Question

The perimeter of a rhombus is 40 cm. If the length of one of its diagonals is 12 cm, what is the length of the other diagonal?
(A) 16 cm
(B) 8 cm
(C) 20 cm
(D) 10 cm

Answer 1

Hint: Assume the length of a side of rhombus be a. Every side of a rhombus is equal to each other. So, the length of each side is a. The perimeter of a rhombus is the summation of all sides of the rhombus. So, the value of the perimeter of the rhombus is equal to 4a. We have the perimeter of the rhombus equal to 40cm. Now, get the value of a. We have a rhombus ABCD in which AD is equal to 10, AC is equal to 12 cm. We know that in a rhombus the diagonals bisect each other at \[90{}^\circ \] . Now, use Pythagoras theorem in the \[\Delta AOD\] and get the value of OD. BD gets bisected at the point O. So, \[BD=2\times OD\] . Now, get the value of the diagonal BD.

Complete step by step solution:
According to the question, we have a rhombus which has a perimeter equal to 40 cm and the length of one of its diagonals is 12 cm.
The perimeter of the rhombus = 40 cm ……………………..(1)
The length of one of its diagonals = 12 cm …………………………(2)
Let us assume the length of a side of the rhombus be a.
We know the property that all sides of a rhombus are equal to each other.
So, the length of each side is a.
We know that the perimeter of a rhombus is the summation of all sides of the rhombus.
The perimeter of the rhombus = a + a + a + a = 4a ………………………(3)
From equation (1), we have the value of the perimeter of the rhombus.
Now, from equation (1) and equation (3), we get
\[\begin{align}
  & 40=4a \\
 & \Rightarrow \dfrac{40}{4}=a \\
 & \Rightarrow 10=a \\
\end{align}\]
So, the length of each side of a rhombus is 10 cm …………………………(4)

We also know the property that the diagonals of a rhombus bisect each other at \[90{}^\circ \] .
It means in a rhombus ABCD the diagonals AC and BD bisect each other at \[90{}^\circ \] .
The length of the diagonal AC is 12 cm. Since the diagonal AC gets bisected at the point O so, the length OA will be half of the diagonal AC.
\[OA=\dfrac{AC}{2}=\dfrac{12}{2}=6\] cm ………………………….(5)
Now, in \[\Delta AOD\] we have,
OA= 6 cm
AD = 10 cm
Now, using Pythagoras theorem,
\[\begin{align}
  & \Rightarrow {{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Height \right)}^{2}} \\
 & \Rightarrow {{\left( AD \right)}^{2}}={{\left( OA \right)}^{2}}+{{\left( OD \right)}^{2}} \\
 & \Rightarrow {{\left( 10 \right)}^{2}}={{\left( 6 \right)}^{2}}+{{\left( OD \right)}^{2}} \\
 & \Rightarrow 100=36+{{\left( OD \right)}^{2}} \\
 & \Rightarrow 100-36={{\left( OD \right)}^{2}} \\
 & \Rightarrow 64={{\left( OD \right)}^{2}} \\
 & \Rightarrow \sqrt{64}=OD \\
 & \Rightarrow 8=OD \\
\end{align}\]
The diagonals AC and BD are bisecting at point O.
So, \[BD=2\times OD=2\times 8=16\] .
Therefore, the length of the diagonal BD is 16 cm.
Hence, option (A) is the correct one.

Note: In this question, one might take \[\dfrac{1}{2}\times {{d}_{1}}\times {{d}_{2}}\] as the formula of the perimeter, where \[{{d}_{1}}\] and \[{{d}_{2}}\] are the diagonals of the rhombus.
 \[\begin{align}
  & Perimeter=\dfrac{1}{2}\times {{d}_{1}}\times {{d}_{2}} \\
 & \Rightarrow 40=\dfrac{1}{2}\times 12\times {{d}_{2}} \\
 & \Rightarrow \dfrac{80}{12}={{d}_{2}} \\
 & \Rightarrow \dfrac{20}{3}={{d}_{2}} \\
\end{align}\]
This calculation for finding the value of the length of other diagonal is wrong because the formula \[\dfrac{1}{2}\times {{d}_{1}}\times {{d}_{2}}\] is for the area of the rhombus not for the perimeter.