Question

1. △rst∼△pqw. find the unknown side lengths. a = 5 16. t △lm is 40 cm². find the area of △abc.

Explanation:

Step1: Set up the proportion for similar - triangles

Since \(\triangle RST\sim\triangle PQW\), the ratios of corresponding sides are equal. Let's set up the proportion for the sides. We know that \(\frac{RS}{PQ}=\frac{RT}{PW}=\frac{ST}{QW}\). We have \(RS = 14\mathrm{cm}\), \(RT = 12\mathrm{cm}\), \(PQ = 9\mathrm{cm}\), and \(QW=7.5\mathrm{cm}\). Let's find \(r\) (corresponding to \(ST\)) using the proportion \(\frac{RS}{PQ}=\frac{ST}{QW}\), i.e., \(\frac{14}{9}=\frac{r}{7.5}\).

Step2: Solve the proportion for \(r\)

Cross - multiply the proportion \(\frac{14}{9}=\frac{r}{7.5}\) to get \(9r=14\times7.5\). Then \(9r = 105\), and \(r=\frac{105}{9}=\frac{35}{3}\approx11.67\mathrm{cm}\). Let's find \(w\) (corresponding to \(RT\)) using the proportion \(\frac{RS}{PQ}=\frac{RT}{PW}\), i.e., \(\frac{14}{9}=\frac{12}{w}\). Cross - multiply to get \(14w=12\times9\), so \(14w = 108\), and \(w=\frac{108}{14}=\frac{54}{7}\approx7.71\mathrm{cm}\).

For the second part about the areas of similar triangles:
If two similar triangles have a ratio of corresponding - side lengths \(k\), the ratio of their areas is \(k^{2}\). Let the ratio of the side lengths of \(\triangle ABC\) and \(\triangle KLM\) be \(k\). We know that if the side lengths of \(\triangle ABC\) and \(\triangle KLM\) are in the ratio \(\frac{AB}{KL}=\frac{8}{10}=\frac{4}{5}\). The ratio of the areas of \(\triangle ABC\) and \(\triangle KLM\) is \(k^{2}=(\frac{4}{5})^{2}=\frac{16}{25}\). Let the area of \(\triangle ABC\) be \(A_{1}\) and the area of \(\triangle KLM\) be \(A_{2} = 40\mathrm{cm}^{2}\). Then \(\frac{A_{1}}{A_{2}}=\frac{16}{25}\), so \(A_{1}=\frac{16}{25}\times40=\frac{16\times40}{25}=\frac{640}{25}=25.6\mathrm{cm}^{2}\).

Answer:

The length \(r=\frac{35}{3}\mathrm{cm}\approx11.67\mathrm{cm}\), \(w = \frac{54}{7}\mathrm{cm}\approx7.71\mathrm{cm}\), and the area of \(\triangle ABC\) is \(25.6\mathrm{cm}^{2}\).