Question

in the figure at the right, df = 28, bc = 7, and db = 9. find the perimeter of \\( \triangle adf \\).
the perimeter of \\( \triangle adf \\) is \\( \square \\).

Explanation:

Step1: Determine the similarity ratio

Since \(BC\) is a mid - segment of \(\triangle ADF\) (by the mid - segment theorem, as \(B\) and \(C\) are mid - points of \(DA\) and \(DF\) respectively), the ratio of similarity between \(\triangle DBC\) and \(\triangle DAF\) is \(\frac{BC}{AF}=\frac{DB}{DA}\). Given \(BC = 7\), and since \(B\) is the mid - point of \(DA\) (\(DB=BA = 9\), so \(DA=DB + BA=18\)), and \(DF = 28\). The ratio of similarity \(k=\frac{BC}{AF}\), and also \(k=\frac{DB}{DA}=\frac{9}{18}=\frac{1}{2}\). Then \(AF = 2BC\), so \(AF=14\).

Step2: Calculate the perimeter of \(\triangle ADF\)

The perimeter of \(\triangle ADF\) is \(P=DA + DF+AF\). We know \(DA = 18\), \(DF = 28\), and \(AF = 14\). Then \(P=18 + 28+14\).

Answer:

\(60\)