Question

In $△ABC$, 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
$△ABC,R=\frac{65}{8},r{r}_1=42,r_1-r=6.5$
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 $\frac{\Delta a}{s(s-a)}=6.5$
$\Rightarrow \frac{a}{\Delta }=\frac{6.5}{42}\dots$ (i)
Also,
$r_1+r_2+r_3-r=4R$
$\Rightarrow r_2+r_3=4R-\left(r_1-r\right)$
$\frac{\Delta }{s-b}+\frac{\Delta }{s-c}=\frac{65}{2}-\frac{65}{10}=26$
$\frac{\Delta (2s-(b+c)}{(s-b)(s-c)}=26$
$\frac{\Delta a}{s(s-a)(s-b)(s-c)}=\frac{26}{s(s-a)}$
$\Rightarrow \frac{a}{\Delta }=\frac{26}{s(s-a)}\dots$ (ii)
From Eqs. (i) and (ii), we get
$\frac{6.5}{42}=\frac{26}{s(s-c)}$
$\Rightarrow s(s-a)=\frac{26\times 42\times 10}{65}=168$