2009 AMC 10B Problem 18

Attempt Problem 18 of the 2009 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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18.

Rectangle

ABCD

has

AB=8

and

BC=6.

Point

M

is the midpoint of diagonal

\overline{AC},

and

E

is on

\overline{AB}

with

\overline{ME}\perp\overline{AC}.

What is the area of

\triangle AME?

\dfrac{65}{8}

\dfrac{25}{3}

9

\dfrac{75}{8}

\dfrac{85}{8}

Answer: D

Concepts:similarity Pythagorean Theorem triangle area

Difficulty rating: 1370

Small hint:

AC=10,

so

AM=5

Big hint:

\triangle AME\sim\triangle ABC,

giving

\dfrac{ME}{AM}=\dfrac{BC}{AB}

Solution:

By the Pythagorean Theorem,

AC=\sqrt{8^2+6^2}=10,

so

AM=5.

Right triangles

AME

and

ABC

share angle

A,

so they are similar with

\dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68,

giving

ME=\dfrac{15}{4}.

Then

\begin{gathered} [\triangle AME]=\dfrac12\cdot AM\cdot ME \\ = \dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}. \end{gathered}

Thus, the correct answer is D.