A) \[2:3\]
B) \[3:4\]
C) \[3:5\]
D) \[2:5\]
Correct Answer: D
Solution :
We know that, Radius of the in-circle of triangle \[=\,\,\,\,\,\,\,\,\frac{\text{2}\times \text{ }Area\text{ }of\text{ }Triangle}{Perimeter}\] and, Radius of Circum-circle of triangle \[=\,\,\,\,\,\frac{abc}{4\times Area\,\,of\,\,Triangle}\] \[=\,\,\,\frac{Incircle\,Radius}{Circumcircle\,Radius}\] Where a, b and c are the sides of the triangle. \[S=\frac{8\times {{\Delta }^{2}}}{abc(a+b+c)}=36\] Area of triangle \[=\sqrt{36(36-18)\,(36-24)\,(36-30)}\] \[\Delta =216\,c{{m}^{2}}\] So, \[\frac{Incircle\,Radius}{Circumcircle\,\,Radius}\] \[=\frac{8\times 216\times 216}{18\times 24\times 30\times 72}=\frac{12}{30}=\underline{\mathbf{2:5}}\]
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