Question

1. respond to each question. (lesson 3 - 10)

a. are triangles abc and def similar?
show or explain your reasoning.
b. if possible, find the length of ef. if not,
explain why the length of ef cannot
be determined.

Explanation:

Step1: Check similarity of triangles

For two triangles to be similar, their corresponding angles must be equal.
In \(\triangle ABC\), \(\angle A = 60^{\circ}\), \(\angle C=20^{\circ}\), so \(\angle B=180^{\circ}-(60^{\circ} + 20^{\circ})=100^{\circ}\)
In \(\triangle DEF\), \(\angle D = 60^{\circ}\), \(\angle E = 100^{\circ}\), so \(\angle F=180^{\circ}-(60^{\circ}+100^{\circ}) = 20^{\circ}\)
Since \(\angle A=\angle D = 60^{\circ}\), \(\angle B=\angle E=100^{\circ}\), \(\angle C=\angle F = 20^{\circ}\), by the AA (Angle - Angle) similarity criterion, \(\triangle ABC\sim\triangle DEF\)

Step2: Find the ratio of similarity

The ratio of similarity \(k\) is determined by the ratio of corresponding sides. Let's take \(\frac{AB}{DE}\). \(AB = 9\), \(DE = 12\), so \(k=\frac{9}{12}=\frac{3}{4}\)

Step3: Calculate the length of \(EF\)

We know that \(\frac{BC}{EF}=k\). Given \(BC = 6\), then \(\frac{6}{EF}=\frac{3}{4}\)
Cross - multiply: \(3\times EF=6\times4\)
\(3EF = 24\)
Divide both sides by 3: \(EF=\frac{24}{3}=8\)

Answer:

a. Yes, triangles \(ABC\) and \(DEF\) are similar. Because \(\angle A=\angle D = 60^{\circ}\), \(\angle B=\angle E = 100^{\circ}\), \(\angle C=\angle F=20^{\circ}\), so by AA similarity criterion, \(\triangle ABC\sim\triangle DEF\)
b. The length of \(EF\) is \(8\)