Math, asked by paarthmakkar, 11 months ago

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

Answers

Answered by mpreet196
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

Answered by VaibhavTezShakya
4

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

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