Question

A Rectangular has area 12a^2 and perimeter 14a what are the dimensions of this Rectangular?​

Answer 1

9514 1404 393

Answer:

  3a by 4a

Step-by-step explanation:

For dimensions L and W, the area and perimeter are ...

  A = LW = 12a^2

  P = 2(L+W) = 14a

Using the second equation, we can find L:

  L +W = 7a . . . . . divide by 2

  L = 7a -W

Substituting into the area formula gives the quadratic ...

  (7a -W)(W) = 12a^2

  W^2 -7aW +12a^2 = 0 . . . . arrange in standard form

  (W -3a)(W -4a) = 0 . . . . . . . factor (find factors of 12 that total 7)

Then we have the two solutions ...

  W = 3a, L = 4a

  W = 4a, L = 3a

The rectangle dimensions are 3a by 4a.