A rectangle has perimeter 62 cm, and its area is 234 sq-cm. Find the dimensions of this rectangle. (For purposes of this problem only, the width is shorter than the length.)

Question

what is the length

what is the width

Answer 1

62=2(L+W)

31= L+W

L=31-W

234 =LW= (31-W)W = 31W-W^2

W^2 -31W +234 = 0

W= 31/2 + or - 5/2 = 16/2 or 36/2 = 13 or 18

L= 31-13 or 31-18 = 18 or 13

Since W<L,

L= 18 cm and W=13 cm

18(13) = 234

Answer 2

62=2(L+W)

31= L+W

L=31-W

234 =LW= (31-W)W = 31W-W^2

W^2 -31W +234 = 0

W= 31/2 + or - 5/2 = 16/2 or 36/2 = 13 or 18

L= 31-13 or 31-18 = 18 or 13

Since W<L,

L= 18 cm and W=13 cm

18(13) = 234