question_answer

The area of the in circle of an equilateral triangle of side \[42\,\,cm\] is\[\left( \text{Take}\,\,\pi =\frac{22}{7} \right)\]

A) \[231\,\,c{{m}^{2}}\]              

B) \[462\,\,c{{m}^{2}}\]  

C) \[22\sqrt{3}c{{m}^{2}}\]         

D)        \[924\,\,c{{m}^{2}}\]

Correct Answer: B

Solution :

Let \[ABC\] be the equilateral triangle of side \[42\,\,cm\] and let \[AD\] be perpendicular from \[A\] on\[BC\]. Since the triangle is equilateral, so \[D\] bisects\[BC\]. \[\therefore \]\[BD=CD=21\,\,cm\] The centre of the inscribed circle will coincide with the centroid of\[\Delta ABC\]. Therefore,\[OD=\frac{1}{3}AD\] In\[\Delta ABC\]             \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\] \[\Rightarrow \]   \[{{42}^{2}}=A{{D}^{2}}+{{21}^{2}}\] \[\Rightarrow \]   \[AD=\sqrt{{{42}^{2}}-{{21}^{2}}}\]             \[=\sqrt{(42+21)(42-21)}\]             \[=\sqrt{63\times 21}=3\times 7\sqrt{3}cm\] \[\therefore \]      \[OD=\frac{1}{3}AD\]             \[=7\sqrt{3}cm\] \[\therefore \] Area of the incircle             \[=\pi {{(OD)}^{2}}\]             \[=\frac{22}{7}\times 7\sqrt{3}\times 7\sqrt{3}\]             \[=22\times 7\times 3=462\,\,c{{m}^{2}}\]