# sum of the areas of two squares is 468 metres square if the difference of their perimeter is 24 M find the sides of the two squares

## Answers

Answered by

5

let the sides of square be x and y

ATQ

4x-4y=24

x-y=6

x^2+y^2=468

(x-y)^2=36

468-2xy=36

-2xy=36-468=--432

xy=216

(x+y)^2=x^2+y^2+2xy=468+432=900

x+y=30

x-y=6

2x=36

x=18m And

y=30-x=30-18=12 m Ans

ATQ

4x-4y=24

x-y=6

x^2+y^2=468

(x-y)^2=36

468-2xy=36

-2xy=36-468=--432

xy=216

(x+y)^2=x^2+y^2+2xy=468+432=900

x+y=30

x-y=6

2x=36

x=18m And

y=30-x=30-18=12 m Ans

Answered by

3

Answer :

Let the sides if the two squares are x and y respectively

A/q,

1st case :

x^2 + y^2 =468 - equation 1

2nd case :

4x - 4y = 24 - equation 2

x - y =24/4 = 6

x = y + 6

putting the value of x from equation 2 to the question 1

x^2 + y^2 = 468

(y + 6)^2 = 468

y^2 + 2*y*6 + 6^2 + y^2 =468

2y^2 + 12y + 36 - 468= 0

2y^2 +12y-432 =0

2 (y^2+ 6y -216)=0

y^2+6y-216=0

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

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

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

y=12 or

y=-18(not possible)

putting Y =0 in equation 2

x=y+6

=12+6=18

so, the sides of the squares are 18 m and 12 m.....

Let the sides if the two squares are x and y respectively

A/q,

1st case :

x^2 + y^2 =468 - equation 1

2nd case :

4x - 4y = 24 - equation 2

x - y =24/4 = 6

x = y + 6

putting the value of x from equation 2 to the question 1

x^2 + y^2 = 468

(y + 6)^2 = 468

y^2 + 2*y*6 + 6^2 + y^2 =468

2y^2 + 12y + 36 - 468= 0

2y^2 +12y-432 =0

2 (y^2+ 6y -216)=0

y^2+6y-216=0

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

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

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

y=12 or

y=-18(not possible)

putting Y =0 in equation 2

x=y+6

=12+6=18

so, the sides of the squares are 18 m and 12 m.....

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