question_answer

Perimeter of a rhombus is 2p unit and sum of length of diagonals is m units, then area of the rhombus is                                                                                                                         [SSC (CGL) Mains 2014]

A) \[\frac{1}{4}{{m}^{2}}p\,\,\text{sq}\,\,\text{units}\]     

B) \[\frac{1}{4}m{{p}^{2}}\,\,\text{sq}\,\,\text{units}\]

C) \[\frac{1}{4}({{m}^{2}}-{{p}^{2}})\,\,\text{sq}\,\,\text{units}\]

D) \[\frac{1}{4}({{p}^{2}}-{{m}^{2}})\,\,\text{sq}\,\,\text{units}\]

Correct Answer: C

Solution :

In a rhombus

\[d_{1}^{2}+d_{2}^{2}=4{{a}^{2}}\]

Here, \[{{d}_{1}}\] and \[{{d}_{2}}\]are diagonals,

and a = Length of edge \[=\frac{2p}{4}=\frac{p}{2}\]

\[\therefore \]      \[d_{1}^{2}+d_{2}^{2}=4\times {{\left( \frac{p}{2} \right)}^{2}}\]

\[\Rightarrow \]   \[d_{1}^{2}+d_{2}^{2}={{p}^{2}}\]

On adding \[2{{d}_{1}}{{d}_{2}}\] both sides, we get

\[d_{1}^{2}+d_{2}^{2}+2{{d}_{1}}{{d}_{2}}={{p}^{2}}+2{{d}_{1}}{{d}_{2}}\]

\[\Rightarrow \]   \[{{({{d}_{1}}+{{d}_{2}})}^{2}}={{p}^{2}}+2{{d}_{1}}{{d}_{2}}\]

\[\Rightarrow \]   \[{{m}^{2}}={{p}^{2}}+2{{d}_{1}}{{d}_{2}}\]

\[[{{d}_{1}}+{{d}_{2}}=m,\text{given}]\]

\[\Rightarrow \]   \[2{{d}_{1}}{{d}_{2}}={{m}^{2}}-{{p}^{2}}\]

On dividing the whole expression by 4, we get

\[\frac{1}{2}{{d}_{1}}{{d}_{2}}=\frac{1}{4}({{m}^{2}}-{{p}^{2}})\]

and       \[\frac{1}{2}{{d}_{1}}{{d}_{2}}=\] Area of rhombus

So, the required area of rhombus \[=\frac{1}{4}({{m}^{2}}-{{p}^{2}})\]sq units