| The medians AD, BE and CF of a \[\Delta ABC\] are of lengths of \[18\,\,cm,\] \[24\,\,cm\] and \[30\,\,cm,\] respectively. The area \[(\text{in}\,\,c{{m}^{2}})\]of the triangle is [SSC (CGL) 2013] |
A) \[96\,\,c{{m}^{2}}\]
B) \[192\,\,c{{m}^{2}}\]
C) \[288\,\,c{{m}^{2}}\]
D) \[374\,\,c{{m}^{2}}\]
Correct Answer: C
Solution :
| Semi-perimeter, |
| \[s=\frac{18+24+30}{2}=\frac{72}{2}=36\,\,cm\] |
| Now, area of triangle \[=\frac{4}{3}\sqrt{s\,\,(s-u)(s-\text{v})(s-w)}\] |
| \[=\frac{4}{3}\sqrt{36\,\,(36-18)(36-24)(36-30)}\] |
| \[=\frac{4}{3}\sqrt{36\times 18\times 12\times 6}\] |
| \[=\frac{4}{3}\times 6\times 3\times 2\times 2\times 3\] |
| \[=24\times 12=288\,\,c{{m}^{2}}\] |
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