Math, asked by taherkanch1, 4 days ago

# 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

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
3

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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