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 \[=\pi {{(OD)}^{2}}\] \[=\frac{22}{7}\times 7\sqrt{3}\times 7\sqrt{3}\] \[=22\times 7\times 3=462\,\,c{{m}^{2}}\]You need to login to perform this action.
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