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The Brainliest Answer!
2016-03-27T14:49:46+05:30
Let ABCD be rhombus
Let AC = 6 and BD = 8
property Diagonals of rhombus bisect each other in right angle.
consider AC and BD meet at O
THen in triangle AOD
AD² = AO² + OD²
= (AC/2)² + (BD/2)²
= 3² + 4²
= 25
AD = √25
AD = 5 side of rhombus
Perimeter = 4 × side = 20 cm

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2016-03-27T14:55:40+05:30
Well diagonals of a rhombus bisect each other If u draw the diagram u would find four triangles. In any triangle u coupd apply the Pythagoras theorem ABCD is rhombus with AC and BD as the diagonal intersecting at O. So AO=CO=3cm and BO=DO=4cm Aplly Pythagoras theorem in triangle AOB where angle AOB = 90 degress since diagonals of rhombus are perpendicular to each other . let the required side be X cm In traingle AOB Square of AB= square of AO + square of BO AB square = 3*3 +4*4 AB square = 25 AB = 5 cm Hence side is 5 cm
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