2005 AMC 10A Problem 25

Attempt Problem 25 of the 2005 AMC 10A 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 2005 AMC 10A solutions, or check the answer key.

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25.

In

\triangle ABC

we have

AB = 25,

BC = 39,

and

AC = 42.

Points

D

and

E

are on

AB

and

AC

respectively, with

AD = 19

and

AE = 14.

What is the ratio of the area of triangle

ADE

to the area of the quadrilateral

BCED?

\dfrac{266}{1521}

\dfrac{19}{75}

\dfrac{1}{3}

\dfrac{19}{56}

1

Answer: D

Concepts:area ratio triangle area

Difficulty rating: 1760

Small hint:

Triangles sharing angle

A

satisfy

\dfrac{[ADE]}{[ABC]} = \dfrac{AD \cdot AE}{AB \cdot AC}

Big hint:

The quadrilateral is the remaining part:

[BCED] = [ABC] - [ADE]

Solution:

Triangles

ADE

and

ABC

share angle

A,

so

\begin{aligned} \dfrac{[ADE]}{[ABC]} &= \dfrac{AD \cdot AE}{AB \cdot AC} \\ &= \dfrac{19 \cdot 14}{25 \cdot 42} \\ &= \dfrac{266}{1050} \\ &= \dfrac{19}{75}. \end{aligned}

Since

[BCED] = [ABC] - [ADE],

we get

\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}.

Thus, the correct answer is D.