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

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==Problem==
 
==Problem==
Placeholder
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Rectangle <math>ABCD</math> has <math>AB=3</math> and <math>BC=4</math>. Point <math>E</math> is the foot of the perpendicular from <math>B</math> to diagonal <math>\overline{AC}</math>. What is the area of <math>\triangle AED</math>?
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<math>\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}</math>
  
 
==Solution==
 
==Solution==
Placeholder
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<asy>
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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);
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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);
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draw(rightanglemark(B,E,C));
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label("$E$",E,N);
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draw(D--E);
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label("$3$",A--B,N);
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label("$4$",B--C,E);
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</asy>
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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>.
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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>
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===Solution 2===
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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>
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===Solution 3===
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We note that the area of <math>ABE</math> must equal area of <math>AED</math> because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of <math>ABE</math> to be <math>\frac{1}{2}*\frac{9}{5}*\frac{12}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</math>
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===Solution 4===
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We know all right triangles are 5-4-3, so the areas are proportional to the square of like sides. Area of <math>ABE</math> is <math> (\dfrac{3}{5})^2</math> of <math>ABC = \frac{54}{25}</math>. Using similar logic in Solution 3, Area of <math>AED</math> is the same as <math>ABE</math>.
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==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 08:06, 24 January 2019

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 $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$

Solution 3

We note that the area of $ABE$ must equal area of $AED$ because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of $ABE$ to be $\frac{1}{2}*\frac{9}{5}*\frac{12}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.$

Solution 4

We know all right triangles are 5-4-3, so the areas are proportional to the square of like sides. Area of $ABE$ is $(\dfrac{3}{5})^2$ of $ABC = \frac{54}{25}$. Using similar logic in Solution 3, Area of $AED$ is the same as $ABE$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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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