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Difference between revisions of "2004 AMC 12A Problems/Problem 8"

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== Solutions ==
 
== Solutions ==
 +
===Solution 0===
 +
Looking, we see that the area of <math>[\triangle EBA]</math> is 16 and the area of <math>[\triangle ABC]</math> is 12. Set the area of <math>[\triangle ADB]</math> to be x. We want to find <math>[\triangle ADE]</math> - <math>[\triangle CDB]</math>. So, that would be <math>[\triangle EBA]-[\triangle ADB]=16-x</math> and <math>[\triangle ABC]-[\triangle ADB]=12-x</math>. Therefore, <math></math>[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x=\boxed{\mathrm{(B)}\ 4}<math>
 +
 +
~ MathKatana
 +
 
=== Solution 1 ===
 
=== Solution 1 ===
−
Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By alternate interior angles and <math>AA\sim</math>, we find that <math>\triangle ADE \sim \triangle CDB</math>, with side length ratio <math>\frac{4}{3}</math>. Their heights also have the same ratio, and since the two heights add up to <math>4</math>, we have that <math>h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}</math> and <math>h_{CDB} = 3 \cdot \frac 47 = \frac {12}7</math>. Subtracting the areas, <math>\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4</math> <math>\Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 4}</math>.
+
Since </math>AE \perp AB<math> and </math>BC \perp AB<math>, </math>AE \parallel BC<math>. By alternate interior angles and </math>AA\sim<math>, we find that </math>\triangle ADE \sim \triangle CDB<math>, with side length ratio </math>\frac{4}{3}<math>. Their heights also have the same ratio, and since the two heights add up to </math>4<math>, we have that </math>h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}<math> and </math>h_{CDB} = 3 \cdot \frac 47 = \frac {12}7<math>. Subtracting the areas, </math>\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4<math> </math>\Rightarrow<math> </math>\boxed{\mathrm{(B)}\ 4}<math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
Let <math>[X]</math> represent the area of figure <math>X</math>. Note that <math>[\triangle BEA]=[\triangle ABD]+[\triangle ADE]</math> and <math>[\triangle BCA]=[\triangle ABD]+[\triangle BDC]</math>.
+
Let </math>[X]<math> represent the area of figure </math>X<math>. Note that </math>[\triangle BEA]=[\triangle ABD]+[\triangle ADE]<math> and </math>[\triangle BCA]=[\triangle ABD]+[\triangle BDC]<math>.
  
−
<math>[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}</math>
+
</math>[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}<math>
  
 
=== Solution 3 (coordbash)===
 
=== Solution 3 (coordbash)===
−
Put figure <math>ABCDE</math> on a graph. <math>\overline{AC}</math> goes from (0, 0) to (4, 6) and <math>\overline{BE}</math> goes from (4, 0) to (0, 8). <math>\overline{AC}</math> is on line <math>y = 1.5x</math>. <math>\overline{BE}</math> is on line <math>y = -2x + 8</math>. Finding intersection between these points,
+
Put figure </math>ABCDE<math> on a graph. </math>\overline{AC}<math> goes from (0, 0) to (4, 6) and </math>\overline{BE}<math> goes from (4, 0) to (0, 8). </math>\overline{AC}<math> is on line </math>y = 1.5x<math>. </math>\overline{BE}<math> is on line </math>y = -2x + 8<math>. Finding intersection between these points,
  
−
<math>1.5x = -2x + 8</math>.  
+
</math>1.5x = -2x + 8<math>.  
  
−
<math>3.5x = 8 </math>
+
</math>3.5x = 8 <math>
  
−
<math> x = 8 \times \frac{2}{7}</math>
+
</math> x = 8 \times \frac{2}{7}<math>
  
−
<math> = \frac{16}{7}</math>
+
</math> = \frac{16}{7}<math>
  
 
This gives us the x-coordinate of D.  
 
This gives us the x-coordinate of D.  
−
So, <math>\frac{16}{7}</math> is the height of <math>\triangle ADE</math>, then area of <math>\triangle ADE</math> is  
+
So, </math>\frac{16}{7}<math> is the height of </math>\triangle ADE<math>, then area of </math>\triangle ADE<math> is  
−
<math>\frac{16}{7} \times 8 \times \frac{1}{2}</math>
+
</math>\frac{16}{7} \times 8 \times \frac{1}{2}<math>
−
<math> = \frac{64}{7}</math>
+
</math> = \frac{64}{7}<math>
  
