The perimeter of a rectangle can be expressed as P = 2l + 2w, where "P" is the perimeter, "l" is the length, and "h" is the height. The area of a rectangle can be expressed as A = lw, where "A" is the area, "l" is the length, and "w" is the width.
According the information given in the problem, the perimeter of the rectangle is 34 cm, and the area of the rectangle is 70 cm^2
Let's plug this into our equations:
P = 2l + 2w
34 = 2l + 2w
A = lw
70 = lw
We can use substitution to replace one variable with an expression with respect to the other. Let's solve the first equation for "l", and then plug that expression in place of "l" in the second equation.
34 = 2l + 2w
34 - 2w = 2l
17 - w = l
70 = lw
70 = (17-w) (w)
70 = 17w - w^2
w^2 - 17w + 70 = 0
Factor and solve
0 = (w - 10) (w - 7)
w - 10 = 0
w = 10
w - 7 = 0
w = 7
Notice how we have two positive values for the width. Let's plug them into one of the prior equations (we can pick any) that contain both a "w" and an "l" to see if it can cancel one of the two possibilities.
70 = lw
70 = 10 l
l = 7
70 = lw
70 = 7w
w = 10
When the width is 10, the length is 7. When the length is 7, the length is 10.
Because the problem indicates that the width is the shorter side and the length is the longer side, then the width would be 7 and the length would be 10.
