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. ∴ Also, \angle AOB = 90^\circ. - Using Pythagous' theorem in right \triangle AOB, we have \begin{aligned} & AB^2 = AO^2 + BO^2 \\ \implies & AB^2 = (7)^2 + (24)^2 \\ \implies & AB^2 = 49 + 576 \\ \implies & AB^2 = 625 \\ \implies & AB = 25 \space cm \\ \end{aligned}
- Now, \begin{aligned} \text{ Perimeter of rhombus } & = 4 \times \text { Length of side } \\ & = 4 \times AB \\ & = 4 \times 25 \\ & = 100 \space cm \end{aligned} Thus, the perimeter of the rhombus is 100 \space cm.