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2004 AMC 12A Problems/Problem 8

Revision as of 19:36, 4 April 2024 by Mathkatanareal (talk | contribs) (→‎Solutions)
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

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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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