Question

Sum of the areas of two squares is 468 m2. If the difference of their
perimeters is 24 m, find the sides of the two squares​

Answer 1

Answer:

QUETION:-

Sum of the areas of two squares is 468 m2. If the difference of theirperimeters is 24 m, find the sides of the two squares.

LET:-

The sides of the two square be x m and y m .therefore thier peremeter will be 4x and 4 y . Respectively and thier areas will be ${x}^{2} and \: {y}^{2}$Respectively.

GIVEN THAT:-

$4x - 4y = 24 \\ x - y = 6 \\ x = y + 6$

ALSO:-

${x}^{2} + {y}^{2} = 468 \\ {(6 + y)}^{2} + {y}^{2} = 468 \\ 36 + {y}^{2} + 12y + {y}^{2} = 468 \\ 2 {y }^{2} + 12y - 432 = 0 \\ {y}^{2} + 6y - 216 = 0 \\ {y}^{2} + 8y - 12y - 216 = 0 \\ y(y + 18) - 12(y + 18) = 0 \\ (y + 18)(y - 12) = 0 \\ y = - 18 \: or \: 12$

However,side of a Square cannot be negative.

Hence the side of squares are 12m and (12+6) m= 18.

Thanks :)

Answer 2

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

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be x2 and y2 respectively.

Given,

4x – 4y = 24

x – y = 6

x = y + 6  

Also, x2 + y2 = 468

⇒ (6 + y2) + y2 = 468

⇒ 36 + y2 + 12y + y2 = 468

⇒ 2y2 + 12y + 432 = 0

⇒ y2 + 6y – 216 = 0

⇒ y2 + 18y – 12y – 216 = 0

⇒ y(y +18) -12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

⇒ y = -18, 12

As we know, the side of a square cannot be negative.

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