AMC 12 Mock Exam 2 Problem 12

Problem 12 of 25

Rectangle $ABCD$ has $AB=12$ and $BC=5$ . Point $E$ lies on side $\overline{CD}$ with $DE=4$ . Segment $\overline{BE}$ meets diagonal $\overline{AC}$ at $F$ , as shown (the figure is not necessarily drawn to scale). What is the area of the shaded triangle $BCF$ ?

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Solution

Since $\overline{AB}\parallel\overline{CE}$ , triangles $FAB$ and $FCE$ are similar, with $CE=12-4=8$ and $AB=12$ . Hence $\frac{CF}{FA}=\frac{CE}{AB}=\frac{8}{12}=\frac23$ , so $CF=\frac25\,CA$ .

Triangles $BCF$ and $BCA$ share the vertex $B$ , and their bases $CF$ and $CA$ lie on the same line, so their areas are in the ratio $\frac{CF}{CA}=\frac25$ . Since $[BCA]=\frac12\cdot12\cdot5=30$ , the area of $\triangle BCF$ is $\frac25\cdot30=\boxed{12}$ .

Tagged: similar triangles in a rectangle · area ratios

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