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2005 AMC 10A Problems/Problem 25

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Problem

In 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?

\mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } ...

Solution

The area of a triangle is \frac{1}{2}bc\sin A.

Using this formula:

[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A

[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A

Since the area of BCED is equal to the area of ABC minus the area of ADE,

[BCED] = 525\sin A - 133\sin A = 392\sin A.

Therefore, the desired ratio is \frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}

See Also

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