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

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m (→‎Solution)
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==Solution==
 
==Solution==
 +
 +
<asy>
 +
pair A,B,C,D,E;
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A=(0,4);
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B=(3,4);
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C=(3,0);
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D=(0,0);
 +
draw(A--B--C--D--cycle);
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label("$A$",A,N);
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label("$B$",B,N);
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label("$C$",C,S);
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label("$D$",D,S);
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E=foot(B,A,C);
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draw(E--B);
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draw(A--C);
 +
draw(rightanglemark(B,E,C));
 +
label("$E$",E,N);
 +
draw(D--E);
 +
</asy>
  
 
First, note that <math>AC=5</math> because <math>ABC</math> is a right triangle. In addition, we have <math>AB\cdot BC=2[ABC]=AC\cdot BE</math>, so <math>BE=\frac{12}{5}</math>. Using similar triangles within <math>ABC</math>, we get that <math>AE=\frac{9}{5}</math> and <math>CE=\frac{16}{5}</math>.
 
First, note that <math>AC=5</math> because <math>ABC</math> is a right triangle. In addition, we have <math>AB\cdot BC=2[ABC]=AC\cdot BE</math>, so <math>BE=\frac{12}{5}</math>. Using similar triangles within <math>ABC</math>, we get that <math>AE=\frac{9}{5}</math> and <math>CE=\frac{16}{5}</math>.

Revision as of 14:01, 13 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); [/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 $\frac{AF}{AB}=\frac{AE}{AB}=\frac{9}{25}$. 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 $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$

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

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