2017 AMC 10B Problems/Problem 15

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 the area of $AED$ because they share the base $AE$ 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 corresponding 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 $ABE$.

Solution 6

Drop an altitude from $E$ to $BC$ and call its foot $X$. We have that $EX \cdot BC=BE \cdot EC$ since both are equal to two times the area of $BEC$. Since $BC=4$, $BE=\frac{12}{5}$, and $EC=\frac{16}{5}$, we can calculate that $EX=\frac{48}{25}$. If $EX$ is extended to meet $AD$ at point $Y$, $EY=3-\frac{48}{25}=\frac{27}{25}$. Therefore, $[AED]=\frac{EY \cdot AD}{2}=\frac{\frac{27}{25} \cdot 4}{2}=\boxed{\textbf{(E)}\frac{54}{25}}$.

- Fasolinka

See Also

 2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 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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