−
Now, the height of <math>\triangle BDC</math> is <math>4-\frac{16}{7} = \frac{12}{7}</math>
+
Now, the height of </math>\triangle BDC<math> is </math>4-\frac{16}{7} = \frac{12}{7}<math>
−
And the area of <math>\triangle BDC</math> is <math>6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}</math>
+
And the area of </math>\triangle BDC<math> is </math>6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}<math>
  
−
This gives us <math>\frac{64}{7} - \frac{36}{7} = 4</math>
+
This gives us </math>\frac{64}{7} - \frac{36}{7} = 4<math>
  
−
Therefore, the difference is <math>4</math>
+
Therefore, the difference is </math>4<math>
  
 
=== Solution 4 ===
 
=== Solution 4 ===
−
We want to figure out <math>Area(\triangle ADE) - Area(\triangle BDC)</math>.
+
We want to figure out </math>Area(\triangle ADE) - Area(\triangle BDC)<math>.
−
Notice that <math>\triangle ABC</math> and <math>\triangle BAE</math> "intersect" and form <math>\triangle ADB</math>.
+
Notice that </math>\triangle ABC<math> and </math>\triangle BAE<math> "intersect" and form </math>\triangle ADB<math>.
  
−
This means that <math>Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)</math> because <math>Area(\triangle ADB)</math> cancels out, which can be seen easily in the diagram.
+
This means that </math>Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)<math> because </math>Area(\triangle ADB)<math> cancels out, which can be seen easily in the diagram.
  
−
<math>Area(\triangle BAE) = 0.5 * 4 * 8 = 16</math>
+
</math>Area(\triangle BAE) = 0.5 * 4 * 8 = 16<math>
  
−
<math>Area(\triangle ABC) = 0.5 * 4 * 16 = 12</math>
+
</math>Area(\triangle ABC) = 0.5 * 4 * 16 = 12<math>
  
−
<math>Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}</math>
+
</math>Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}$
  
 
==Video Solution==
 
==Video Solution==

Revision as of 18:36, 4 April 2024

The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.

Problem

In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$, $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$, $BC=6$, $AE=8$, and $\overline{AC}$ and $\overline{BE}$ intersect at $D$. What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$?

[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]

$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solutions

Solution 0

Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ - $[\triangle CDB]$. So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$. Therefore, $$ (Error compiling LaTeX. Unknown error_msg)[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x=\boxed{\mathrm{(B)}\ 4}$~ MathKatana

=== Solution 1 === Since$ (Error compiling LaTeX. Unknown error_msg)AE \perp AB$and$BC \perp AB$,$AE \parallel BC$. By alternate interior angles and$AA\sim$, we find that$\triangle ADE \sim \triangle CDB$, with side length ratio$\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to$4$, we have that$h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$and$h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$. Subtracting the areas,$\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$$ (Error compiling LaTeX. Unknown error_msg)\Rightarrow$$ (Error compiling LaTeX. Unknown error_msg)\boxed{\mathrm{(B)}\ 4}$.

=== Solution 2 === Let$ (Error compiling LaTeX. Unknown error_msg)[X]$represent the area of figure$X$. Note that$[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$and$[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$.$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}$=== Solution 3 (coordbash)=== Put figure$ABCDE$on a graph.$\overline{AC}$goes from (0, 0) to (4, 6) and$\overline{BE}$goes from (4, 0) to (0, 8).$\overline{AC}$is on line$y = 1.5x$.$\overline{BE}$is on line$y = -2x + 8$. Finding intersection between these points,$1.5x = -2x + 8$.$3.5x = 8 $$ (Error compiling LaTeX. Unknown error_msg) x = 8 \times \frac{2}{7}$$ (Error compiling LaTeX. Unknown error_msg) = \frac{16}{7}$This gives us the x-coordinate of D. So,$\frac{16}{7}$is the height of$\triangle ADE$, then area of$\triangle ADE$is$\frac{16}{7} \times 8 \times \frac{1}{2}$$ (Error compiling LaTeX. Unknown error_msg) = \frac{64}{7}$Now, the height of$\triangle BDC$is$4-\frac{16}{7} = \frac{12}{7}$And the area of$\triangle BDC$is$6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}$This gives us$\frac{64}{7} - \frac{36}{7} = 4$Therefore, the difference is$4$=== Solution 4 === We want to figure out$Area(\triangle ADE) - Area(\triangle BDC)$. Notice that$\triangle ABC$and$\triangle BAE$"intersect" and form$\triangle ADB$.

This means that$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$because$Area(\triangle ADB)$cancels out, which can be seen easily in the diagram.$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle ABC) = 0.5 * 4 * 16 = 12$$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}$

Video Solution

https://youtu.be/DlA71MBSviU

Education, the Study of Everything


See also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 12 Problems and Solutions
2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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