Sum of the areas of two …

Let perimeter of first square = x metres
Let perimeter of second square = (x +24) metres
Length of side of first square = {tex}\frac { x } { 4 }{/tex}metres {Perimeter of square = 4 × length of side}
Length of side of second square = {tex}\left( \frac { x + 24 } { 4 } \right){/tex}metres
Area of first square = side × side = {tex}\frac { x } { 4 } \times \frac { x } { 4 } = \frac { x ^ { 2 } } { 16 } m ^ { 2 }{/tex}
Area of second square = {tex}\left( \frac { x + 24 } { 4 } \right) ^ { 2 } m ^ { 2 }{/tex}
According to given condition:
{tex}\frac { x^{ 2 } } { 16 } + \left( \frac { x + 24 } { 4 } \right) ^ { 2 } = 468{/tex} {tex}\Rightarrow \frac { x ^ { 2 } } { 16 } + \frac { x ^ { 2 } + 576 + 48 x } { 16 } = 468{/tex}  
{tex}\Rightarrow \frac { x ^ { 2 } + x ^ { 2 } + 576 + 48 x } { 16 } = 468{/tex}  {tex}\Rightarrow{/tex} 2x² + 576 + 48x = 468 × 16
{tex}\Rightarrow{/tex} 2x² +48x + 576 = 7488 {tex}\Rightarrow{/tex} 2x² + 48x - 6912 = 0
{tex}\Rightarrow{/tex} x² + 24x - 3456 = 0
Comparing equation x² + 24x - 3456 = 0 with standard form ax² + bx + c = 0,
We get a = 1, b = 24 and c = -3456
Applying Quadratic Formula {tex}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{/tex}
{tex}x = \frac { - 24 \pm \sqrt { ( 24 ) ^ { 2 } - 4 ( 1 ) ( - 3456 ) } } { 2 \times 1 }{/tex}

_{{tex}\Rightarrow x = \frac { - 24 \pm \sqrt { 576 + 13824 } } { 2 }{/tex}} 
{tex}\Rightarrow x = \frac { - 24 \pm \sqrt { 14400 } } { 2 } = \frac { - 24 \pm 120 } { 2 }{/tex}

{tex}\Rightarrow x = \frac { - 24 + 120 } { 2 } , \frac { - 24 - 120 } { 2 }{/tex} 
{tex}\Rightarrow{/tex} x = 48, -72
Perimeter of square cannot be in negative. Therefore, we discard x = -72
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres
{tex}\Rightarrow{/tex} Side of First square {tex}= \frac { \text { Perimeter } } { 4 } = \frac { 48 } { 4 } = 12 \mathrm { m }{/tex}
And, Side of second Square {tex}= \frac { \text { Permeter } } { 4 } = \frac { 72 } { 4 } = 18 \mathrm { m }{/tex}