In any triangle ABC, if a: b: c = 2: 3: 4, th | vedclass

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(D) Given $a: b: c = 2: 3: 4$. Let $a = 2k, b = 3k, c = 4k$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2k+3k+4k}{2} = \frac{9k}{2}$. Area $\Delta =

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$16: 5$

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(D) Given $a: b: c = 2: 3: 4$. Let $a = 2k, b = 3k, c = 4k$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2k+3k+4k}{2} = \frac{9k}{2}$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{9k}{2}(\frac{9k}{2}-2k)(\frac{9k}{2}-3k)(\frac{9k}{2}-4k)} = \sqrt{\frac{9k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2} \cdot \frac{k}{2}} = \frac{3k^2\sqrt{15}}{4}$. We know $R = \frac{abc}{4\Delta}$ and $r = \frac{\Delta}{s}$. Thus, $\frac{R}{r} = \frac{abc \cdot s}{4\Delta^2} = \frac{(2k)(3k)(4k) \cdot (9k/2)}{4 \cdot (\frac{9k^2\sqrt{15}}{4})^2} = \frac{24k^3 \cdot 9k/2}{4 \cdot \frac{81k^4 \cdot 15}{16}} = \frac{108k^4}{\frac{81k^4 \cdot 15}{4}} = \frac{108 \cdot 4}{81 \cdot 15} = \frac{432}{1215} = \frac{16}{45}$. Wait, re-evaluating: $\frac{R}{r} = \frac{abc}{4 \cdot \frac{\Delta^2}{s}} = \frac{abc}{4 \cdot s(s-a)(s-b)(s-c)/s} = \frac{abc}{4(s-a)(s-b)(s-c)}$. $\frac{R}{r} = \frac{(2k)(3k)(4k)}{4(\frac{5k}{2})(\frac{3k}{2})(\frac{k}{2})} = \frac{24k^3}{4 \cdot \frac{15k^3}{8}} = \frac{24k^3}{\frac{15k^3}{2}} = \frac{48}{15} = \frac{16}{5}$. Therefore, $R: r = 16: 5$.

In any triangle $ABC$, if $a: b: c = 2: 3: 4$, then $R: r =$

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