Question

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

Hint: - Perimeter of a rectangle is $2\left( {l + b} \right)$ and the area of the rectangle is$l \times b$.

Perimeter of rectangular field $= 2\left( {l + b} \right) = 82m$
$\Rightarrow l + b = 41m$
Let the length of the rectangular field be $Xm$ , then the breadth will be $\left( {41 - X} \right)m$.
$\because$ Area of the rectangular field $= l \times b$
$\Rightarrow X\left( {41 - X} \right) = 400 \\ \Rightarrow 41X - {X^2} = 400 \\ \Rightarrow {X^2} - 41X + 400 = 0 \\$
Now, after splitting the middle term we get:
$\Rightarrow {X^2} - 16X - 25X + 400 = 0 \\ \Rightarrow X\left( {X - 16} \right) - 25\left( {X - 16} \right) = 0 \\ \Rightarrow \left( {X - 16} \right)\left( {X - 25} \right) = 0 \\ \therefore {X_1} = 16{\text{ \& }}{X_2} = 25 \\$
For${X_1} = 16m$, the length of the rectangle is 16m and the breadth of the rectangle is 25m.
For${X_2} = 25m$, the length of the rectangle is 25m and the breadth of the rectangle is 16m.
Hence, both sides of the rectangle are 16m and 25m.

Note: - In any of the questions while solving quadratic equations in between the problem, never ignore any of the result as it may lead to half answer or inappropriate answer.