# The area of a rectangular field is equal to 300 square meters. Its perimeter is equal to 70 meters. Find the length and witdh of this rectangle

*print*Print*list*Cite

Let us assume that the width of the field is "W" and the length of the field is "L"

We are given that the area of the rectangular field is 300 m^2

But we know that the area = L*W

==> L*W = 300 ..................(1)

Also, we are given that the perimeter of the field is 70 m.

But we know that the perimeter = 2L+2W

==> 2L + 2W = 70

We will divide by 2.

==> L + W = 35......................(2)

Now we will use the substitution method to solve for L and W.

==> L = 35 - w

==> L*W = 300

==> (35-W)*W = 300

==> 35w - w^2 = 300

==> w^2 - 35w +300 = 0

Now we will factor.

==> (w - 15)(w-20) = 0

==> w = 15 == L = 20

==> w = 20 ==> L = 15

But we know that L > w

==> L = 20 and W = 15

**Then, the length of the field = 20 meter, and the width of the field = 15 meter.**

We have that the area of a rectangular field is equal to 300 square meters and its perimeter is equal to 70 meters.

Let the length and the width of the field be L and W

Perimeter = 2(W + L) = 70

=> W + L = 35

=> W = 35 - L

Area = W * L = 300

=> (35 - L)* L = 300

=> 35 L - L^2 = 300

=> L^2 - 35 L + 300 = 0

=> L^2 - 15 L - 20 L + 300 = 0

=> L( L-15) - 20( L- 15) = 0

=> (L - 20) ( L -15) = 0

So L can be 20 and 15

W = 15 and 20

**As length is greater than the width, we have the length equal to 20 and the width equal to 15.**

Area A = l*w, where l is length and w = width of the rectangle.

So l*w = 300 m^2 (given).

Perimeter P = 2(l+w) is the formula.

=> 2(l+w) = 70 m.

Therefore l+w = 70/2 = 35 m.

Therefore lw = 300 and l+w = 35.

So l and w are the roots of x^2-(l+w)x+lw = 0

x^2-35x+300 = 0.

(x-20)(x-15) = 0.

x-20 = 0, or x-15 = 0.

So x1 = l = 20 and x2 = w = 15.

**So length and breadth of the rectangular field are 20 m and 15 respectively.**