Question

Sum of the areas of two squares is 468 m. If the difference of their perimeters is 24 m.Find the sides of the two squares

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Answer 1

Step-by-step explanation:

18 and 12 m are the answers okkkkkkkk

Answer 2

$\large \mapsto \boxed{ \sf \red{Question : - }}$

•Sum of the areas of two squares is 468 m. If the difference of their perimeters is 24 m.Find the sides of the two squares.

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$\large \mapsto \boxed{ \sf\purple{Concept : - }}$

In this question the concept of using quadratic formula :-

$\sf \bullet \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

has been clarified.

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$\large \mapsto \boxed { \sf \blue{Solution : - }}$

•Let the sides of the smaller square be x m

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Then, perimeter of the smaller square =4x m

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$\bf \bullet \underline{ \: According \: to \: question : -}$

Perimeter of the larger square =(4x+24)m

Therefore, side of the larger square

$\sf = \dfrac{4x + 24}{4} m$

$\sf = \dfrac{4(x + 6)}{4} m$

$\sf = (x + 6)m$

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Again,

•Area of the smaller square =x²cm²

•Area of the larger square =(x+6)²cm²

A/Q

$\implies \sf{x}^{2} + (x + 6)^{2} = 468$

$\implies \sf {x}^{2} + {x}^{2} + 12x + 36 = 468$

$\implies \sf2 {x}^{2} + 12x - 432 = 0$

$\implies \sf {x}^{2} + 6x - 216 = 0$

Which is a quadratic equation in x .

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➾Comparing the above quadratic equation with ax² +bx+c =0 we get,

•a=1

•b=6

•c=-216

Therefore,

b²-4ac =(6)²-4(1)(-216)

⠀⠀⠀⠀ =36 +864

⠀⠀⠀ ⠀$\sf =900\geq 0$

➾So, the given equation can be solved for x .

Using quadratic formula, we get:-

$\sf \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\sf \: x = \dfrac{ - 6 \pm \sqrt{900 } }{2(1)}$

$\sf x = \dfrac{ - 6 \pm30}{2}$

$\sf x = \dfrac{ - 6 + 30}{2} ,\: \dfrac{ - 6 - 30}{ 2}$

$\sf \: x = 12 \: , \: - 18$

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Since, x cannot be negative, being the length of side of the smaller square.

Therefore , the length of the side of the smaller square is 12 m.

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➾The length of side of the larger square =x+6

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