The 4•x by 5•x inner dimensions of the 53 inches outer perimeter picture frame, the 1 inch wide frame and the 4 × 5 inch picture to be enlarged, gives;
a. The outer perimeter equation is presented as follows;
18•x + 8 = 53 inches
b. The scale factor required for the enlargement of the picture is 2.5
The dimensions of the photo = 4 × 5 inch
Width of the frame = 1 inch
Outer perimeter of the frame = 53 inches
The width of the inside of the picture frame = 4•x
The height of the inside of the picture frame = 5•x
a. The equation for the outer perimeter of the picture frame is therefore;
2 × (4•x + 1 + 1) + 2 × (5•x + 1 + 1) = 53
Which gives;
2 × (4•x + 2) + 2 × (5•x + 2) = 53
8•x + 4 + 10•x + 4 = 53
A simplified equation for the outer perimeter of the picture frame is therefore;
b. Solving the above equation, we have;
18•x + 8 = 53
18•x = 53 - 8 = 45
18•x = 45
Therefore;
x = 45 ÷ 18 = 2.5
x = 2.5
The width and height of the picture frame are therefore;
Width = 4•x = 4 × 2.5 = 10
Height = 5•x = 5 × 2.5 = 12.5
Which gives;
Width = 10 inches
Height = 12.5 inches
The enlargement factor, sf, is given by the ratio of a side of the picture frame to a side of the picture as follows;
sf = 12.5/5 = 2.5
Learn more about linear scale factors here:
https://brainly.com/question/19381630
#SPJ1
