Question

Sum of the areas of two squares is 468m^2.if the difference of their perimeter is 24m, find the sides of the two squares

Answer 1

Step-by-step explanation:

Let x and y be the sides of two squares.

${x}^{2} + {y}^{2} = 468 \: \: (1) \\ 4x - 4y = 24 \\ x - y = 6 \: \: \\ x = 6 + y \\ put \: x \: in \: (1) \\ {(6 + y)}^{2} + {y}^{2} = 468 \\ 36 + {y}^{2} + 12y + {y}^{2} = 468 \\ 2 {y}^{2} + 12y - 432 = 0$

solving these equations to get the required values of x and y

Answer 2

Step-by-step explanation:

Let the sides of two square be X and y.

AC/1---

x^2+y^2=468. (1)

AC/2-----

4x-4y=24

Now divide by 4 on both sides----

x-y=6

X=6+y

now put the value of X in (1)---

therefore,( 6+y)^2+ y^2=6

36+y^2+12y+y^2=468

36+12y+2y^2=468

y^2+6y=216

y^2+6y-216=0

Now, by splitting middle term method-----

y^2+(18-12)y-216=0

y^2+18y-12y-216=0

y(y+18)-12(y+18)=0

(y+18)(y-12)=0

since, y+18 is not possible

therefore, sides of squares are 18and 12