Question

practice question 1
segment ab is parallel to segment ab.
what is the length of segment ab?
segment ab is parallel to segment ab.
what is the length of segment bb?
(image of a triangle with points a, b, c, a, b; lengths: aa = 3, ac = 6, ab = 5, bc = 7)

Explanation:

Step1: Use the similarity of triangles

Since \(A'B'\parallel AB\), \(\triangle CA'B'\sim\triangle CAB\) (by the AA similarity criterion as \(\angle CA'B'=\angle CAB\) and \(\angle CB'A'=\angle CBA\) because of parallel lines).
The ratio of corresponding sides of similar triangles is equal. So, \(\frac{A'B'}{AB}=\frac{CA'}{CA}\).
We know that \(CA' = 6\), \(A'A=3\), so \(CA=CA'+A'A=6 + 3=9\), and \(A'B' = 5\).

Step2: Solve for \(AB\)

Substitute the values into \(\frac{A'B'}{AB}=\frac{CA'}{CA}\), we get \(\frac{5}{AB}=\frac{6}{9}\).
Cross - multiply: \(6AB=5\times9\).
Then \(AB=\frac{5\times9}{6}=\frac{45}{6}=\frac{15}{2}=7.5\).

For the length of \(B'B\):

Step1: Use the ratio of similar triangles for another pair of sides

Since \(\triangle CA'B'\sim\triangle CAB\), \(\frac{CB'}{CB}=\frac{CA'}{CA}\). Let \(B'B=x\), then \(CB=CB'+B'B=7 + x\).
We know that \(\frac{7}{7 + x}=\frac{6}{9}\).

Step2: Solve for \(x\)

Cross - multiply: \(6(7 + x)=7\times9\).
Expand: \(42+6x = 63\).
Subtract 42 from both sides: \(6x=63 - 42=21\).
Divide both sides by 6: \(x=\frac{21}{6}=3.5\).

Answer:

The length of segment \(AB\) is \(7.5\).
The length of segment \(B'B\) is \(3.5\).