question_answer

In \[\Delta \,ABC,\]if the length of medians AD, BE and CF are 9 cm, 12 cm and 15cm then what will be the area of triangle in\[(c{{m}^{2}})\].

A)  \[288\text{ }c{{m}^{2}}\]                  

B)  \[\text{96 }c{{m}^{2}}\]

C)  \[\text{144 }c{{m}^{2}}\]                  

D)  \[\text{72 }c{{m}^{2}}\]

Correct Answer: D

Solution :

ABC is a triangle and AD, BE and CF are median of triangle. Given, \[{{S}_{1}}=9\,cm.,\,\,{{S}_{2}}=12\,cm.,\,{{S}_{3}}=15\,cm.\] \[{{S}_{m}}=\frac{{{S}_{1}}+{{S}_{2}}+{{S}_{3}}}{2}=\frac{9+12+15}{2}\] \[=\,\,\,\,\frac{36}{2}=18\] Area of triangle \[=\frac{4}{3}\,\,\sqrt{{{S}_{m}}({{S}_{m}}-{{S}_{1}})\,({{S}_{m}}-{{S}_{2}})\,({{S}_{m}}-{{S}_{3}})}\] \[=\frac{4}{3}\,\,\sqrt{18(18-9)\,(18-12)\,(18-15)}\] \[=\frac{4}{3}\,\,\sqrt{18\times 9\times 6\times 3}\] \[=\frac{4}{3}\,\,\sqrt{6\times 3\times 6\times 3\times 3\times 3}\] \[=\frac{4}{3}\times 6\times 3\times 3=\mathbf{72}\,\mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}\]