A) \[O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}\]
B) \[O{{D}^{2}}+O{{E}^{2}}+O{{F}^{2}}\]
C) \[A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}}\]
D) \[A{{E}^{2}}+B{{F}^{2}}+C{{D}^{2}}\]
Correct Answer: D
Solution :
In \[\Delta ODB\]and \[\Delta ODC,\] using Pythagoras theorem. \[O{{B}^{2}}=O{{D}^{2}}+B{{D}^{2}}\]and \[O{{C}^{2}}-O{{D}^{2}}+C{{D}^{2}}\] \[\therefore \] \[O{{B}^{2}}-O{{C}^{2}}=B{{D}^{2}}-C{{D}^{2}}\] ?.(i) Similarity, we have \[O{{C}^{2}}-O{{A}^{2}}=C{{E}^{2}}-A{{E}^{2}}\] ?..(ii) And \[O{{A}^{2}}-O{{B}^{2}}=C{{E}^{2}}-A{{E}^{2}}\] ?..(iii) Adding (i), (ii) and (iii), we get \[B{{D}^{2}}+C{{E}^{2}}+A{{F}^{2}}-C{{D}^{2}}-A{{E}^{2}}-B{{F}^{2}}=0\] \[\Rightarrow \] \[B{{D}^{2}}+C{{E}^{2}}+A{{F}^{2}}=C{{D}^{2}}+A{{E}^{2}}+B{{F}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
