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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AnswEr:

• The sides of the squares are 12 m and 18 m.

Given Information:

• Sum of the areas of two squares is 468 m².
• The difference of their perimeters is 24 m.

Need To Find:

• The sides of the two squares = ?

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.

Now Given Here,

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.

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Answer 2

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Sum of the areas of two squares is 468 m²

∵ x² + y² = 468 . ………..(1) .[ ∵ area of square = side²] → The difference of their perimeters is 24 m.

∵ 4x – 4y = 24 [ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24

⇒ x – y = 24/4 .

⇒ x – y = 6 .

∴ y = x – 6 ……….(2)

From equation (1) and (2),

∵ x² + ( x – 6 )² = 468

⇒ x² + x² – 12x + 36 = 468

⇒ 2x² – 12x + 36 – 468 = 0

⇒ 2x² – 12x – 432 = 0

⇒ 2( x² – 6x – 216 ) = 0

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0

⇒ x( x – 18 ) + 12( x – 18 ) = 0

⇒ ( x + 12 ) ( x – 18 ) = 0

⇒ x + 12 = 0 and x – 18 = 0

⇒ x = – 12m [ rejected ] and x = 18m

∴ x = 18 m

Put the value of ‘x’ in equation (2),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12 m

Hence, sides of two squares are 18m and 12m respectively

Hope it's Helpful....:)