If \(a, b\) and \(c\) are the sides of \(\triangle A B C\) for which \(r_1=8, r_2=12\) and \(r_3=24\), then…
If \(a, b\) and \(c\) are the sides of \(\triangle A B C\) for which \(r_1=8, r_2=12\) and \(r_3=24\), then the ordered triad \((a, b, c)=\)
\((12,20,16)\)
\((12,16,20)\)
\((16,12,20)\)
\(20,16,12)\)
Solution
It is given that
\(\begin{aligned}
r_1 & =8=\frac{\Delta}{s-a} \quad \ldots (i) \\
& r_2=12=\frac{\Delta}{s-b} \quad \ldots (ii) \\
\text { and } \quad r_3 & =24=\frac{\Delta}{s-c} \quad \ldots (iii)
\end{aligned}\)
From Eqs. (i) and (ii), we get
\(\begin{gathered}
\quad \frac{s-b}{s-a}=\frac{2}{3} \Rightarrow 3 s-3 b=2 s-2 a \\
\Rightarrow \quad 5 a+c=5 b \quad \ldots (iv)
\end{gathered}\)
From Eqs. (ii) and (iii), we get
\(\begin{array}{cc}
& \frac{s-c}{s-b}=\frac{1}{2} \\
\Rightarrow & 2 s-2 c=s-b \\
\Rightarrow \quad & a+3 b=3 c \quad \ldots (v)
\end{array}\)
and from Eqs. (i) and (iii), we get
\(\begin{array}{cc}
& \frac{s-c}{s-a}=\frac{1}{3} \\
\Rightarrow & 3 s-3 c=s-a \\
\Rightarrow \quad & 2 a+b=2 c \quad \ldots (vi)
\end{array}\)
On solving Eqs. (iv), (v) and (vi), we get
\((a, b, c)=(12,16,20)\)
Hence, option (2) is correct.