Question

In $\triangle ABC$, if $R=\frac{65}{8}, r r_{1}=42$ and $r_{1}-r=6.5$, then $s ( s - a )=$

• A. 147
• B. 126
• C. 105
• D. 168

Answer 1

Answer: (D)

In
$\triangle A B C, R=\frac{65}{8}, r r_{1}=42, r_{1}-r=6.5$
We know that,
$r=\frac{\Delta}{s}, r_{1}=\frac{\Delta}{s-a}, r_{2}=\frac{\Delta}{s-b}, r_{3}=\frac{\Delta}{s-c} $
$\because r r_{1}=42$
$ \Rightarrow \frac{\Delta^{2}}{s(s-a)}=42 $ and $ \frac{\Delta}{s-a}-\frac{\Delta}{s}=6.5 $
$\frac{\Delta^{2}}{s(s-a)}=42 $ and $ \frac{\Delta a}{s(s-a)}=6.5 $
$\Rightarrow \frac{a}{\Delta}=\frac{6.5}{42} \ldots$ (i)
Also,
$r_{1}+r_{2}+r_{3}-r=4 R $
$\Rightarrow r_{2}+r_{3}=4 R-\left(r_{1}-r\right) $
$\frac{\Delta}{s-b}+\frac{\Delta}{s-c}=\frac{65}{2}-\frac{65}{10}=26 $
$\frac{\Delta(2 s-(b+c)}{(s-b)(s-c)}=26 $
$\frac{\Delta a}{s(s-a)(s-b)(s-c)}=\frac{26}{s(s-a)} $
$\Rightarrow \frac{a}{\Delta}=\frac{26}{s(s-a)} \ldots $ (ii)
From Eqs. (i) and (ii), we get
$\frac{6.5}{42}=\frac{26}{s(s-c)} $
$\Rightarrow s(s-a)=\frac{26 \times 42 \times 10}{65}=168$