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

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## 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.

Video Solution
Ex 4.3 | Question 11

## 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 $$= 12 \,\rm{m}.$$

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