Question

The area of the rhombus is \[216{\text{ }}sq.{\text{ }}cm.\]If its one diagonal is \[24{\text{ }}cm,\] find the perimeter of the rhombus.
A) \[60{\text{ }}cm\]
B) \[80{\text{ }}cm\]
C) \[100{\text{ }}cm\]
D) \[120{\text{ }}cm\]

Answer 1

Hint: To solve this question, we will start with finding the second diagonal by using the area of the rhombus. Then by using Pythagoras theorem, we will get the one side of rhombus. After that we use the perimeter formula of rhombus, we will get our required answer.

Complete step-by-step answer:
We have been given area of rhombus \[ = {\text{ }}216c{m^2}\]
And, we have been given one diagonal \[ = {\text{ }}24{\text{ }}cm.\]
Let us construct a figure of a rhombus to get a better idea of it.

We know that, area of rhombus $ = \dfrac{1}{2} \times {d_1} \times {d_2}$
$\begin{gathered}
  So,216 = \dfrac{1}{2} \times 24 \times {d_2} \\
  432 = 24 \times {d_2} \\
  {d_2} = \dfrac{{432}}{{24}} = 18cm \\
\end{gathered} $
Thus, we get second diagonal \[ = {\text{ }}18cm\]
We know that diagonals of rhombus bisect each other at right angles.
So, from the figure, we get
$\because $ BD \[ = {\text{ }}24{\text{ }}cm\]
$\therefore $DO \[ = {\text{ }}12{\text{ }}cm\]
And, $\because $ AC \[ = {\text{ }}18{\text{ }}cm\]
$\therefore $ AO \[ = {\text{ }}9{\text{ }}cm\]
Then, In △AOD, from Pythagoras, theorem we get
\[\begin{array}{*{20}{l}}
  {A{O^2} + D{O^2} = A{D^2}} \\
  {{{(9)}^2} + {{(12)}^2} = A{D^2}} \\
  {81 + 144 = A{D^2}} \\
  {225 = A{D^2}} \\
  { \Rightarrow AD = 15cm}
\end{array}\]
 $\therefore $side of rhombus (AD) \[ = {\text{ }}15cm\]
Now, the perimeter of rhombus \[ = {\text{ }}4{\text{ }} \times {\text{ }}side{\text{ }} = {\text{ }}4{\text{ }} \times {\text{ }}15{\text{ }} = {\text{ }}60cm\]
Thus, perimeter of rhombus is\[\;60cm.\]

Note: In the solution, we have mentioned Pythagoras theorem, it is a theorem which gives the relation between sides of a right-angled triangle.