question_answer

                                                                                                        From any point inside an equilateral triangle, the lengths of perpendiculars 015 the sides are 'a? cm ?b? cm and ?c? cms. Its area (in\[\mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}\]) is

A)  \[\frac{\sqrt{2}}{3}\left( a+b+c \right)\]             

B)  \[\frac{\sqrt{3}}{3}{{\left( a+b+c \right)}^{2}}\]

C)  \[\frac{\sqrt{3}}{3}\left( a+b+c \right)\]             

D)  \[\frac{\sqrt{2}}{3}{{\left( a+b+c \right)}^{2}}\]

Correct Answer: B

Solution :

(b): \[OD=a\text{ }cm,OE=b\text{ }cm.\] \[OF=c\] cm. \[BC=AC=AB\] Area of \[\Delta ABC\]\[=Area\text{ }of\left( \Delta BOC+\Delta COA+\Delta BOA \right)\] \[=\frac{1}{2}\times BC\times a+\frac{1}{2}AC\times b+\frac{1}{2}\times AB\times c\] \[\frac{1}{2}BC\left( a+b+c \right)\]       ....(i) \[\left( \therefore AB=BC=CA \right)\] Again, Area of \[\Delta ABC\] \[=\frac{\sqrt{3}}{4}\times B{{C}^{2}}\] \[\therefore \frac{\sqrt{3}}{4}\times B{{C}^{2}}=\frac{1}{2}BC(a+b+c)\] \[\Rightarrow \]\[BC=\frac{2}{\sqrt{3}}\left( a+b+c \right)\] \[\therefore \]Required area \[=\frac{1}{2}\times \frac{2}{\sqrt{3}}{{\left( a+b+c \right)}^{2}}\] \[=\frac{\sqrt{3}}{\sqrt{3}\times \sqrt{3}}={{\left( a+b+c \right)}^{2}}\] \[=\frac{\sqrt{3}}{3}={{\left( a+b+c \right)}^{2}}\] sq. units