Arectangle has an area of 35.70 cm2. when both the length and width of the rectangle are increased by 1.60 cm, the area of the rectangle becomes 58.58 cm2. calculate the length of the shorter of the two sides of the initial rectangle.

  4.2 cm

Step-by-step explanation:

The given information can be expressed in the two relations ...

  xy = 35.70 . . . . . . the area of the original x by y rectangle is 35.70 cm^2

  (x+1.6)(y+1.6) = 58.58 . . . . . the area of the larger rectangle is 58.58 cm^2

Simplifying the latter equation, we get ...

  xy +1.6y +1.6x +2.56 = 58.58

  Substituting for xy using the first equation, this is ...

  35.70 +2.56 + 1.6(x +y) = 58.58

  1.6(x +y) = 20.32 . . . . . . . subtract 38.26

  x +y = 12.7 . . . . . . . . . . . . divide by 1.6

Now, we can use this to say ...

   y = 12.7 -x

and we can substitute this into the first equation to get

  x(12.7 -x) = 35.70

In standard quadratic form, this is ...

  x^2 -12.7x +35.70 = 0

Then the smaller value of x is found from the quadratic formula as ...

  x = (-(-12.7) -√(12.7^2 -4·1·35.70))/2 = (12.7-√18.49)/2

  x = 4.2

The short side of the initial rectangle is 4.2 cm.

can someone answer this question plz

step-by-step explanation: