Question

Sum of the areas of two squares is 468m2 if the difference of their perimeter is 24meter find the sides of the two squares

Answer 1

Answer:

side of first sq=x and second sq=y

area will be x^2 and y^2 resp.

and perimeter will be 4x and 4y resp.

acc to ques

${x}^{2} + {y}^{2} = 468 \\ 4x - 4y = 24 \\ using \: second \: eq \\ 4x = 24 + 4y \\ x = \frac{24 + 4y}{4} \\ x = \frac{24}{4} + \frac{4y}{4} \\ x = 6 + y \\ putting \: in \: first \\ {(6 + y)}^{2} + {y}^{2} = 468 \\ 36 + {y}^{2} + 12y + {y}^{2} = 468 \\ 2 {y}^{2} + 12y - 432 = 0 \\ 2 {y }^{2} + 36y - 24y - 432 \\ 2y(y + 18) - 24(y + 18) \\ (2y - 24)(y + 18) = 0 \\ 2y - 24 = 0 \\ 2y = 24 \\ y = 12 \\ or \\ y + 18 = 0 \\ y = - 18 \\ as \: the \: side cannot \: be \: negative\: 12 \: is \: your \: answer$

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