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 |
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