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

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m (→‎Solution)
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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>
 
<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 1==
  
 
<asy>
 
<asy>
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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>
 
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===
+
==Solution 2==
 
From similar triangles, we know that <math>AE=\frac{9}{5}</math> (see Solution 1). Furthermore, we also know that <math>AD=4</math> from the rectangle. Using the sine formula for area, we have
 
From similar triangles, we know that <math>AE=\frac{9}{5}</math> (see Solution 1). Furthermore, we also know that <math>AD=4</math> from the rectangle. Using the sine formula for area, we have
 
<cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath>
 
<cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath>
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~coolwiz
 
~coolwiz
  
−
===Solution 3===
+
==Solution 3==
 
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>
 
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>
  
−
===Solution 4===
+
==Solution 4==
 
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>
 
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>
−
===Solution 5===
+
 
 +
==Solution 5==
 
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 4, Area of <math>AED</math> is the same as <math>ABE</math>.
 
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 4, Area of <math>AED</math> is the same as <math>ABE</math>.
  

Revision as of 18:02, 17 January 2021

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 1

[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

From similar triangles, we know that $AE=\frac{9}{5}$ (see Solution 1). Furthermore, we also know that $AD=4$ from the rectangle. Using the sine formula for area, we have \[\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).\] But, note that $\sin(m\angle EAD)=\sin(m\angle CAD)=\frac{O}{H}=\frac{3}{5}$. Thus, we see that \[[AED]=\frac{1}{2}\cdot \frac{9}{5}\cdot 4\cdot\frac{3}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.\] ~coolwiz

Solution 3

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 4

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 5

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 4, Area of $AED$ is the same as $ABE$.

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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