2006 AMC 10B Problems/Problem 23

Revision as of 20:31, 18 July 2006 by Xantos C. Guin (talk | contribs) (Added problem and solution)

Problem

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?

2006amc10b23.gif

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$

Solution

To help in finding the area of the quadrilateral, draw an auxillary line from the top vertex to the intersection of the two lines.

Doing so, and labling the points yeilds this as the resulting diagram:

2006amc10b23solution.gif

Since triangles $AFC$ and $EFC$ share an altitude, and their respective areas are equal, their bases must be equal. So $AF=EF$.

Since triangles $AFB$ and $EFB$ share an altitude, and their respective bases are equal, their areas must be equal. So $x+3=y$.

Since triangles $DFA$ and $CFA$ share an altitude, and their respective areas are in the ratio $3:7$, their bases must be in the same ratio. So $\frac{DF}{3}=\frac{CF}{7}$

Since triangles $DFB$ and $CFB$ share an altitude, and their respective bases are in the ratio $3:7$, their areas must be in the same ratio. So $\frac{x}{3}=\frac{y+7}{7}$

Substituting $x+3$ for $y$:

$\frac{x}{3}=\frac{x+3+7}{7}$

$\frac{x}{3}=\frac{x+10}{7}$

$7x=3x+30$

$4x=30$

$x=\frac{15}{2}$

$y=\frac{21}{2}$

Therefore, the shaded area is $x+y = \frac{15}{2} + \frac{21}{2} = 18 \Rightarrow D$

See Also

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