A) \[\frac{1}{4}{{m}^{2}}p\,\text{sq}\,\text{unit}\]
B) \[\frac{1}{4}\,m{{p}^{2}}\,\text{sq}\,\text{unit}\]
C) \[\frac{1}{4}({{m}^{2}}-{{p}^{2}})\,\text{sq}\,\text{unit}\]
D) \[\frac{1}{4}({{p}^{2}}-{{m}^{2}})\,sq\,unit\]
Correct Answer: C
Solution :
| [c] in a rhombus \[\frac{1}{4}({{p}^{2}}-{{m}^{2}})\,sq\] Here, \[{{d}_{1}}\] and \[{{d}_{2}}\]are diagonals and a = length of edge \[=\frac{2p}{4}=\frac{p}{2}\] \[d_{1}^{2}+d_{2}^{2}=4\times {{\left( \frac{p}{2} \right)}^{2}}\] \[d_{1}^{2}+d_{2}^{2}={{p}^{2}}\] Adding \[2{{d}_{1}}{{d}_{2}}\]on both side \[d_{1}^{2}+d_{2}^{2}+2{{d}_{1}}{{d}_{2}}={{p}^{2}}+2{{d}_{1}}{{d}_{2}}\] \[{{({{d}_{1}}+{{d}_{2}})}^{2}}-{{p}^{2}}+2{{d}_{1}}{{d}_{2}}\] \[{{m}^{2}}={{p}^{2}}+2{{d}_{1}}{{d}_{2}}\] \[2{{d}_{1}}{{d}_{2}}={{m}^{2}}-{{p}^{2}}\] Dividing the whole expression by 4 \[\frac{1}{2}{{d}_{1}}{{d}_{2}}=\frac{1}{4}({{m}^{2}}-{{p}^{2}})\]and\[\frac{1}{2}{{d}_{1}}{{d}_{2}}=\]area of Rhombus So, the required area of rhombus \[=\frac{1}{4}({{m}^{2}}-{{p}^{2}})\,\text{sq}\,\text{unit}\] |
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