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Extensions, Bisections and Dissections in a Rectangle

Alignments to Content Standards: G-SRT.B.5


In rectangle $ABCD$, $|AB|=6$, $|AD|=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended 2 units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of $BFDG$?

IM Commentary

This task involves a reasonably direct application of similar triangles, coupled with a moderately challenging procedure of constructing a diagram from a verbal description.

Depending on its intended use, the teacher may wish to help students draw the picture of the rectangle and the extended side. It is important to remember the convention that the vertices of the quadrilaterals are labelled consecutively although the answer does not depend on whether they are labelled clockwise (the usual convention) or counterclockwise. Once an accurate picture is obtained, there are multiple approaches for computing the area. Two approaches are presented in the solutions, one using the formula for the area of a trapezoid, the other using triangles. In either case, the vital step in the argument is calculating the length of $\overline{BF}$, for which similar triangles are employed.

This task was adapted from problem #19 on the 2012 American Mathematics Competition (AMC) 10B Test.

For the 2012 AMC 10B, which was taken by 35,086 students, the multiple choice answers for the problem had the following distribution:

Choice Answer Percentage of Answers
(A) ${\frac{133}{2}}$ 3.01
(B) ${67}$ 4.26
(C)* ${\frac{135}{2}}$ 37.32
(D) ${68}$ 5.84
(E) ${\frac{137}{2}}$ 3.94
Omit -- 45.60

Of the 35,086 students: 17,169, or 49%, were in 10th grade; 9,928 or 28%, were in 9th grade; and the remainder were below 9th grade.


Solution: 1

Below is a picture of rectangle $ABCD$ and the point $E$ and trapezoid $BFDG$:


Note that $\triangle EBF$ and $\triangle EAD$ are both right triangles since $ABCD$ is a rectangle. They share angle $E$ and so by the AA criterion for similarity, $\triangle EBF$ is similar to $\triangle EAD$. Since corresponding sides of similar triangles are proportional to one another, we conclude that $$ \frac{|BF|}{|AD|} = \frac{|BE|}{|AE|}. $$ Plugging in the values that have been given we get $$ \frac{|BF|}{30} = \frac{2}{8} $$ and we conclude that $|BF|=\frac{15}{2}$. Segments $\overline{BF}$ and $\overline{DG}$ belong to opposite sides of rectangle $ABCD$ which are parallel and so $BFDG$ is a trapezoid. Its area can be computed using the formula $\frac{1}{2}h(b_1 +b_2)$ where $b_1, b_2$ are the lengths of the (parallel) bases and $h$ is the height:

\begin{align} \frac{1}{2}h(b_1+b_2) &= \frac{1}{2}\cdot |AB|\cdot (|BF|+|GD|) \\ &= \frac{1}{2}\cdot 6\cdot \left(\frac{15}{2}+15\right) \\ &= \frac{135}{2}. \end{align}

Solution: 2 Subtracting triangles from a rectangle

Alternatively, the area of the trapezoid $BFDG$ is the area of rectangle $ABCD$ minus the sum of the areas of right triangles $BAG$ and $DCF$. We are given $|AB| = 6$ and so $|CD| = 6$ as well. We also know that $|AG| = 15$ since $G$ is the midpoint of $\overline{AD}$. Using the similarity argument from the first paragraph, $|BF| = \frac{15}{2}$. Since $|CF| + |BF| = 30$ we have $|CF| = \frac{45}{2}$.

The area of rectangle $ABCD$ is $6 \times 30 = 180$ and the area of triangle $BAG$ is $\frac{1}{2} \times 15 \times 6 = 45$. The area of triangle $DCF$ is $\frac{1}{2} \times 6 \times \frac{45}{2} = \frac{135}{2}$. Combining all of this information, the area of trapezoid $BFDG$ is $$ 180 - 45 - \frac{135}{2} = \frac{135}{2}. $$