Question

The perimeter of a square filed is 40 m more than another square field. Three times the area of the smaller filed is $50{m^2}$ more than the area of the larger field. Find the sides of both the fields.

Answer 1

Hint:
We will first let that the side of the small square is \[x\]. Use the conditions given in the question to form the quadratic equation and then use quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , to find the values of the $x$, which will in turn be the sides of both the fields. After getting the side of the smaller square we can easily calculate the side of the bigger square by adding 40 in the value of $x$. While calculating the value of $x$, we have first find the quadratic equation and then substitute the values in the formula.

Complete step by step solution:
Let us take the smaller side of the smaller square to be $x$. Then find the area of the smaller square, which is the square of the side of the smaller square.
$ \Rightarrow A = {x^2}$

Now, it is given in the question that the perimeter of a square filed is 40 m more than another square field, thus the side of the bigger square is $x + 40$. Then find the area of the bigger square, which is the square of the side of the bigger square.
$ \Rightarrow A' = {\left( {x + 40} \right)^2}$

Now, it is given that three times the area of the smaller filed is $50{m^2}$ more than the area of the larger field.
$
   \Rightarrow 3{x^2} = {\left( {x + 40} \right)^2} + 50 \\
   \Rightarrow 3{x^2} + {x^2} + 1600 + 80x + 50 \\
   \Rightarrow 2{x^2} = 1650 + 80x \\
   \Rightarrow {x^2} = 825 + 40x \\
   \Rightarrow {x^2} - 40x - 825 = 0 \\
 $

Thus, we have obtained a quadratic equation, whose roots are given by the formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Here,
$
  a = 1 \\
  b = - 40 \\
  c = - 825 \\
 $

Substitute these values into the quadratic formula.
$
  \therefore x = \dfrac{{ - \left( { - 40} \right) \pm \sqrt {{{\left( { - 40} \right)}^2} - 4\left( 1 \right)\left( { - 825} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{40 \pm \sqrt {1600 + 3300} }}{2} \\
   \Rightarrow x = \dfrac{{40 \pm \sqrt {4800} }}{2} \\
   \Rightarrow x = \dfrac{{40 \pm \sqrt {1600 \times 3} }}{2} \\
   \Rightarrow x = \dfrac{{40 \pm 40\sqrt 3 }}{2} \\
   \Rightarrow x = 20 \pm 20\sqrt 3 \\
   \Rightarrow x = 20 + 20\sqrt 3 {\text{ or }}x = 20 - 20\sqrt 3 \\
 $

It is to be noted that $x$ denotes the side of the square which can never be negative.
Thus,
$x = 20 + 2\sqrt 3 $ is a valid value.

Find the value of sides for both the fields by substituting these values.
${x^2} = {\left( {20 + 2\sqrt 3 } \right)^2}$
And of the larger field;
$
  {\left( {x + 40} \right)^2} = {\left( {20 + 20\sqrt 3 + 40} \right)^2} \\
   = {\left( {60 + 20\sqrt 3 } \right)^2} \\
 $

Thus, these are the required values of the sides of both the fields.

Note:
Take care of the sign of the coefficients to be used in the quadratic formula. Keep note of which variable is denoting what, and is the value obtained after solving the quadratic equation a valid choice for that variable? Also after you are done with finding the value of $x$, make the necessary substitution to find the required values.