Answer:
The vertices are:
A' = (-3, -3)
B' = (3, -3)
C' = (3, 3)
D' = (-3, 3)
The ratio of area of larger square to smaller square is 9:1
Step-by-step explanation:
Given:
A 2 x 2 square is centered at the origin.
So, the center of the square is (0, 0)
Since it is 2 x 2 square, the side of the square is 2 units.
So, the vertices of the 2 x 2 square are A (-1, -1), B(1, -1), C(1. 1), D(-1, 1)
The above square is dilated by a factor of 3.
Let's name the dilated square A'B'C'D'
To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.
A(-1, -1) = 3(-1, -1) = A'(-3, -3)
B(1, -1) = 3(1, -1) = B'(3, -3)
C(1, 1) = 3(1, 1) = C'(3, 3)
D(-1, 1) = 3(-1, 1) = D'(-3, 3)
To find the area of the small square
the side of the small square is 2 units
so the are of the small square is [tex]2^2[/tex] = 4 square units
To find the area of the larger square
lets find the side AB of the square using distance formula
=>[tex]\sqrt{(x_2 -x_1)^2 +(y_2-y_1)^2}[/tex]
=>[tex]\sqrt{(3 - (-3))^2 +(-3 - (-3))^2}[/tex]
=>[tex]\sqrt{(3 +3)^2 +(-3 +3)^2}[/tex]
=>[tex]\sqrt{(6)^2 +(0)^2}[/tex]
=>[tex]\sqrt{36}[/tex]
=>6
AB =6 units
In a square all the sides will be equal
Now the area of the larger square will be
[tex]6^2[/tex]
36 square units
The ratio of larger square to smaller square is
=>36 : 4
=>9 : 1
