Question

1. in the diagram, segment ab is parallel to segment ab. \\(\overline{ab} \parallel \overline{ab}\\)

a. what is the length of segment aa?
b. what is the length of segment bb?
(diagram: triangle with point c, b, b on the base, a and a above; lengths: from c to b is 6, c to a is 12, a to some point is 8, a to b is 20)

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{AB} \parallel \overline{A'B'}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle CA'B' \sim \triangle CAB\). The ratio of corresponding sides should be equal. Let's find the ratio of \(CA'\) to \(CA\) and \(CB'\) to \(CB\).

First, \(CA' = 12\), \(A'A\) is what we need to find, so \(CA = CA' + A'A = 12 + A'A\)? Wait, no, looking at the diagram (assuming \(CB' = 6\), \(B'B\) is unknown, \(A'B' = 8\), \(AB = 20\)? Wait, maybe the sides: \(CB' = 6\), \(CB = 6 + B'B\), \(CA' = 12\), \(CA = 12 + A'A\), \(A'B' = 8\), \(AB = 20\). Since the triangles are similar, \(\frac{CA'}{CA} = \frac{CB'}{CB} = \frac{A'B'}{AB}\).

Let's use \(\frac{A'B'}{AB} = \frac{8}{20} = \frac{2}{5}\). So the ratio of similarity is \(\frac{2}{5}\). Therefore, \(\frac{CA'}{CA} = \frac{2}{5}\). Let \(CA = x\), then \(\frac{12}{x} = \frac{2}{5}\). Solving for \(x\): \(2x = 60\) → \(x = 30\). Then \(A'A = CA - CA' = 30 - 12 = 18\).

Step2: Find \(B'B\)

Using the ratio \(\frac{CB'}{CB} = \frac{2}{5}\). Let \(CB = y\), \(CB' = 6\), so \(\frac{6}{y} = \frac{2}{5}\). Solving: \(2y = 30\) → \(y = 15\). Then \(B'B = CB - CB' = 15 - 6 = 9\). Wait, but let's check with the other ratio. Alternatively, since \(\frac{A'B'}{AB} = \frac{2}{5}\), the ratio of \(CA'\) to \(CA\) is \(2:5\), so the difference is \(3\) parts. \(CA' = 12\) (2 parts), so 1 part is \(6\), so \(A'A = 3 \times 6 = 18\) (correct). Similarly, \(CB' = 6\) (2 parts), 1 part is \(3\), so \(B'B = 3 \times 3 = 9\)? Wait, no, 2 parts is 6, so 1 part is 3, then \(CB = 5\) parts = \(15\), so \(B'B = 15 - 6 = 9\). Yes.

Part a:

Step1: Determine Similar Triangles

Since \(\overline{AB} \parallel \overline{A'B'}\), \(\triangle CA'B' \sim \triangle CAB\) (AA similarity, as \(\angle C\) is common and \(\angle CA'B' = \angle CAB\) (corresponding angles)).

Step2: Find the Similarity Ratio

The ratio of corresponding sides \(A'B'\) and \(AB\) is \(\frac{8}{20} = \frac{2}{5}\). So the similarity ratio is \(\frac{2}{5}\).

Step3: Calculate \(CA\)

Let \(CA = x\). Then \(\frac{CA'}{CA} = \frac{2}{5}\) → \(\frac{12}{x} = \frac{2}{5}\). Solving: \(2x = 60\) → \(x = 30\).

Step4: Find \(A'A\)

\(A'A = CA - CA' = 30 - 12 = 18\).

Part b:

Step1: Use Similarity Ratio for \(CB\)

Let \(CB = y\). The ratio \(\frac{CB'}{CB} = \frac{2}{5}\) → \(\frac{6}{y} = \frac{2}{5}\). Solving: \(2y = 30\) → \(y = 15\).

Step2: Find \(B'B\)

\(B'B = CB - CB' = 15 - 6 = 9\).

Answer:

a. The length of segment \(A'A\) is \(\boldsymbol{18}\).
b. The length of segment \(B'B\) is \(\boldsymbol{9}\).