Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 10B Problems/Problem 15"

m (→‎Solution)
(→‎Solution)
Line 30: Line 30:
  
 
Let <math>F</math> be the foot of the perpendicular from <math>E</math> to <math>AB</math>. Since <math>EF</math> and <math>BC</math> are parallel, <math>\Delta AFE</math> is similar to <math>\Delta ABC</math>. Therefore, we have <math>\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}</math>. Since <math>AB=3</math>, <math>AF=\frac{27}{25}</math>. Note that <math>AF</math> is an altitude of <math>\Delta AED</math> from <math>AD</math>, which has length <math>4</math>. Therefore, the area of <math>\Delta AED</math> is <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math>
 
Let <math>F</math> be the foot of the perpendicular from <math>E</math> to <math>AB</math>. Since <math>EF</math> and <math>BC</math> are parallel, <math>\Delta AFE</math> is similar to <math>\Delta ABC</math>. Therefore, we have <math>\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}</math>. Since <math>AB=3</math>, <math>AF=\frac{27}{25}</math>. Note that <math>AF</math> is an altitude of <math>\Delta AED</math> from <math>AD</math>, which has length <math>4</math>. Therefore, the area of <math>\Delta AED</math> is <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math>
 +
 +
===Solution 2===
 +
Alternatively, we can use coordinates. Denote <math>D</math> as the origin. We find the equation for <math>AC</math> as <math>y=-\frac{4}{3}x+4</math>, and <math>BE</math> as <math>y=-\frac{3}{4}x+\frac{7}{4}</math>. Solving for <math>x</math> yields <math>\frac{27}{25}</math>. Our final answer then becomes <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:08, 20 August 2017

Problem

Rectangle $ABCD$ has $AB=3$ and $BC=4$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}$

Solution

[asy] pair A,B,C,D,E; A=(0,4); B=(3,4); C=(3,0); D=(0,0); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,S); label("$D$",D,S); E=foot(B,A,C); draw(E--B); draw(A--C); draw(rightanglemark(B,E,C)); label("$E$",E,N); draw(D--E); label("$3$",A--B,N); label("$4$",B--C,E); [/asy]

First, note that $AC=5$ because $ABC$ is a right triangle. In addition, we have $AB\cdot BC=2[ABC]=AC\cdot BE$, so $BE=\frac{12}{5}$. Using similar triangles within $ABC$, we get that $AE=\frac{9}{5}$ and $CE=\frac{16}{5}$.

Let $F$ be the foot of the perpendicular from $E$ to $AB$. Since $EF$ and $BC$ are parallel, $\Delta AFE$ is similar to $\Delta ABC$. Therefore, we have $\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}$. Since $AB=3$, $AF=\frac{27}{25}$. Note that $AF$ is an altitude of $\Delta AED$ from $AD$, which has length $4$. Therefore, the area of $\Delta AED$ is $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$

Solution 2

Alternatively, we can use coordinates. Denote $D$ as the origin. We find the equation for $AC$ as $y=-\frac{4}{3}x+4$, and $BE$ as $y=-\frac{3}{4}x+\frac{7}{4}$. Solving for $x$ yields $\frac{27}{25}$. Our final answer then becomes $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_15&oldid=87131"