Question

Find the dimensions of a rectangular park whose perimeter is 60m and area 200m square

Answer 1

Given,

Perimeter of the rectangle = 60 m

Area of the rectangle = 200 m sq

To Find,

The dimensions of the rectangle.

Solution,

Since, we are given with the area and perimeter

2(l + b) = 60

(l + b) = 30

l = 30-b

Now,

l*b = 200

Substituting the value of l in above equation

b(30-b)= 200

30b - b² = 200

b = 20 or 10

So, length will be 10 or 20.

Hence the dimensions of the rectangle is 10 m and 20 m.

Answer 2

For a rectangle of length $l$ and breadth $b$

The perimeter of a rectangle $P=2(l+b)$

The area of a rectangle $A=lb$

Given:

perimeter$=60m$; area$=100m^{2}$

Explanation:

The equation for perimeter: $P=2(l+b)$

$\implies 60=2(l+b)$

$\implies l+b=\frac{60}{2}$

$\implies l+b=30$..............(1)

The equation for area: $A=lb$

$\implies 200=lb$

$\implies l=\frac{200}{b}$............(2)

Substituting equation (2) in (1)

$\implies \frac{200}{b} +b=30$

$\implies \frac{200+b^{2} }{b}=30$

$\implies 200+b^{2} =30b$

$\implies b^{2}-30b+200 =0$

$\implies b^{2}-20b-10b+200 =0$

$\implies b(b-20)-10(b-20) =0$

$\implies (b-10)(b-20) =0$

$\implies b=10,20$

Substituting in (2)

$\implies l=20,10$

Since length is greater than breadth,

The length of the rectangle $l=20m$, breadth $b=10m$