
 
<math>[BCED] = 525\sin A - 133\sin A = 392\sin A</math>.
  
−
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}=\boxed{\textbf{(D) }\frac{19}{56}}</math>
  
  
−
Note: <math>BC=39</math> was not used in this problem
+
Note: <math>BC=39</math> was not used in this problem.
−
 
 
  
 
== Solution 4 ==
 
== Solution 4 ==
Line 82: Line 117:
 
<cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</cmath>
 
<cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</cmath>
 
<cmath>\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}</cmath>
 
<cmath>\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}</cmath>
−
Finally,  
+
Finally, after some calculations,
−
<cmath>\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}</cmath>
+
<cmath>\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}</cmath>.
−
after some calculations.
 
  
 
~ Nafer
 
~ Nafer
  
−
~ LaTeX changes by dolphin7
+
~ 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].
 +
 
 +
<cmath>\text{Diagram:}</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("$A$", 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 <math>\triangle ABC</math> be <math>x</math>. <math>\triangle ABE</math> and <math>\triangle BEC</math> share a height, and the ratio of their bases are <math>1:2</math>, so the area of <math>\triangle ABE</math> is <math>\frac{x}{3}</math>.
 +
 
 +
Similarly, <math>\triangle AED</math> and <math>\triangle DEB</math> share a height, and the ratio of their bases is <math>19:6</math>, so the ratio of <math>\frac{[AED]}{[AEB]}=\frac{19}{25}</math>. Therefore,
 +
<cmath>[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</cmath>
 +
<cmath>[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x</cmath>
 +
The ratio <math>\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}</math> which is answer choice <math>\boxed{\textbf{(D) } \frac{19}{56}}</math>.
 +
 
 +
 
 +
~JH. L
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}}
−
 
+
[[Category:Triangle Area Ratio Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg

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
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