Question

given △abc with parallel lines dividing \\( \overline { a b } \\) 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.)
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 \\( \frac { 1 } { 3 } \\)
(type an integer or a simplified fraction)
e. the ratio of the area of trapezoid fgca 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 parallel lines divide \(AB\) into three congruent segments, let the length of \(BD = DF=FA = x\). Let the height of \(\triangle DBE\) be \(h_1\), the height of \(\triangle BFG\) (with base \(FG\)) be \(h_2\), and the height of \(\triangle ABC\) be \(h\). By the similarity of triangles (\(\triangle DBE\sim\triangle BFG\sim\triangle ABC\)), if the ratio of sides \(BD:BA = 1:3\), then the ratio of heights of \(\triangle DBE\) and \(\triangle ABC\) is \(1:3\) (say \(h_1=\frac{1}{3}h\)), and the ratio of heights of \(\triangle BFG\) and \(\triangle ABC\) is \(2:3\) (say \(h_2 = \frac{2}{3}h\)).
The area of \(\triangle DBE=\frac{1}{2}\times DE\times h_1\), the area of \(\triangle BFG=\frac{1}{2}\times FG\times h_2\), and the area of \(\triangle ABC=\frac{1}{2}\times AC\times h\). Also, by the property of similar - triangles \(DE:FG:AC = 1:2:3\). Let \(DE = k\), then \(FG = 2k\) and \(AC=3k\).
The area of trapezoid \(DEGF=S_{\triangle BFG}-S_{\triangle DBE}\), and the area of trapezoid \(FGCA=S_{\triangle ABC}-S_{\triangle BFG}\).
\(S_{\triangle DBE}=\frac{1}{2}\times k\times\frac{1}{3}h=\frac{1}{6}kh\), \(S_{\triangle BFG}=\frac{1}{2}\times2k\times\frac{2}{3}h=\frac{2}{3}kh\), \(S_{\triangle ABC}=\frac{1}{2}\times3k\times h=\frac{3}{2}kh\)
\(S_{DEGF}=\frac{2}{3}kh-\frac{1}{6}kh=\frac{4kh - kh}{6}=\frac{1}{2}kh\), \(S_{FGCA}=\frac{3}{2}kh-\frac{2}{3}kh=\frac{9kh - 4kh}{6}=\frac{5}{6}kh\)

Step2: Calculate the ratio of the area of trapezoid \(FGCA\) to the area of \(\triangle ABC\)

The ratio \(r=\frac{S_{FGCA}}{S_{\triangle ABC}}\)
\(S_{FGCA}=\frac{5}{6}kh\) and \(S_{\triangle ABC}=\frac{3}{2}kh=\frac{9}{6}kh\)
\(r = \frac{\frac{5}{6}kh}{\frac{9}{6}kh}=\frac{5}{9}\)

Answer:

\(\frac{5}{9}\)