Question

25 pts to who ever can help with this question:

A. Find the perimeter of the square
B. Find the perimeter of the rectangle
C. Find the sum of the perimeters of the two figures if s=5
D. The perimeter of the rectangle is greater than the perimeter of the square. Find the difference between the perimeters of the two figures if s=7

Answer 1

A. 5s+2

B. 8s+8

C. 75

D. 27

Explanation:

Is difficult to read the exact expressions. If a mistake is present then fix the number and use the same method.

A. The perimeter of the square is all 4 equal sides added together. The length of a side is
`(5s+2)/(4)`. Since there are four sides then:

`P= (5s+2)/(4)+(5s+2)/(4)+(5s+2)/(4)+(5s+2)/(4) = 5s+2`

Adding 1/4+1/4+1/4+1/4=1. This applies here leaving 5s+2 as the answer.

B. The perimeter of the rectangle is all four sides added together. A rectangle has two pairs of equal side so P = 2l+2w.

`P = 2(s+9) + 2(3s-5)\\P = 2s+18 +6s-10\\P=8s+8`

C. If s=5 then the sum of the two perimeters is:

5s+2+8s+8

5(5)+2+8(5)+8

25+2+40+8

75

D. Subtract the two perimeters by computing s=7:

8s+8 -(5s+2)

8s+8-5s-2

8(7)+8-5(7)-2

56+8-35-2

64-35-2

29-2

27