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

1 Answer

14 votes

9514 1404 393

Answer:

3a by 4a

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.

User Kiran Reddy
by
7.8k points

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