Answer:
[tex]\overline {R'S'} = 3 {\overline {RS} }[/tex]
Step-by-step explanation:
The scale factor of dilation of triangle ΔOPQ, S.F. = 3
The center of dilation = Point Q
Therefore;
The length of the segment, [tex]{\overline {S'Q'} }[/tex] = 3 × The length of the segment, [tex]{\overline {SQ} }[/tex]
Similarly;
The length of the segment, [tex]{\overline {R'Q'} }[/tex] = 3 × The length of the segment, [tex]{\overline {RQ} }[/tex]
By Pythagoras theorem, we have;
[tex]{\overline {R'Q'} }^2 = {\overline {R'S'} }^2 + {\overline {S'Q'} }^2[/tex]
Therefore;
[tex]{\overline {R'S'} }^2 = {\overline {R'Q'} }^2 - {\overline {S'Q'} }^2 = \left ({3 \times \overline {RQ} } \right) ^2 - \left (3 \times {\overline {SQ} } \right) ^2 = 9 \times \left (\overline {RQ} } ^2 - {\overline {SQ} } ^2 \right) = 9 \times {\overline {RS} }^2[/tex]
[tex]\therefore \sqrt{ {\overline {R'S'} }^2} = \sqrt{ 9 \times {\overline {RS} }^2}= 3 \times {\overline {RS} }[/tex]
[tex]\overline {R'S'} = 3 \times {\overline {RS} }[/tex].
