Question

given △abc with parallel lines dividing (overline{ab}) into three congruent segments as shown, compare the following areas.
a. triangle dbe and trapezoid degf
b. triangle dbe and trapezoid fgca
c. trapezoids degf and fgca
d. trapezoid degf and triangle abc
e. trapezoid fgca and triangle abc
f. triangle abc and trapezoid deca
(type an integer or a simplified fraction )
b. the ratio of the area of triangle dbe to the area of trapezoid fgca is (\frac{1}{5})
(type an integer or a simplified fraction )
c. the ratio of the area of trapezoid degf to the area of trapezoid fgca is (\frac{3}{5})
(type an integer or a simplified fraction )
d. the ratio of the area of trapezoid degf to the area of triangle abc is (square)
(type an integer or a simplified fraction )

Explanation:

Step1: Use the property of similar triangles

Since the lines dividing \(\overline{AB}\) are parallel, \(\triangle DBE\sim\triangle FBG\sim\triangle ABC\). Let the height of \(\triangle DBE\) be \(h_1\), the height of \(\triangle FBG\) be \(h_2\), and the height of \(\triangle ABC\) be \(h_3\). If the length of each segment of \(\overline{AB}\) is \(a\), then the ratio of the sides of \(\triangle DBE\) to \(\triangle ABC\) is \(\frac{1}{3}\), and the ratio of the sides of \(\triangle FBG\) to \(\triangle ABC\) is \(\frac{2}{3}\).
The area of a triangle \(A = \frac{1}{2}bh\). For similar triangles, if the ratio of side - lengths is \(k\), the ratio of areas is \(k^{2}\). So, \(A_{DBE}:A_{ABC}=(\frac{1}{3})^{2}=\frac{1}{9}\), \(A_{FBG}:A_{ABC}=(\frac{2}{3})^{2}=\frac{4}{9}\)

Step2: Calculate the area of trapezoid \(DEGF\)

\(A_{DEGF}=A_{FBG}-A_{DBE}\). Since \(A_{DBE}=\frac{1}{9}A_{ABC}\) and \(A_{FBG}=\frac{4}{9}A_{ABC}\), then \(A_{DEGF}=\frac{4}{9}A_{ABC}-\frac{1}{9}A_{ABC}=\frac{3}{9}A_{ABC}\)

Answer:

\(\frac{1}{3}\)