Question

item 37
an isosceles triangle, \\( \triangle a b c \\) with \\( a b \cong \overline{c b} \\) is shown. points \\( d, e \\), and \\( h \\) are the midpoints of \\( \overline{a b}, \overline{c b} \\), and \\( \overline{a c} \\), respectively. points \\( f \\) and \\( g \\) of \\( \overline{d h} \\) and \\( \overline{e h} \\), respectively. what is the ratio of the area of \\( \triangle a b c \\) to the area of \\( \triangle f g h \\)?

a. \\( 4: 1 \\)
b. \\( 8: 1 \\)
c. \\( 16: 1 \\)
d. \\( 64: 1 \\)

Explanation:

Step1: Use the mid - segment theorem

By the mid - segment theorem, \(DE\parallel AC\), \(DE=\frac{1}{2}AC\), \(DH\parallel BC\), \(DH = \frac{1}{2}BC\), \(EH\parallel AB\), \(EH=\frac{1}{2}AB\). Since \(AB = CB\) (given \(\triangle ABC\) is isosceles), and \(D\), \(E\), \(H\) are mid - points. Then \(\triangle DEH\sim\triangle ABC\) with a similarity ratio of \(\frac{1}{2}\). The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding side lengths. So, \(A_{\triangle DEH}=\frac{1}{4}A_{\triangle ABC}\).

Step2: Consider the smaller triangle \(\triangle FGH\)

Since \(F\) is the mid - point of \(DH\) and \(G\) is the mid - point of \(EH\), then \(\triangle FGH\sim\triangle DEH\) with a similarity ratio of \(\frac{1}{2}\). Using the area - similarity ratio formula (\(A = k^{2}\times A_{0}\), where \(k\) is the similarity ratio and \(A_{0}\) is the area of the larger triangle), we have \(A_{\triangle FGH}=\frac{1}{4}A_{\triangle DEH}\).

Step3: Find the relationship between \(A_{\triangle ABC}\) and \(A_{\triangle FGH}\)

Substitute \(A_{\triangle DEH}=\frac{1}{4}A_{\triangle ABC}\) into \(A_{\triangle FGH}=\frac{1}{4}A_{\triangle DEH}\). Then \(A_{\triangle FGH}=\frac{1}{4}\times\frac{1}{4}A_{\triangle ABC}=\frac{1}{16}A_{\triangle ABC}\).

Answer:

C. \(16:1\)