A rectangle field with area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field?

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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...

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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

 

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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