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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

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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

bonniegknight | High School Teacher | (Level 1) eNoter

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Let L represent length, let W represent Width.

Perimeter is equal to 2L+2W for any rectangle, or P=2L+2W

Area is equal to L*W for any rectangle, or A=L*W

Here, the perimeter is 80 meters and the area is 300m^2.  How can we use this information to find out answers?  Let's think and figure it out.

L*W=300m^2

2L+2W=80m >>>>>>>> 2(L+W)=80>>>>>>> (L+W)=80/2 =40

L+W=40 then L=40-W  OR W=40-L  Choose one or the other to use in the equation for area.

A=L*W>>> 300m^2 = L*(40-L)

300m^2=40L-L^2 >>>>>>rewrite>>> L^2-40L+300=0 >>>now solve for L by factoring this equation to get (L-30)(L-10)=0 Therefore L=30 or L = 10.  It would make more sense that the length would be the longest dimension so we will say that L=30.  Then W would have to be 10.  The rectangle is 30m long and 10m wide.  Just to be sure, let's multiply L*W to see if we get 300m^2, which we know is the area of this rectangle.

Is 10*30 = 300?  Yes, it is.  Therefore 10m*30m=300m^2

Now lets check the Perimeter: 80 = 2L+2W OR 80 = 2*30 + 2*10  >>>> 60+20=80  Is that correct?  I think it is.  Now you have your solution.  The length of this rectangle is 30 meters and the width of it is 10 meters.  That wasn't too difficult, was it?

neela | High School Teacher | (Level 3) Valedictorian

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Perimeter of arectangle P = 2(l+w), where l = length and w = width of the rectangle.

Given the perimeter  of the rectangular field , P = 80 m. So the l = P/2 -w = 80/2 -w.

Therefore the area of the rectanglar field = l*w = (80/2 - w)w . But are is given to be 300 sq m.

So the required equation is :

(80/2-w)w = 300.

(40-w)w = 300.

40w - w^2 = 300.

40w - w^2 -300 = 0

Multiply by (-1) and write as below:

w^2 -40w +300 = 0.

w^2 -30w -10 w +300 = 0.

w(w-30) -10(w-30) = 0.

(w-30) (w-10) = 0.

Therefore w -30 = 0, Or w -10 = 0.

So w-30 = 0 gives w = 30, and w-10 = 0 gives w = 10.

Therefore w = 10 m  and  l = (80/2 -10) = 30 m

So  length = 30 meter and width = 10 m.