Question

Triangle ABC was dilated and translated to form similar triangle A'B'C'.

Answer 1

we have that

$A(0,2)\ B(2,2)\ C(2,0)\\A'(-4,-1)\ B'(1,-1)\ C'(1,-6)$

we know that

Triangles ABC and A'B'C' are similar

Step $1$

Find the distance AB and distance A'B' with the formula

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

a) distance AB

$dAB=\sqrt{(2-2)^{2}+(2-0)^{2}}$

$dAB=\sqrt{(0)^{2}+(2)^{2}}$

$dAB=2\ units$

b) distance A'B'

$dA'B'=\sqrt{(-1+1)^{2}+(1+4)^{2}}$

$dA'B'=\sqrt{(0)^{2}+(5)^{2}}$

$dA'B'=\sqrt{25}\ units$

$dA'B'=5\ units$

Step $2$

Find the scale factor

$scale\ factor= \frac{measure\ A'B'}{measure\ AB} \\ \\ scale\ factor=\frac{5}{2} \\ \\ scale\ factor=2.5$

therefore

the answer is

the scale factor of the dilation is equal to $2.5$

Answer 2

Answer:

Scale factor of dilation is:

2.5

Step-by-step explanation:

" The scale factor, r, determines how much bigger or smaller the dilation image will be compared to the preimage "

we know that

Triangles ABC and A'B'C' are similar as ΔA'B'C' is formed by dilation of ΔABC with some scale factor.

The coordinates of:

A are (0,2)

B are (2,2)

C are (0,2)

A' are (-4,-1)

B' are (1,-1)

and C' are (1,-6)

Find the length AB and length A'B' with the distance formula

The length of a lie segment is same as a distance between the end points of the line segment.

AB has end points A and B.

The distance between two points A(a,b) and B(c,d) is calculated as:

$\sqrt{(c-a)^2+(d-b)^2}$

a) length AB

i.e. distance between (0,2) and (2,2) is:

$\sqrt{(2-0)^2+(2-2)^2} =\sqrt{4}=2 units$

b) length A'B'

i.e. distance between A'(-4,-1) and B'(1,-1) is:

$\sqrt{(1-(-4))^2+(-1-(-1))^2}=\sqrt{(1+4)^2+0}=\sqrt{5^2}=5 units$

Now we Find the scale factor of the dilation?

Scale factor is:

$\dfrac{Length A'B'}{Length AB}\\\\\\=\dfrac{5}{2}\\\\=2.5$

Hence, the scale factor of the dilation is equal to 2.5