Question

The area of a rectangular field is 150 sq. units. If its perimeter is 50 units, then its dimensions are
A.7, 5
B.3, 50
C.5,30
D.10,15

Answer 1

Hint: We can assume 2 variables for length and breadth. We can write the area and perimeter in terms of the variables of length and breadth. By the given conditions we can form 2 equations and we can get the length and breadth by solving these equations.

Complete step by step answer:

Let l be the length of the rectangle and b be the breadth of the rectangle.
Then area of the rectangle is given by, ${\text{A = $l \times$ b}}$. But the area is given as 150 sq. units. So, we get,
${\text{$l \times b$ = 150}}$ ... (1)
The perimeter of the rectangle is given by, ${\text{P = 2}}\left( {{\text{l + b}}} \right)$. But the perimeter is given as 50 units. So we get,
${\text{50 = 2}}\left( {{\text{l + b}}} \right)$
On dividing throughout with 2, we get
${\text{l + b = }}\dfrac{{{\text{50}}}}{{\text{2}}}{\text{ = 25}}$ … (2)
Equation (1) can be written as,
${\text{l = }}\dfrac{{{\text{150}}}}{{\text{b}}}$.. (3)
On Substituting for l in equation (2), we get,
$\dfrac{{{\text{150}}}}{{\text{b}}}{\text{ + b = 25}}$
On Multiplying throughout with b, we get,
${\text{150 + }}{{\text{b}}^{\text{2}}}{\text{ = 25b}}$
$ \Rightarrow {{\text{b}}^{\text{2}}}{\text{ - 25b + 150 = 0}}$
We can solve the quadratic equation to get the value of b
$
   \Rightarrow {{\text{b}}^{\text{2}}}{\text{ - 10b - 15b + 150 = 0}} \\
   \Rightarrow {\text{b}}\left( {{\text{b - 10}}} \right){\text{ - 15}}\left( {{\text{b - 10}}} \right){\text{ = 0}} \\
   \Rightarrow \left( {{\text{b - 10}}} \right)\left( {{\text{b - 15}}} \right){\text{ = 0}} \\
  $
$ \Rightarrow {\text{b = 10,15}}$
We can find the length by substituting the value of b in equation (3)
When b=10 units, ${\text{l = }}\dfrac{{{\text{150}}}}{{{\text{10}}}}{\text{ = 15 units}}$
When b=15 units, ${\text{l = }}\dfrac{{{\text{150}}}}{{{\text{15}}}}{\text{ = 10 units}}$
So, the dimensions of the rectangular fields are 10,15
Therefore, the correct answer is option D.

Note: The two equations in 2 variables can be solved by making it a quadratic equation in any one variable. The quadratic equation can be solved by any method. We get 2 values each for breadth and length. This means that the area the rectangle will be the same even if the length and breadth are interchanged