# A rectangle field with area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field? Our task is to find out the length and width of the field.

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

Area of rectangle is given by  A=L*B .

Perimeter of rectangle is given by  P=2(L+B).

Using information given by...

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Our task is to find out the length and width of the field.

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

Area of rectangle is given by  A=L*B .

Perimeter of rectangle is given by  P=2(L+B).

Using information given by the question we have the following 2 equations:

L*B = 300  ......(1)

2(L+B) = 80 ......(2)

Since we have two unknowns and two equations, we can solve the simultaneous equation.

From (1):

L=300/B ........(3)

Substitute (3) into (2):

2(300/B+B)=80

Divide by 2 on both sides:

300/B + B = 40

Multiply by B on both sides:

300 + B^2 = 40B

Rearranging,

B^2 -40B + 300 = 0

Factorizing,

(B-30)(B-10) = 0

B = 30 or 10

Checking:

Substitute the values of B into (1), we realise that

If B=30, L=10

If B=10, L=30

Hence the length and breadth are 30cm and 10 cm or vice versa

Approved by eNotes Editorial Team Given that the area of the rectangular field A is :

A = 300 m^2

Also given that the perimeter P is"

p = 80 m

Let  us assume that the sides are x and y

==> A = x*y

==> P = x+ y

==> x*y = 300 .........(1)

==>2 x+2 y = 80

==>2( x+ y) = 80

==> x+ y= 40

==> x= 40 - y..............(2)

Now substitute in (1):

==> x*y = 300

==> (40-y) *y = 300

==> 40y - y^2 = 300

==> y^2 - 40y + 300 = 0

==> ( y- 30) ( y+10)  = 0

==> y= 30 ==> x = 50

Then the length = 30 m  and the width = 30 m

Approved by eNotes Editorial Team