In $\triangle A B C$, if $r_1: r_2=7: 8$ and $r_1: r_3=3: 4$, then $a: b: c=$

In $\triangle A B C$, if $r_1: r_2=7: 8$ and $r_1: r_3=3: 4$, then $a: b: c=$
  1. $24: 21: 28$
  2. $8: 7: 6$
  3. $13: 14: 15$
  4. $7: 8: 6$

Solution

Given: $r_1: r_2=7: 8 \& r_1: r_3=3: 4$ Now, $r_1: r_2: r_3=7: 8: 28 / 3=21: 24: 28$ $ \begin{aligned} & \Rightarrow \frac{\Delta}{\mathrm{s}-\mathrm{a}}: \frac{\Delta}{\mathrm{s}-\mathrm{b}}: \frac{\Delta}{\mathrm{s}-\mathrm{c}}=21: 24: 28 \\ & \Rightarrow \frac{1}{\mathrm{~s}-\mathrm{a}}: \frac{1}{\mathrm{~s}-\mathrm{b}}: \frac{1}{\mathrm{~s}-\mathrm{c}}=21: 24: 28 \end{aligned} $ Let $\frac{1}{s-a}=21 \mathrm{k}, \frac{1}{s-b}=24 \mathrm{k}, \frac{1}{s-c}=28 \mathrm{k}$ $ \begin{aligned} & \Rightarrow \mathrm{s}-\mathrm{a}=\frac{1}{21 \mathrm{k}}, \mathrm{s}-\mathrm{b}=\frac{1}{24 \mathrm{k}}, \mathrm{s}-\mathrm{c}=\frac{1}{28 \mathrm{k}} \\ & \Rightarrow 3 \mathrm{~s}-(\mathrm{a}+\mathrm{b}+\mathrm{c})=\frac{1}{21 \mathrm{k}}+\frac{1}{24 \mathrm{k}}+\frac{1}{28 \mathrm{k}} \\ & \Rightarrow 3 \mathrm{~s}-2 \mathrm{~s}=\frac{1}{8 \mathrm{k}} \Rightarrow \mathrm{s}=\frac{1}{8 \mathrm{k}} \\ & \mathrm{a}=\frac{1}{8 \mathrm{k}}-\frac{1}{21 \mathrm{k}}=\frac{13}{21 \times 8 \mathrm{k}}, \mathrm{b}=\frac{1}{8 \mathrm{k}}-\frac{1}{24 \mathrm{k}}=\frac{2}{24 \mathrm{k}} \\ & \mathrm{c}=\frac{1}{8 \mathrm{k}}-\frac{1}{28 \mathrm{k}}=\frac{5}{56 \mathrm{k}} \end{aligned} $ $ \begin{aligned} & a: b: c=\frac{13}{21 \times 8 k}: \frac{2}{24 k}: \frac{5}{56 k} \\ & \Rightarrow a: b: c=13: 14: 15 \end{aligned} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

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