The length of one diagonal of rhombus is less than the second diagonal by 4cm the area of rhombus is 30 Sq. cm. find the length of diagonals
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Sol. Let the length of second diagonal of a rhombus be ‘x’ cm.
∴ the length of first diagonal of a rhombus = x – 4
Area of rhombus = ½ × Product of length of diagonals
Area of rhombus = ½ (x) (x – 4)
According to given condition,
½ (x) (x – 4) = 30
∴ x ( x – 4) = 60
∴ x2 – 4x = 60
∴ x2 – 4x – 60 = 0
∴ x2 – 10x + 6x – 60 = 0
∴ x(x – 10) + 6(x – 10) = 0
∴ (x – 10) (x + 6 ) = 0
∴ x – 10 = 0 or x + 6 = 0
∴ x = 10 or x = - 6
∵ The length of diagonal of the rhombus cannot be negative.
∴ x = 10
∴ x – 4 = 10 – 4 = 6
∴ The length of first diagonal of a rhombus is 6 cm and second diagonal is 10 cm.
∴ the length of first diagonal of a rhombus = x – 4
Area of rhombus = ½ × Product of length of diagonals
Area of rhombus = ½ (x) (x – 4)
According to given condition,
½ (x) (x – 4) = 30
∴ x ( x – 4) = 60
∴ x2 – 4x = 60
∴ x2 – 4x – 60 = 0
∴ x2 – 10x + 6x – 60 = 0
∴ x(x – 10) + 6(x – 10) = 0
∴ (x – 10) (x + 6 ) = 0
∴ x – 10 = 0 or x + 6 = 0
∴ x = 10 or x = - 6
∵ The length of diagonal of the rhombus cannot be negative.
∴ x = 10
∴ x – 4 = 10 – 4 = 6
∴ The length of first diagonal of a rhombus is 6 cm and second diagonal is 10 cm.
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Let the length of second diagonal of a rhombus be ‘x’ cm.
∴ the length of first diagonal of a rhombus = x – 4
Area of rhombus = ½ × Product of length of diagonals
Area of rhombus = ½ (x) (x – 4)
According to given condition,
½ (x) (x – 4) = 30
∴ x ( x – 4) = 60
∴ x2 – 4x = 60
∴ x2 – 4x – 60 = 0
∴ x2 – 10x + 6x – 60 = 0
∴ x(x – 10) + 6(x – 10) = 0
∴ (x – 10) (x + 6 ) = 0
∴ x – 10 = 0 or x + 6 = 0
∴ x = 10 or x = - 6
∵ The length of diagonal of the rhombus cannot be negative.
∴ x = 10
∴ x – 4 = 10 – 4 = 6
∴ The length of first diagonal of a rhombus is 6 cm and second diagonal is 10 cm.
∴ the length of first diagonal of a rhombus = x – 4
Area of rhombus = ½ × Product of length of diagonals
Area of rhombus = ½ (x) (x – 4)
According to given condition,
½ (x) (x – 4) = 30
∴ x ( x – 4) = 60
∴ x2 – 4x = 60
∴ x2 – 4x – 60 = 0
∴ x2 – 10x + 6x – 60 = 0
∴ x(x – 10) + 6(x – 10) = 0
∴ (x – 10) (x + 6 ) = 0
∴ x – 10 = 0 or x + 6 = 0
∴ x = 10 or x = - 6
∵ The length of diagonal of the rhombus cannot be negative.
∴ x = 10
∴ x – 4 = 10 – 4 = 6
∴ The length of first diagonal of a rhombus is 6 cm and second diagonal is 10 cm.
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