# Ex.4.3 Q11 Quadratic Equations Solutions - NCERT Maths Class 10

## Question

Sum of the areas of two squares \(468 \,\rm{m}^2.\) If the difference of perimeters is \(24\,\rm{m,}\) find the sides of two squares.

## Text Solution

**What is Known?**

i) Sum of the areas of two squares is \(468\,\rm{m}^2.\)

ii) The difference of perimeters is \(24\,\rm{m.}\)

**What is Unknown?**

Sides of two squares.

**Reasoning:**

Let the side of first square is *\(x\)* and side of the second square is* *\(y.\)

Area of the square \(= \rm{side} \times \rm{side}\)

Perimeter of the square \(= 4 \times \rm{side}\)

Therefore, the area of the first and second square are \({x^2}\) and \({y^2}\) respectively. Also, the perimeters of the first and second square are \(4x\) and \(4y\) respectively. Applying the known conditions:

\[\begin{align}&{\rm{(i)}}\quad{x^2} + {y^2} = 468 \quad \ldots ..\left( 1 \right)\\\\

&{\rm{(ii)}}\quad 4x - 4y = 24 \quad \ldots .{\rm{ }}\left( 2 \right)\end{align}\]

**Steps:**

\[\begin{align}{x^2} + {y^2} &= 468\\4x - 4y &= 24\\4(x - y) &= 24\\x - y &= 6\\x &= 6 + y\end{align}\]

Substitute \(x = y + 6\) in equation (1)

\[\begin{align}{(y + 6)^2} + {y^2} &= 468\\{y^2} + 12y + 36 + {y^2} &= 468\\2{y^2} + 12y + 36 &= 468\\2({y^2} + 6y + 18)& = 468\\

{y^2} + 6y + 18 &= 234\\{y^2} + 6y + 18 - 234 &= 0\\{y^2} + 6y - 216 &= 0

\end{align}\]

Solving by factorization method

\[\begin{align}{y^2} + 18y - 12y - 216 &= 0\\y\left( {y + 18} \right) - 12\left( {y + 18} \right)& = 0\\\left( {y + 18} \right)\left( {y - 12} \right) &= 0\\y + 18 = 0 &\qquad y - 12= 0\\y = - 18 &\qquad y = 12\end{align}\]

*\(y\) *can’t be negative value as it represents the side of the square.

Side of the first square \(x = y + 6 = 12 + 6 = 18\,\rm{m}\)

Side of the second square \(= 12 \,\rm{m}.\)