A) \[R{{X}^{2}}+P{{Y}^{2}}=5X{{Y}^{2}}\]
B) \[R{{X}^{2}}+P{{Y}^{2}}=X{{Y}^{2}}+P{{R}^{2}}\]
C) \[4(R{{X}^{2}}+P{{Y}^{2}})=5P{{R}^{2}}\]
D) \[R{{X}^{2}}+P{{Y}^{2}}=3\text{ (}P{{Q}^{2}}+Q{{R}^{2}})\]
Correct Answer: C
Solution :
\[R{{X}^{2}}+P{{Y}^{2}}=(Q{{X}^{2}}+Q{{R}^{2}})+(P{{Q}^{2}}+Q{{Y}^{2}})\] \[=Q{{X}^{2}}+Q{{Y}^{2}}+P{{Q}^{2}}+Q{{R}^{2}}\] \[{{\left( \frac{1}{2}PQ \right)}^{2}}+{{\left( \frac{1}{2}QR \right)}^{2}}+P{{Q}^{2}}+Q{{R}^{2}}\] [\[\because \] X and Y are mid-points of PQ and QR respectively] \[=\,\,\,\,\,\,\frac{5}{4}\,P{{Q}^{2}}+\frac{5}{4}Q{{R}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,4(R{{X}^{2}}+P{{Y}^{2}})=5(P{{Q}^{2}}+Q{{R}^{2}})\] \[\Rightarrow \,\,\,\,\,\,\,4(R{{X}^{2}}+P{{Y}^{2}})=\] \[[\because \,\,\,\,\,P{{Q}^{2}}+Q{{R}^{2}}=P{{R}^{2}}]\]You need to login to perform this action.
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