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

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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, x^2 + y^2 = 468

⇒ (6 + y^2) + y^2 = 468

⇒ 36 + y^2 + 12y + y^2 = 468

⇒ 2y^2 + 12y + 432 = 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 = -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.

Answer 2

Answer:

18 m.

Step-by-step explanation:

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, x^2 + y^2 = 468

⇒ (6 + y^2) + y^2 = 468

⇒ 36 + y^2 + 12y + y^2 = 468

⇒ 2y^2 + 12y + 432 = 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 = -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.