Question:
The width of the enlarged rectangle is equal to the product of 9 in and $6106_w9_h37.png$, or 13.5 in. The length of the enlarged rectangle is equal to the product of 15 in and $6106_w9_h37.png$, or 22.5 in. Perimeter= 2w + 2l . Thus, the perimeter is equal to $5484_w243_h18.png$