Question

∆PQR has vertices at P(2, 4), Q(3, 8) and R(5, 4). A dilation and series of translations map ∆PQR to ∆ABC, whose vertices are A(2, 4), B(5.5, 18), and C(12.5, 4). What is the scale factor of the dilation in the similarity transformation?

A.
2

B.
2.5

C.
4

D.
3.5

Answer 1

Given : ∆PQR has vertices at P(2, 4), Q(3, 8) and R(5, 4). A dilation and series of translations map ∆PQR to ∆ABC, whose vertices are A(2, 4), B(5.5, 18), and C(12.5, 4).

To Find : Scale factor of the dilation in the similarity transformation  

A.     2

B.     2.5

C.     4

D.     3.5

Solution:

P(2, 4), Q(3, 8) and R(5, 4)

PQ = √(3-2)² + (8 - 4)²   = √17

PR = √(5-2)² + (4 - 4)²   = 3

QR = √(5-3)² + (4 - 8)²   = 2√5

A(2, 4), B(5.5, 18), and C(12.5, 4)

AB = √(5.5-2)² + (18 - 4)²   = 3.5√17

AC = √(12.5-2)² + (4 - 4)²   = 10.5

BC = √(12.5-5.5)² + (4 - 18)²   =  7√5

AB/PQ  = AC/PR  = BC/QR   =   3.5

scale factor of the dilation in the similarity transformation = 3.5

Learn More:

Triangle STV was dilated with the origin as the center of dilation to ...

https://brainly.in/question/17620960

Suppose figure A underwent a dilation to produce figure B. Are ...

https://brainly.in/question/26054879

Answer 2

${\underline{\underline{\textsf\color{green}{required ~diagram }}}}$

Step-by-step explanation:

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

•∆PQR has vertices at P(2, 4), Q(3, 8) and R(5, 4).

•A dilation and series of translations map ∆PQR to ∆ABC, whose vertices are A(2, 4), B(5.5, 18), and C(12.5, 4).⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

☆What is the scale factor of the dilation in the similarity transformation??

⠀⠀⠀⠀

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

Here,

$\huge{\triangle }$PQR,⠀

p=(2,4)

Q=(3,8)

R=(5,4)

⠀⠀⠀⠀

$\boxed{\underline{\underbrace{\mathtt\color{gold}{Distance={\sqrt{(y_2 -y_1)^2+(x_2 -x-1)^2}}}}}}$

So,

PQ =$\tt{\sqrt{(3-2)^2+(8-4)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{1+16} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{17} }$ ⠀

PR= $\tt{\sqrt{(5-2)^2+(4-4)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{9+0} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{9}=3 }$ ⠀

QR=$\tt{\sqrt{(5-3)^2+(4-8)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{4+16} }$ ⠀

$\rightsquigarrow$ =$\tt{\sqrt{20} =2\sqrt{5} }$ ⠀

Then,

$\huge{\triangle }$ ABC,⠀

A=(2,4)

B=(5.5,18)

C=(12.5,4)

$\boxed{\underline{\underbrace{\mathtt\color{gold}{Distance={\sqrt{(y_2 -y_1)^2+(x_2 -x-1)^2}}}}}}$

So,

AB=$\tt{\sqrt{(5.5-2)^2+(18-4)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{3.5\sqrt{17} }$ ⠀

AC=$\tt{\sqrt{(12.5-2)^2+(4-4)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{10.5 }$ ⠀

BC=$\tt{\sqrt{(12.5-5.5)^2+(4-18)^2} }$ ⠀

$\rightsquigarrow$ =$\tt{7\sqrt{5} }$ ⠀

HERE,

$\tt{\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}=3.5 }$ ⠀

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

⠀⠀⠀⠀

So,

scale factor of the dilation in the similarity transformation$\implies { 3.5 }$ ⠀