a) perimeter of R is 48in. what is the perimeter of R'
b) if the area of R is 122sq.in. what is the area of R'
b) if the area of R is 122sq.in. what is the area of R'
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Since the rectangles are similar, that would mean the length of R would also be proportional to the length of R'. Okay so let's say L is the length of R. 4L would be the length of R'. With that in mind, let's move onto the problems.
a) 192 in.
Remember that Perimeter is 2l + 2w (l= length, w= width). They say that the perimeter of R is 48 in. Which means 2L + 2a = 48. The perimeter of R' would be 2(4L) + 2(4a) or 8L + 8a. In reference to R perimeter, that's 4 times the perimeter so it would be 48 * 4 = 192.
b) 1952 sq.in
Remember that Area is lw. Area of R is La = 122. Area of R' would be 4L*4a or 16La. In relation to R area, it's 16 times the area so 122 * 16 = 1952
a) 192 in.
Remember that Perimeter is 2l + 2w (l= length, w= width). They say that the perimeter of R is 48 in. Which means 2L + 2a = 48. The perimeter of R' would be 2(4L) + 2(4a) or 8L + 8a. In reference to R perimeter, that's 4 times the perimeter so it would be 48 * 4 = 192.
b) 1952 sq.in
Remember that Area is lw. Area of R is La = 122. Area of R' would be 4L*4a or 16La. In relation to R area, it's 16 times the area so 122 * 16 = 1952