question_answer

P and O are the mid-points of the sides CA and CB respectively of a \[\Delta \,ABC,\] right angled at C, then find :

(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];