diagonals of a rhombus are 25 and 42cm find its area and perimeter.
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Heya !!
Let ABCD is a rhombus , in which AC and BD are it's two diagonals.
AB = AD = BC = DC are its sides.
Let AC = 25 cm and BD = 42 cm.
We know that the diagonals of a rhombus bisect each other at right angles.
Therefore,
OA = 1/2 × AC = 1/2 × 25 = 25/2 = 12.5
OA = 12.5 cm
And,
OB = 1/2 × BD = 1/2 × 42 = 21 cm.
By Pythagoras theorem , we have :
AB = √(OA)² + (OB)²
AB = √(12.5)² + (21)²
AB = √156.25 + 441
AB = √ 597.25
AB = 24.43 ( approx ).
Therefore,
Perimeter of rhombus = 4 × 24.43 = 97.72 cm
And,
Area of rhombus = 1/2 × ( AC × BD ) = 1/2 × (25 × 42 ) = 1050/2 = 525 cm².
Let ABCD is a rhombus , in which AC and BD are it's two diagonals.
AB = AD = BC = DC are its sides.
Let AC = 25 cm and BD = 42 cm.
We know that the diagonals of a rhombus bisect each other at right angles.
Therefore,
OA = 1/2 × AC = 1/2 × 25 = 25/2 = 12.5
OA = 12.5 cm
And,
OB = 1/2 × BD = 1/2 × 42 = 21 cm.
By Pythagoras theorem , we have :
AB = √(OA)² + (OB)²
AB = √(12.5)² + (21)²
AB = √156.25 + 441
AB = √ 597.25
AB = 24.43 ( approx ).
Therefore,
Perimeter of rhombus = 4 × 24.43 = 97.72 cm
And,
Area of rhombus = 1/2 × ( AC × BD ) = 1/2 × (25 × 42 ) = 1050/2 = 525 cm².
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