A) \[\frac{15}{4}\]
B) \[\frac{11}{5}\]
C) \[\frac{16}{7}\]
D) \[\frac{16}{3}\]
Correct Answer: C
Solution :
Let sides of the triangle are \[4x,5x,6x.\] \[s=\frac{4x+5x+6x}{2}=\frac{15}{2}x\] \[\Delta =\sqrt{\frac{15}{2}x\left( \frac{15}{2}x-4x \right)\left( \frac{15}{2}x-5x \right)\left( \frac{15}{2}x-6x \right)}\] \[=\sqrt{\frac{15}{2}x\times \frac{7}{2}x\times \frac{5}{2}x\times \frac{3}{2}x}\] \[=\frac{15\sqrt{7}{{x}^{2}}}{4}\] Circumradius, \[R=\frac{4x\times 5x\times 6x}{4\times \frac{15\sqrt{7}{{x}^{2}}}{4}}\] \[=\frac{8}{\sqrt{7}}x\] Inradius, \[r=\frac{\frac{15\sqrt{7}}{4}{{x}^{2}}}{\frac{15}{2}x}\] \[=\frac{\sqrt{7}}{2}xax\] \[\frac{R}{r}=\frac{\frac{8x}{\sqrt{7}}}{\frac{\sqrt{7x}}{2}}=\frac{16}{7}\]
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