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 \\( \frac { 1 } { 3 } \\)
(type an integer or a simplified fraction.)
b. the ratio of the area of triangle dbe to the area of trapezoid fgca 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 \(BD = DF=FA=a\). Let the height of \(\triangle DBE\) be \(h_1\), the height of \(\triangle BFG\) be \(h_2\) and the height of \(\triangle ABC\) be \(h_3\).
For similar triangles \(\triangle DBE\sim\triangle BFG\sim\triangle ABC\). If the ratio of side - lengths of similar triangles is \(k\), the ratio of their areas is \(k^{2}\).
For \(\triangle DBE\) and \(\triangle ABC\), the ratio of side - lengths \(k_1=\frac{BD}{BA}=\frac{a}{3a}=\frac{1}{3}\), so \(A_{\triangle DBE}=\frac{1}{9}A_{\triangle ABC}\) (using the formula \(A = \frac{1}{2}bh\) and for similar triangles \(A_1/A_2=(b_1/b_2)^2\) where \(b\) is the base and the ratio of heights is the same as the ratio of bases).
For \(\triangle BFG\), the ratio of side - lengths \(k_2=\frac{BF}{BA}=\frac{2a}{3a}=\frac{2}{3}\), so \(A_{\triangle BFG}=\frac{4}{9}A_{\triangle ABC}\)

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

We know that \(A_{FGCA}=A_{\triangle ABC}-A_{\triangle BFG}\)
Substitute \(A_{\triangle BFG}=\frac{4}{9}A_{\triangle ABC}\), then \(A_{FGCA}=A_{\triangle ABC}-\frac{4}{9}A_{\triangle ABC}=\frac{5}{9}A_{\triangle ABC}\)

Step3: Find the ratio of \(A_{\triangle DBE}\) to \(A_{FGCA}\)

We know that \(A_{\triangle DBE}=\frac{1}{9}A_{\triangle ABC}\)
The ratio \(\frac{A_{\triangle DBE}}{A_{FGCA}}=\frac{\frac{1}{9}A_{\triangle ABC}}{\frac{5}{9}A_{\triangle ABC}}\)

Answer:

\(\frac{1}{5}\)