Question

If the perimeter of a rectangle and a square, each is equal to 80 cm and the difference of their areas is 100 sq cm, the sides of the rectangle are:
a. 25 cm, 15 cm
b. 28 cm, 12 cm
c. 30 cm, 10 cm
d. 35 cm, 15 cm

Answer 1

Hint: In this question, we will use the perimeter to find out the side length of the square. Then, we shall calculate the area of the square, and then by equating the areas of the square and the rectangle according to the conditions in the question, we will calculate the dimensions of the rectangle.

Complete step-by-step solution
Let us consider the side of the square to be “a” cm long.

Since the perimeter of a square is given by 4a, we have:
$\begin{align}
  & 4a=80 \\
 & \Rightarrow a=\dfrac{80}{4} \\
 & \Rightarrow a=20 \\
\end{align}$
$\therefore $ The side length of square = 20 cm.
Now, we know that:
Area of Square = ${{a}^{2}}$
 $\begin{align}
  & ={{20}^{2}} \\
 & =400 \\
\end{align}$
$\therefore $ The area of square = $400c{{m}^{2}}...\left( 1 \right)$
According to the question, we have:
Area of Square - Area of Rectangle = \[100c{{m}^{2}}\]
Substituting the value of the area of square in the above equation, we get:
Area of Rectangle = \[300c{{m}^{2}}\]
Using formula for area of rectangle, we get:
 $\begin{align}
  & \Rightarrow l\times b=300 \\
 & \Rightarrow \,b=\dfrac{300}{l}\quad ...(2) \\
\end{align}$
Also, we know that perimeter of rectangle = $2(l+b)$
$\Rightarrow $ The perimeter of rectangle = 80 cm
$\Rightarrow 2(l+b)=80$
Substituting the value of b from eq. (2) in the above equation, we get:
\[~2\left( l+\dfrac{300}{l} \right)=80\]
This will give us a quadratic equation in l as:
$\begin{align}
  & {{l}^{2}}-40l+300=0 \\
 &\Rightarrow {{l}^{2}}-10l-30l+300\,=0 \\
 &\Rightarrow l(l-10)-30(l-10)=0 \\
 &\Rightarrow (l-30)(l-10)=0 \\
 &\Rightarrow l-30=0\quad or\quad l-10=0 \\
 & \therefore \quad l=30,\,\,10 \\
\end{align}$
Now, we will find the breadth using the relation $\Rightarrow l\times b=300$ . For l = 30, we get
$\begin{align}
  & \Rightarrow l\times b=300 \\
 & \Rightarrow \,b=\dfrac{300}{30} \\
 & \Rightarrow b=10 \\
\end{align}$
For l = 10, we get
$\begin{align}
  & \Rightarrow l\times b=300 \\
 & \Rightarrow \,b=\dfrac{300}{10} \\
 & \Rightarrow b=30 \\
\end{align}$
$\therefore $ The dimensions of the rectangle are 30 cm and 10 cm.
Hence, option (c) is the correct answer.
Hence, the option (c) is the correct answer.

Note: In such questions, the variable dimension for solving the equation can be either length or breadth. Either way, the equation will have two roots; hence the answer will be the same. The above quadratic equation can also be solved using the Sridharacharya formula, given as:
${{l}^{2}}-40l+300=0$
On comparing it with the standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get:
$a=1,\,\,b=-40,\ c=300$
Substituting the values of a, b and c in the formula $l=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we get:
\[\begin{align}
  &\Rightarrow l=\dfrac{-(-40)\pm \sqrt{{{(-40)}^{2}}-4\times 1\times 300}}{2\times 1} \\
 &\Rightarrow l=\dfrac{40\pm \sqrt{1600-1200}}{2} \\
 &\Rightarrow l=\dfrac{40\pm \sqrt{400}}{2} \\
 &\Rightarrow l=\dfrac{40\pm 20}{2} \\
 &\Rightarrow l=20\pm 10 \\
 &\Rightarrow l=30,\ 10 \\
\end{align}\]