A) \[1:2\]
B) \[1:3\]
C) \[1:4\]
D) \[1:9\]
Correct Answer: C
Solution :
| [c] Let side of equilateral triangle = x units |
| \[\therefore \] Altitude \[=\frac{\sqrt{3}}{2}=x\] units |
| Radius of in circle \[=\frac{1}{3}\times \frac{\sqrt{3}}{2}\times \text{units}\] |
| \[=\frac{\sqrt{3}}{6}\times \text{units}\] |
| Radius of circumcircle |
| \[=\frac{2}{3}\times \frac{\sqrt{3}}{2}\times \text{units=}\frac{\sqrt{3}}{3}\times \text{units}\] |
| \[\therefore \frac{\text{Area of incircle}}{\text{Area of circumcircle}}=\frac{\pi {{\left( \frac{\sqrt{3}}{6}x \right)}^{2}}}{\pi {{\left( \frac{\sqrt{3}}{3}x \right)}^{2}}}\frac{9}{36}=\frac{1}{4}=1:4\] |
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