Question

sum of the area of the two squares is 500metre square. if the difference of their perimeter is 40 . find the sides of the two squares ​

Answer 1

$\huge\underbrace\mathfrak{Answer}$

let x be the side of the first square

y be the side of the second square

$x² + y² = 500 ------1$

and

$4x - 4y = 40$

$= > x - y = 10$

$= > x = 10 + y$

sub x = 10 + y in (1)

$(10 + y)² + y² = 500$

$= > 100 + 20y + y² + y² = 500$

$= > 2y² +20y + 100 - 500 = 0$

$= > 2y² + 20y - 400 = 0$

$= > y² + 10y - 200 = 0$

$= > {y}^{2} - 10y + 20y - 200 = 0$

$= > y(y - 10) + 20(y - 10) = 0$

$= > (y + 20)(y - 10) = 0$

$y = 10$

as y can't be in negative

$x = 10 + y$

$= > x = 10 + 10$

$= > x = 20$

∴ The Sides of the squares are 20 and 10

Answer 2

Answer:

let x be the side of the first square

y be the side of the second square

x² + y² = 500 ------1x²+1y²=500−−−−−−1

and

4x - 4y = 40

= > x - y = 10

= > x = 10 + y

sub x = 10 + y in (1)

(10 + y)² + y² = 500

= > 100 + 20y + y² + y² = 500

= > 2y² +20y + 100 - 500 = 0

= > 2y² + 20y - 400 = 0

= > y² + 10y - 200 = 0

= > {y}^{2} - 10y + 20y - 200 = 0

= > y(y - 10) + 20(y - 10) = 0

= > (y + 20)(y - 10) = 0=>

as y can't be in negative

x = 10 + y

= > x = 10 + 10

= > x = 20

∴ The Sides of the squares are 20 and 10