Question

Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

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Answer 1

Answer:

The sides of the two squares are 14 m & 8 m.

Step-by-step-explanation:

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

Now, we know that,

Area of square = ( Side )²

∴ Area of first square = x² m

Area of other square = y² m

From the first condition,

x² + y² = 260

⇒ x² = 260 - y²

⇒ x² = - y² + 260 - - - ( 1 )

Now, we know that,

Perimeter of square = 4 * Side

∴ Perimeter of first square = 4x m

Perimeter of other square = 4y m

From the second condition,

4x - 4y = 24

⇒ x - y = 6 - - - [ Dividing each term by 4 ]

⇒ x = 6 + y

⇒ x = y + 6 - - - ( 2 )

⇒ x² = ( y + 6 )² - - - [ Squaring both sides ]

⇒ x² = y² + 2 * y * 6 + 6² - - - [ ( a + b )² = a² + 2ab + b² ]

⇒ x² = y² + 12y + 36

⇒ - y² + 260 = y² + 12y + 36 - - - [ From ( 1 ) ]

⇒ y² + 12y + 36 = - y² + 260

⇒ y² + 12y + 36 + y² - 260 = 0

⇒ 2y² + 12y - 224 = 0

⇒ y² + 6y - 112 = 0 - - - [ Dividing each term by 2 ]

⇒ y² + 14y - 8y - 112 = 0

⇒ y ( y + 14 ) - 8 ( y + 14 ) = 0

⇒ ( y + 14 ) ( y - 8 ) = 0

⇒ ( y + 14 ) = 0 OR ( y - 8 ) = 0

⇒ y + 14 = 0 OR y - 8 = 0

⇒ y = - 14 OR y = 8

As the length of square can't be negative, y = - 14 is unacceptable.

∴ y = 8 m

Now,

x = y + 6 - - - ( 2 )

⇒ x = 8 + 6

⇒ x = 14 m

∴ The sides of the two squares are 14 m & 8 m.