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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.Hey peeps ! Please drop some thanks. |
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Answer» 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. |
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