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

Revision as of 18:37, 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

@@ Line 35: / Line 35: @@
 == 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>
+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

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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