Question

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), and C′(15, 0). The origin is the center of dilation. What is the scale factor of the dilation? A. 1.5 B. 2 C. 2.5 D. 3

Answer 1

ANSWER

The scale factor is
$2.5$

EXPLANATION

The given triangle has vertices,

$A(0,0),B(0,4),\:and\:C(6, 0)$.

The vertices of the image triangle is,

$A'(0,0),B'(0,10),\:and\:C'(15, 0)$.

The scale factor is given by

$k = \frac{image \: length}{object \: length}$

So we can use any of the corresponding sides to determine the scale factor,

$k = \frac{|A'B'|}{ |AB|}$

$k = \frac{ |10 - 0| }{ |4 - 0|}$

$k = \frac{ |10| }{ |4 |} = \frac{10}{4} = 2.5$

Or

$k = \frac{|A'C'|}{ |AC|}$

$k = \frac{ |15 - 0| }{ |6 - 0|}$

$k = \frac{ |15| }{ |6|} = \frac{15}{6} = 2.5$


Or

$k=\frac{|B'C'|}{|BC|}$

$k = \frac{\sqrt{(15 - 0)^2+(0-10)^2 }}{\sqrt{(6 - 0)^2+(0-4)^2}}$

$k = \frac{ 5\sqrt{13}}{2\sqrt{13}} = \frac{5}{2} = 2.5$

The correct answer is C

Answer 2

Answer:

2.5

Step-by-step explanation:

just got it right