A) 8 cm
B) 12cm
C) 48cm
D) 52cm
Correct Answer: C
Solution :
Directions (i-ii) Given, radius of the circle = 32 cm Let the side of the equilateral \[\Delta ABC\]be a cm. Let h be the height of the triangle.
We know that in an equilateral triangle, centroid and circumcentre coincide. \[\therefore \,\,\,AO=\frac{2}{3}\,h\,cm\] [\[\because \]centroid divides the median in the ratio 2 : 1] which is equal to the radius of circle. \[\therefore \,\,\,\,\frac{2}{3}h=32\,\,\Rightarrow h=48\,cm\]
(i) Now, we draw a perpendicular from vertex A to side BC which bisects BC at D. In right angled \[\Delta ADB\], \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\] [by Pythagoras theorem] \[\Rightarrow \,\,{{a}^{2}}={{\left( \frac{a}{2} \right)}^{2}}+{{h}^{2}}\Rightarrow {{h}^{2}}={{a}^{2}}-\frac{{{a}^{2}}}{4}=\frac{3{{a}^{2}}}{4}\] \[\Rightarrow \,\,{{\left( 48 \right)}^{2}}=\frac{3{{a}^{2}}}{4}\] [from Eq.(i)] \[\Rightarrow \,\,\,{{a}^{2}}=3072\] \[\Rightarrow \,\,\,a=\sqrt{3072}\] [taking positive square root] \[=32\sqrt{3}\,\,cm\] (i) \[32\sqrt{3}\,cm\] (ii) 48cm
You need to login to perform this action.
You will be redirected in
3 sec
