Question

The area of a rhombus is \[28c{m^2}\] and one of its diagonal is 4cm. What is its perimeter?
\[\begin{array}{l}
& (a){\rm{ 4}}\sqrt {53} cm\\
&(b){\rm{ 36cm}}\\
&{\rm{(c) 2}}\sqrt {53} cm\\
&{\rm{(d) 52cm}}
\end{array}\]

Answer 1

Hint: We know that rhombus is a quadrilateral having all four sides equal. The area of a rhombus is equal to half of the product of its diagonal and its perimeter is equal to 4 times each side of the rhombus.
Mathematically,
Area\[ = \dfrac{1}{2} \times \left( {{d_1} \times {d_2}} \right)\], Perimeter\[ = 4a\]
Where, \[{d_1},{d_2}{\rm{ and\ a}}\] are the diagonals and sides of a rhombus.

Complete step-by-step answer:
We have been given the area of a rhombus as \[28c{m^2}\] and one of its diagonal is 4cm.

Let, the rhombus ABCD of side length 'a' cm, the diagonal \[{\rm{AC}} = {\rm{4cm }},{\rm{ BD}} = {\rm{lcm}}\] and its area \[ = 28\,c{m^2}\] . Also, the diagonal interest at O.

We know that area of rhombus \[ = \dfrac{1}{2} \times {d_1} \times {d_2}\]
Where, \[{d_1}{\rm{ and\ }}{d_2}\] are diagonals.
\[\begin{array}{l} \Rightarrow \dfrac{1}{2} \times 4 \times l = 28\\ \Rightarrow 2 \times l = 28\\ \Rightarrow l = \dfrac{{28}}{2}\\ \Rightarrow l = 14cm.\\ \Rightarrow BD = 14cm\end{array}\]
Since, we know that the diagonal gets bisected at the point of intersection.
\[\begin{array}{l} \Rightarrow AO = OC = 2cm\\ \Rightarrow BO = OD = 7cm\end{array}\]
In \[\Delta AOB\], we have
\[{\rm{AO}} = {\rm{2cm\ and\ OB}} = {\rm{7cm}}\].
Using Pythagora's theorem, which states that, the sum of square of perpendicular and base of a right angled triangle is equal to square of its hypotenuse.
\[\begin{array}{l} \Rightarrow A{B^2} = A{O^2} + B{O^2}\\ \Rightarrow {2^2} + {7^2}\\ \Rightarrow 4 + 49\\ \Rightarrow A{B^2} = 53\end{array}\]
Taking square root both sides, we get:
\[AB = \sqrt {53} \,cm\]
We know, perimeter of the rhombus is equal to 4 times its each side.
\[\begin{array}{l} \Rightarrow Perimeter = 4 \times \sqrt {53} \\ \Rightarrow 4\sqrt {53} cm\end{array}\]
Therefore, the correct option is ‘a’.

Note: In this type of question, it is better to draw a suitable diagram according to the question, as you can easily visualize what you have to find next, so that you can get the required value. You can also use the direct formula in terms of area (A), diagonals \[\left( {{d_1}} \right)\] as follows:
Perimeter \[ = 4\sqrt {{{\left( {\dfrac{A}{{{d_1}}}} \right)}^2} + {{\left( {\dfrac{{{d_1}}}{2}} \right)}^2}} \]