Answer:
A'B' = 36.28 units
Step-by-step explanation:
Given the scale factor is 4.5, we'll multiply the coordinates of each point by 4.5 to get the coordinates of the corresponding points in the new triangle.
The coordinates of [tex]\sf A' [/tex] will be:
[tex]\sf A'(x', y') = (1 \times 4.5, 2 \times 4.5) [/tex]
[tex]\sf A'(x', y') = (4.5, 9) [/tex]
The coordinates of [tex]\sf B' [/tex] will be:
[tex]\sf B'(x', y') = (5 \times 4.5, 9 \times 4.5) [/tex]
[tex]\sf B'(x', y') = (22.5, 40.5) [/tex]
The coordinates of [tex]\sf C' [/tex] will be:
[tex]\sf C'(x', y') = (7 \times 4.5, 3 \times 4.5) [/tex]
[tex]\sf C'(x', y') = (31.5, 13.5) [/tex]
Now, we can use the distance formula to find the length of the segment [tex]\sf A'B' [/tex].
The distance formula is given by:
[tex] \Large\boxed{\boxed{\sf d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}} [/tex]
Where
- [tex]\sf (x_1, y_1) [/tex] and [tex]\sf (x_2, y_2) [/tex] are the coordinates of the two points.
For [tex]\sf A'B' [/tex], the coordinates are [tex]\sf A'(4.5, 9) [/tex] and [tex]\sf B'(22.5, 40.5) [/tex].
Substitute the value and simplify:
[tex]\sf d = \sqrt{(22.5 - 4.5)^2 + (40.5 - 9)^2} [/tex]
[tex]\sf d = \sqrt{(18)^2 + (31.5)^2} [/tex]
[tex]\sf d = \sqrt{324 + 992.25} [/tex]
[tex]\sf d = \sqrt{1316.25} [/tex]
[tex]\sf d \approx 36.28 \textsf{(in nearest hundredth)}[/tex]
Therefore, the length of the segment [tex]\sf A'B' [/tex] is approximately 36.28 units.
