Question

width (shorter side) length (longer side) perimeter area first rectangle 2 (in) 5 (in) 14 (in) 10 (in²)? second rectangle 8 20 56 160 2. suppose a photographer for the school newspaper took this picture. the editors want to resize the photo to fit in a specific space on a page. can the original photo be changed to a similar rectangle with the given measurements (in inches)? if so, what is the scale factor? 8 in by 12 in 2 in by 3 in 6 in by 9 in 12 in by 15 in

Explanation:

Step1: Calculate the ratio of width to length for each rectangle

For the original photo (assuming width \(w = 4\) and length \(l=6\)), the ratio of width to length is \(\frac{w}{l}=\frac{4}{6}=\frac{2}{3}\).
For the rectangle \(8\) in by \(12\) in: \(\frac{8}{12}=\frac{2}{3}\).
For the rectangle \(2\) in by \(3\) in: \(\frac{2}{3}\).
For the rectangle \(6\) in by \(9\) in: \(\frac{6}{9}=\frac{2}{3}\).
For the rectangle \(12\) in by \(15\) in: \(\frac{12}{15}=\frac{4}{5}\).

Step2: Determine the scale factor

For the rectangle \(8\) in by \(12\) in: Let the scale factor be \(k\). If we consider the original width \(w = 4\) and new width \(w_{new}=8\), then \(k=\frac{8}{4} = 2\) (or using length: original \(l = 6\), new \(l_{new}=12\), \(k=\frac{12}{6}=2\)).
For the rectangle \(2\) in by \(3\) in: Using width \(k=\frac{2}{4}=\frac{1}{2}\) (using length \(k=\frac{3}{6}=\frac{1}{2}\)).
For the rectangle \(6\) in by \(9\) in: Using width \(k=\frac{6}{4}=\frac{3}{2}\) (using length \(k=\frac{9}{6}=\frac{3}{2}\)).

Answer:

The rectangles \(8\) in by \(12\) in, \(2\) in by \(3\) in, and \(6\) in by \(9\) in are similar to the original photo (with width \(4\) in and length \(6\) in). The scale factors are \(2\), \(\frac{1}{2}\), and \(\frac{3}{2}\) respectively. The rectangle \(12\) in by \(15\) in is not similar to the original photo.