(i) \[4A{{C}^{2}}+B{{C}^{2}}\] |
(ii) \[4B{{C}^{2}}+A{{C}^{2}}\] |
(iii) \[4(A{{Q}^{2}}+B{{P}^{2}})\] |
A)
i-\[4A{{Q}^{2}}\] ii-\[4B{{P}^{2}}\] iii-\[5A{{B}^{2}}\]
B)
i-\[5A{{Q}^{2}}\] ii-\[5B{{P}^{2}}\] iii-\[~4A{{B}^{2}}\]
C)
i-\[4A{{Q}^{2}}\] ii-\[5B{{P}^{2}}\] iii-\[5A{{B}^{2}}\]
D)
i-\[5A{{Q}^{2}}\] ii-\[4B{{P}^{2}}\] iii-\[~4A{{B}^{2}}\]
Correct Answer: A
Solution :
Construction : Join PQ, BP and AQ. (i) \[4A{{C}^{2}}+B{{C}^{2}}=4A{{C}^{2}}+{{(2QC)}^{2}}\] [\[\because \] Q is mid-point of BC] \[=4A{{C}^{2}}+4Q{{C}^{2}}\] \[=4(A{{C}^{2}}+Q{{C}^{2}})\] \[4A{{Q}^{2}}\] [ \[\because \] AQC is right angled triangle] (ii) \[4B{{C}^{2}}+A{{C}^{2}}=4B{{C}^{2}}+{{(2CP)}^{2}}\] [\[\because \] Q is mid-point of AC] \[=4B{{C}^{2}}+4C{{P}^{2}}=4(B{{C}^{2}}+C{{P}^{2}})=4B{{P}^{2}}\] [\[\because \] PBC is right angled triangle]; (iii) \[4[A{{C}^{2}}+Q{{C}^{2}}]=4{{(AQ)}^{2}}\] [from (i) part] \[\Rightarrow \] \[4A{{C}^{2}}+B{{C}^{2}}=4A{{Q}^{2}}\] ...(i) \[4[B{{C}^{2}}+C{{P}^{2}}]=4{{(BP)}^{2}}\] [from (ii) part] \[\Rightarrow \] \[4B{{C}^{2}}+A{{C}^{2}}=4B{{P}^{2}}\] ...(ii) Now, adding (i) and (ii), we get \[4(A{{Q}^{2}}+B{{P}^{2}})=5(A{{C}^{2}}+B{{C}^{2}})=5A{{B}^{2}}\] [\[\because \] ABC is right angled triangle];You need to login to perform this action.
You will be redirected in
3 sec