In a triangle ABC if the sides are in the ratio 4:5:7 then r:R=?

[admins]

In a triangle ABC if the sides are in the ratio 4:5:7 then r:R=?

 

[anisingh259]

The ratio of the r:R is solved in the image given below:

$\rightarrow \quad r$ (iuradius) :-
-PAGE NO:
72

$$
\left[s=\frac{a+b+c}{2}\right]
$$


$$
\begin{gathered}
x=r s . \\
K=\sqrt{s(s-a)(s-b)(s-c)}
\end{gathered}
$$

let the sides of the $\Delta$ be: $-4 x, 5 x, 7 x$.

$$
\begin{aligned}
& \sqrt{8 x(4 x) \cdot(3 x) \cdot(1 x)}=r 8 x . \\
\rightarrow & \gamma=\sqrt{\frac{3}{2}} x .
\end{aligned}
$$

$\rightarrow$ radius of outer circle of $\Delta:-(R)$

$$
\begin{aligned}
R & =\frac{a b c}{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}} \begin{aligned}
R & =\frac{4 x \cdot 5 x \cdot 7 x}{\sqrt{16 x \cdot 8 x \cdot 6 x \cdot 2 x}}=\frac{4 \cdot 5 \cdot 7 \times x^2}{16 x^2 \sqrt{3 \cdot 2}} \\
R & =\frac{35 x}{4 \sqrt{3 \cdot 2}} \\
\frac{r}{R} & =\frac{\sqrt{3 / 2} x}{35 x} \\
R & =\frac{4 \cdot \sqrt{3 / 2}}{35} \\
R & =\frac{12}{3}
\end{aligned}
\end{aligned}
$$

 

[solution.livedoubt]

Given, sides of triangle are in the ratio 4:5:7.

So, \(a=4k,b=5k,c=7k\)

Semi-perimeter is\[ s=\frac{\left(a+b+c\right)}{2}=8k \]
Area of triangle is \[\Delta=\sqrt{(\mathrm{s})(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\]\[ =\sqrt{8k\left(8k-4k\right)\left(8k-5k\right)\left(8k-7k\right)} \]\[=\sqrt{8k\left(4k\right)\left(3k\right)\left(k\right)}=4\sqrt6k^2\]
Now, \[\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{4.5.7.k^3}{4.4\sqrt6k^2}=\frac{5\left(7k\right)}{4\sqrt6}\]
and \[\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{4\sqrt6k^2}{8k}=\frac{\sqrt6k}{2}\]
Thus,
\[\frac{\mathrm{r}}{\mathrm{R}}=\frac{\sqrt6k}{2}\times\frac{4\sqrt6}{5\left(7k\right)}=\frac{12}{35}\]
Therefore, \[ r:R=12:35 \]