Question

Sum of the areas of two squares is 468m². If the difference of their perimeters is 24 m. formulate the quadratic equation to find the sides of the two squares.​

Answer 1

Explanation:

Let the side if one of the squares be s.

Hence, its perimeter is 4s.

Given, the other square has a perimeter greater than the first one by 24 m. Thus, its perimeter is 4s + 24.

Hence, length of its side is s + 6.

Areas of the two squares are

$s {}^{2}$

and

$(s + 6) {}^{2} = s {}^{2} + 12s + 36$

But it is given that the sum of their areas is 468 square meters.

Hence, we can form the equation:

$s {}^{2} + (s {}^{2} + 12s + 36) = 468$

$2s {}^{2} + 12s + 36 - 468 = 0$

$2s {}^{2} + 12s - 432 = 0$

This is the quadratic equation this formed for finding the value of s.

Answer: 2s^2 + 12s - 432 = 0

I hope this helps. :D

Answer 2

Let,

the side of the first square be 'x'm and that of the second be 'y' m.

Area of,

the first square = x² sq m.

the second square = y² sq m.

Their perimeters would be 4x and 4y respectively.

Given,

$\implies$$\tt{4y - 4x = 24}$

$\implies$$\tt{4( y - x) = 24}$

$\implies$$\tt{y - x =}$ $\tt{\frac{24}{6}}$

$\tt{ y - x = 6 ----(1)}$

$\tt{ (y)^{2} - (x)^{2} = 468-----(2)}$

From (1), $\tt{y = x + 6}$

Substituting for y in (2),

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

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

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

$\implies$ $\tt{2(x^{2}+6x+18)=468}}$

$\implies$ $\tt{x^{2} + 6x+18= \frac{468}{2}}$

$\implies$ $\tt{x^{2} + 6x+18= 234}}$

$\implies$ $\tt{x^{2} + 6x+18-234= 0}}$

$\implies$ $\tt{x^{2} + 6x -216= 0\:(quadratic\:equation)}}$

$\implies$ $\tt{x^{2} +18x - 12a -216= 0}}$

$\implies$ $\tt{x(x +18) - 12(x +18)= 0}}$

$\implies$ $\tt{(x-12)(x +18)= 0}}$

∴ x = 12 & -18

So, the side of the first square is 12 m. and the side of the second square is 18 m.

Hope it helps:)