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# The perimeter of a rectangular field is 82 m and its area is $400\ m^2$. Find the breadth of the rectangle.

Given:

The perimeter of a rectangular field $= 82\ m$

Area of the rectangular field $=400\ m^2$.

To do:

We have to find the breadth of the rectangle.

Solution:

Let the length of the rectangle be $x$ m and the breadth of the rectangle be $y$ m.

This implies,

$2(x+y)=82\ m$

$x+y=\frac{82}{2}\ m$

$x+y=41\ m$------(1)

According to the question,

$x \times y=400$

$x(41-x)=400$

$41x-x^2=400$

$x^2-41x+400=0$

Solving for $x$ by factorization method, we get,

$x^2-25x-16x+400=0$

$x(x-25)-16(x-25)=0$

$(x-25)(x-16)=0$

$x-25=0$ or $x-16=0$

$x=25$ or $x=16$

The breadth of the rectangle is $25\ m$ or $16\ m$.

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