2005 AMC 10A Problem 25

Below is the professionally curated solution for Problem 25 of the 2005 AMC 10A, 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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Concepts:area ratio triangle area

Difficulty rating: 1760

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 $BCED?$

$\dfrac{266}{1521}$

$\dfrac{19}{75}$

$\dfrac{1}{3}$

$\dfrac{19}{56}$

$1$

Solution:

Triangles $ADE$ and $ABC$ share angle $A,$ so $\dfrac{[ADE]}{[ABC]} = \dfrac{AD \cdot AE}{AB \cdot AC} = \dfrac{19 \cdot 14}{25 \cdot 42} = \dfrac{266}{1050} = \dfrac{19}{75}.$ Since $[BCED] = [ABC] - [ADE],$ we get $\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}.$

Thus, the correct answer is D.

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