If the area of the rhombus is 336 cm2 and one of its diagonals is 14 cm, find its perimeter.
Answer:
100 cm
- Let ABCD be the given rhombus with
AC=14 cm and BD = x cm.
We know, Area of rhombus =12× Product of its diagonals ⟹336=12×AC×BD⟹336=12×14×x⟹2×33614=x⟹48=x Thus, the length of diagonal BD=48 cm. - Let the diagonals AC and BD bisect at a point O.
We know that the diagonals of a rhombus bisect each other at right angles.
So, AO=12AC and BO=12BD. ∴AO=12×14=7 cm and BO=12×48=24 cm. Also, ∠AOB=90∘. - Using Pythagous' theorem in right △AOB, we have AB2=AO2+BO2⟹AB2=(7)2+(24)2⟹AB2=49+576⟹AB2=625⟹AB=25 cm
- Now, Perimeter of rhombus =4× Length of side =4×AB=4×25=100 cm Thus, the perimeter of the rhombus is 100 cm.