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

Answer
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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.
Last updated date: 24th Sep 2023
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