If in a \[\Delta ABC,\] BE and CF are two medians perpendicular to each other and if AB = 19 and AC = 22 cm, then the length of BC is [SSC (CGL) Pre 2015] |
A) 26 cm
B) 20.5 cm
C) 13 cm
D) 19.5 cm
Correct Answer: C
Solution :
In \[\Delta BOF,\] |
\[{{(BO)}^{2}}+{{(OF)}^{2}}={{\left( \frac{19}{2} \right)}^{2}}\] |
\[\Rightarrow \] \[{{\left( \frac{2}{3}BE \right)}^{2}}+{{\left( \frac{1}{3}CF \right)}^{2}}={{\left( \frac{19}{2} \right)}^{2}}\] |
\[\Rightarrow \] \[\frac{4}{9}{{(BE)}^{2}}+\frac{1}{9}{{(CF)}^{2}}={{\left( \frac{19}{2} \right)}^{2}}\] (i) |
In \[\Delta COE,\]\[{{(CO)}^{2}}+{{(OE)}^{2}}={{(11)}^{2}}\] |
\[\Rightarrow \] \[{{\left( \frac{2}{3}CF \right)}^{2}}+{{\left( \frac{1}{3}BE \right)}^{2}}=121\] |
\[\Rightarrow \] \[\frac{4}{9}{{(CF)}^{2}}+\frac{1}{9}{{(BE)}^{2}}=121\] (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 |
\[2\,{{(BE)}^{2}}+2\,{{(CF)}^{2}}=-\,2\,{{(BC)}^{2}}=119+484-\frac{361}{2}\] |
\[\Rightarrow \] \[2\,(B{{E}^{2}}+C{{F}^{2}})-2B{{C}^{2}}=603-\frac{361}{2}\] |
\[\Rightarrow \] \[2\times \frac{1521}{4}-2\,{{(BC)}^{2}}=603-\frac{361}{2}\] |
\[\Rightarrow \] \[2\,{{(BC)}^{2}}=\frac{1521}{2}+\frac{361}{2}-603\] |
\[\Rightarrow \] \[2\,{{(BC)}^{2}}=338\]\[\Rightarrow \]\[{{(BC)}^{2}}=169\] |
\[\therefore \] \[BC=13\,cm\] |
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