Question

sum of the areas of two squares is 468 meter square. If the difference of their perimeter is 24 meter. Find the sides of two square.
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Answer 1

Given :

• sum of areas of two squares = 468 m²
• difference of their perimeters = 24 m

To Find :

• sides of two squares ?

Solution :

• let the length of the smaller square be x
• so, perimeter of the smaller square = 4x

now, perimeter of larger square is 24 more than the perimeter of smaller square

• ➪ perimeter of larger square = 4x + 24

and we know that,

$\bold{side = \frac{perimeter}{4} }$

so, length of side of larger square = $\frac{4x + 24}{4} = x + 6$

it is given that, sum of areas of two squares is 468 m²

and we know that,$\boxed{ \bold{area \: of \: square = {side}^{2} }}$

$\bold{: \implies {(6 + x)}^{2} + {x}^{2} = 468 } \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies 36 + {x}^{2} + 12x + {x}^{2} = 468 } \\ \\ \bold{: \implies \: 2 {x}^{2} + 12x + 36 = 468} \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: 2( {x}^{2} + 6x + 18) = 468 } \: \: \: \: \: \: \\ \\ \bold{: \implies \: {x}^{2} + 6x + 18 = \cancel\frac{468}{2} } \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \: {x}^{2} + 6x + 18 = 234 } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \: {x}^{2} + 6x + 18 - 234 = 0} \: \: \: \\ \\ \bold{: \implies \: {x}^{2} + 6x -216 = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies {x}^{2} +18x - 12x - 216 = 0 } \\ \\ \: \: \: \bold{: \implies \: x(x + 18) - 12(x + 18} ) = 0 \\ \\ \bold{: \implies \: (x + 18)(x - 12) = 0} \: \: \: \: \: \: \: \: \: \:$

now,

x + 18 = 0 or x - 12 = 0

➪ x = -18 or x = 12

(The measure of side of a square is a non-negative quantity).

Thus, the measure of sides of the smaller square is 12 m

Thus, the measure of sides of the smaller square is 12 m and the measure of sides of the larger square is 12m + 6m = 18 m

• 12 m and 18 m