Question

given △abc with parallel lines dividing 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
a. the ratio of the area of triangle dbe to the area of trapezoid degf is
(type an integer or a simplified fraction.)

Explanation:

Step1: Use the similarity of triangles

Since the parallel lines divide \(AB\) into three congruent segments, let the length of \(BD = a\), \(DF=a\), \(FA = a\). Let the height of \(\triangle DBE\) be \(h_1\), the height of \(\triangle BFG\) be \(h_2\) (\(h_2=2h_1\)), and the height of \(\triangle ABC\) be \(h_3\) (\(h_3 = 3h_1\)).
By the similarity of triangles (\(\triangle DBE\sim\triangle BFG\sim\triangle ABC\)), if the ratio of sides of \(\triangle DBE\) and \(\triangle BFG\) is \(k_1=\frac{BD}{BF}=\frac{1}{2}\), and the ratio of sides of \(\triangle DBE\) and \(\triangle ABC\) is \(k_2=\frac{BD}{BA}=\frac{1}{3}\).
The area of a triangle \(A=\frac{1}{2}bh\). For similar triangles, if the ratio of corresponding sides is \(k\), the ratio of their areas is \(k^{2}\). So \(A_{\triangle DBE}=\frac{1}{2}x_1h_1\), \(A_{\triangle BFG}=\frac{1}{2}x_2h_2\) with \(x_2 = 2x_1\) and \(h_2=2h_1\), \(A_{\triangle BFG}=4A_{\triangle DBE}\), \(A_{\triangle ABC}=9A_{\triangle DBE}\).

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

The area of trapezoid \(A_{DEGF}=A_{\triangle BFG}-A_{\triangle DBE}\). Since \(A_{\triangle BFG} = 4A_{\triangle DBE}\), then \(A_{DEGF}=4A_{\triangle DBE}-A_{\triangle DBE}=3A_{\triangle DBE}\)

Answer:

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