Segment A'B' is parallel to segment AB.
So, The length of segment AB is 7.5 units
The length of segment B'B is 3.5 units
Given :
Segment A'B' is parallel to segment AB. Few sides of the triangle is given.
Apply basic proportionality theorem
When A'B' is parallel to segment AB then sides are proportional
[tex]\frac{CB'}{B'B} =\frac{CA'}{A'A}[/tex]
Substitute the values
[tex]\frac{7}{B'B} =\frac{6}{3} \\cross \; multiply\\7(3)=B'B(6)\\21=B'B(6)\\Divide \; by \; 6\\B'B= \frac{21}{6} \\B'B=3.5[/tex]
Now we find AB by making a proportion
[tex]\frac{CA'}{CA} =\frac{BA'}{BA} \\\frac{6}{9} =\frac{5}{AB} \\Cross \; multiply\\6(AB)=5 \cdot 9\\6AB=45\\Divide \; by \; 6\\AB=7.5[/tex]
The length of segment AB is 7.5 units
The length of segment B'B is 3.5 units
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