A) 26 cm
B) 20.5 cm
C) 13 cm
D) 19.5 cm
Correct Answer: C
$ \Rightarrow $ $ {{( \frac{2}{3}BE )}^{2}}+{{( \frac{1}{3}CF )}^{2}}={{( \frac{19}{2} )}^{2}} $
$ \Rightarrow $ $ \frac{4}{9}{{(BE)}^{2}}+\frac{1}{9}{{(CF)}^{2}}={{( \frac{19}{2} )}^{2}} $ (i) In $ \Delta COE, $ $ {{(CO)}^{2}}+{{(OE)}^{2}}={{(11)}^{2}} $
$ \Rightarrow $ $ {{( \frac{2}{3}CF )}^{2}}+{{( \frac{1}{3}BE )}^{2}}=121 $
$ \Rightarrow $ $ \frac{4}{9}{{(CF)}^{2}}+\frac{1}{9}{{(BE)}^{2}}=121 $ (ii) On adding Eqs. (i) and (ii), we have $ \frac{5}{9}{{(BE)}^{2}}+\frac{5}{9}{{(CF)}^{2}}=121+\frac{361}{4} $
$ \Rightarrow $ $ {{(BE)}^{2}}+{{(CF)}^{2}}=\frac{845}{4}\times \frac{9}{5}=\frac{1521}{4} $ (iii) By Apollonius theorem, $ {{(AB)}^{2}}+{{(BC)}^{2}}=2,[{{(BE)}^{2}}+{{(AE)}^{2}}] $
$ \Rightarrow $ $ 19^{2}+{{(BC)}^{2}}=2,{{(BE)}^{2}}+2\times {{(11)}^{2}} $
$ \Rightarrow $ $ 2,{{(BE)}^{2}}-{{(BC)}^{2}}=361-242 $
$ \Rightarrow $ $ 2,{{(BE)}^{2}}-{{(BC)}^{2}}=119 $ (iv) Again by Apollonius theorem, $ {{(AC)}^{2}}+{{(BC)}^{2}}=2,[{{(CF)}^{2}}+{{(AF)}^{2}}] $
$ \Rightarrow $ $ 22^{2}+{{(BC)}^{2}}=2,{{(CF)}^{2}}+2\cdot \frac{19^{2}}{4} $
$ \Rightarrow $ $ 2,{{(CF)}^{2}}-{{(BC)}^{2}}=484-\frac{361}{2} $ (v) Now, on adding Eqs. (iv) and (v), we get $ 2,{{(BE)}^{2}}+2,{{(CF)}^{2}}=-,2,{{(BC)}^{2}}=119+484-\frac{361}{2} $
$ \Rightarrow $ $ 2,(BE^{2}+CF^{2})-2BC^{2}=603-\frac{361}{2} $
$ \Rightarrow $ $ 2\times \frac{1521}{4}-2,{{(BC)}^{2}}=603-\frac{361}{2} $
$ \Rightarrow $ $ 2,{{(BC)}^{2}}=\frac{1521}{2}+\frac{361}{2}-603 $
$ \Rightarrow $ $ 2,{{(BC)}^{2}}=338 $
$ \Rightarrow $ $ {{(BC)}^{2}}=169 $
$ \therefore $ $ BC=13,cm $
BETA
