Question

Tanisha has 51 m of fencing to build a

three-sided fence around a rectangular
plot of land that sits on a riverbank. (The
fourth side of the enclosure would be the
river.) The area of the land is 304 square
meters. List each set of possible
dimensions (length and width) of the
field.
Possible dimensions #1:
meters by
meters.
Possible dimensions #2:
meters by
meters.

Answer 1

Two sets of dimensions are possible:

L(m) W(m) Area(m^2) Fence (m) Perimeter

Set 1 19 16 304 51

Set 2 32 9.5 304 51

Explanation:

Let's set L and W for the Length and Width of the perimeter. We know that:

Area = L*W

Area = 304 m^2

Perimeter = 51 m

Let's say that the side bounded by the river is the length side. No fence is required for that portion, so the fencing for the perimeter is given by:

Perimeter, P = L + 2W

P = 51 m

We can write: L+2W=51 m

Rearrange to isolate one of the two variables. Lets pick L:

L+2W=51

L=51-2W

L = (51-2W)

Now use this in the area equation:

304 m^2 = L*W

304 m^2 = (51-2W)*W

304 m^2 = (51-2W)*W

304 m^2 = 51W-2W^2

Arrange this in the standard form of a quadratice equation and solve:

-2W^2+51W-304 m^2 = 0

W = 16 and 9.5 meters

W = 16 m

Since L=51-2W

L=51-2*(16)

L = 51-32

L = 19 m

Area = L*W

L*W = 304 m^2

(19m)W = 304 m^2

W = 16 m

Check:

Does a length (L) of 19 m and a width (W) of 16 m give an area of 304 m^2 and a fencing perimeter of 51 m if one of the L sides is the river?

Area = (19 m)*(16 m) = 304 m^2 YES

Per. = 2*(16)+ 19 = 51 m YES

W = 9.5 m

Ding the same calculation for a width of 9.5 m also satisfies all the equations, as per above.

Both values of W are valid, so the twos sets of possible dimensions are:

L(m) W(m) Area(m^2) Fence (m) Perimeter

Set 1 19 16 304 51

Set 2 32 9.5 304 51