Question

Answer question no. 11 fast

Answer 1

Hola mate..

Here's your answer..

Answer :

Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2 respectively.

It is given that 4x − 4y = 24 [Difference of perimeter]

or x − y = 6

x = y + 6

Also, x2 + y2 = 468 [sum of squares is 468]

(6+y)2 +y2 = 468

36 + y2 +12y + y2 = 468

2y2 +12y – 432 = 0

y2 + 6y – 216 = 0

y2 +18y – 12y – 216 = 0

y(y+18)(y – 12) = 0

y = - 18 or 12.

However, side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

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Answer 2

let area of one square is x^2 and other square is y^2

sum of areas =468

difference of their perimeter =4(x-y)=24

x-y=6

x=y+6

put x in first equation

(y+6)^2+y^2=468

2y^2+36+12y=468

y^2+6y-216=0

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

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

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

y=12m and y cannot be 18

×=y+6

x=18m