2009 AMC 10B Problems/Problem 18

Problem

Rectangle $ABCD$ has $AB=8$ and $BC=6$ . Point $M$ is the midpoint of diagonal $\overline{AC}$ , and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$ . What is the area of $\triangle AME$ ?

$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E) } \frac{85}{8}$

Solution

$[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); [/asy]$

By the Pythagorean theorem we have $AC=10$ , hence $AM=5$ .

The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar.

The ratio of their sides is $\frac{AM}{AB} = \frac 58$ , hence the ratio of their areas is $\left( \frac 58 \right)^2 = \frac{25}{64}$ .

And as the area of triangle $ABC$ is $\frac{6\cdot 8}2 = 24$ , the area of triangle $AME$ is $24\cdot \frac{25}{64} = \boxed{ \frac{75}8 }$ .

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2009 AMC 10B Problems/Problem 18

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