Question

The dimensions of the wrapper of a regular size Hershey bar are 10 in x 4 in. If we dilate the wrapper by a scale factor of 2, 3, and 4 then how much greater is the area of the image to the area of the original wrapper. Compare the scale factors of the areas.

a) 2 times greater
b) 9 times greater
c) 16 times greater
d) 24 times greater

Answer 1

The area of the image is 48 square inches greater than the area of the original wrapper. Comparing the scale factors of the areas, the area of the image is 3 times greater than the area of the original wrapper.

Step-by-step explanation:

The dimensions of the larger square can be found by multiplying the dimensions of the original square by the scale factor, which is 2. So, the side length of the larger square is 4 inches x 2 = 8 inches. The area of the larger square is found by squaring the side length, which is 8 inches x 8 inches = 64 square inches. The area of the original square is 4 inches x 4 inches = 16 square inches. The difference in area between the two squares is 64 square inches - 16 square inches = 48 square inches. Therefore, the area of the image is 48 square inches greater than the area of the original wrapper. Comparing the scale factors of the areas, we can see that the area of the image is 48 square inches / 16 square inches = 3 times greater than the area of the original wrapper.