The distance between points S' and S is 2.5 units.
If a line is drawn parallel to any one side of a triangle so that it intersects the other two sides in two distinct points, then the other two sides of the triangle are divided in the same ratio.
Given that,
line segment ST is dilated to create line segment S'T' using the dilation rule DQ.
Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5 units, SS' = x units.
We need to find the value of x, the distance between points S' and S.
Since the line ST is dilated to S'T' with center of dilation Q, so the triangles STQ and S'T'Q must be similar.
We know that the corresponding sides of two similar triangles are proportional.
So, from ΔSTQ and ΔS'T'Q, we get
[tex]\frac{SQ}{S'Q} =\frac{TQ}{T'Q}[/tex]
[tex]\frac{SQ}{SQ+S'S} =\frac{TQ}{TQ+T'Q}[/tex]
[tex]\frac{2}{2+x} =\frac{1.2}{1.2+1.5}[/tex]
[tex]\frac{2}{2+x} =\frac{1.2}{2.7}[/tex]
[tex]\frac{2}{2+x} =\frac{12}{27}[/tex]
[tex]\frac{1}{2+x} =\frac{6}{27}[/tex]
12+6x = 27
6x = 15
x = 2.5
Hence, Thus, the required value of x is 2.5 units.
To learn more about proportionality theorem from the given link:
https://brainly.com/question/2475257
#SPJ4
