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

Revision as of 14:02, 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); 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 $\frac{AF}{AB}=\frac{AE}{AB}=\frac{9}{25}$ . 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 $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$

2017 AMC 10B (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AMC 10B Problems/Problem 15"

Revision as of 14:02, 13 August 2017

Problem

Solution

See Also

@@ Line 23: / Line 23: @@
 label("$E$",E,N);
 draw(D--E);
+label("$3$",A--B,N);
+label("$4$",B--C,E);
 </asy>