| If two medians BE and CF of a \[\Delta ABC,\] intersect each other at G and if \[BG=CG,\]\[\angle BGC=60{}^\circ ,\]\[BC=8\,\,cm,\]then area of the \[\Delta ABC\]is |
A) \[96\sqrt{3}\,\,c{{m}^{2}}\]
B) \[48\sqrt{3}\,\,c{{m}^{2}}\]
C) \[48\,\,c{{m}^{2}}\]
D) \[64\sqrt{3}\,\,c{{m}^{2}}\]
Correct Answer: B
Solution :
| Given, \[BG=GC,\]\[BC=8\,\,cm\] |
| Let \[\angle GBC=\angle GCB=x\] |
| \[\angle BGC+\angle GCB+\angle CBG=180{}^\circ \] |
| \[\Rightarrow \]\[60{}^\circ +x+x=180{}^\circ \] |
| \[\Rightarrow \]\[2x=120{}^\circ \]\[\Rightarrow \]\[x=60{}^\circ \] |
| \[\therefore \]\[\Delta BCG\]Is an equilateral triangle as all the angles are \[60{}^\circ .\] |
| Area of \[\Delta BCG=\frac{\sqrt{3}}{4}\times {{8}^{2}}=\frac{\sqrt{3}}{4}\times 8\times 8=16\sqrt{3}\,\,c{{m}^{2}}\] |
| \[\therefore \]Area of |
| \[\Delta ABC=3\times \Delta BCG=3\times 16\sqrt{3}=48\sqrt{3}\,\,c{{m}^{2}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
