Question

In triangle $ABC$, if $r_1=36,r_2=18$ and $r_3=12$, then $a+b=$

• A. 36
• B. 24
• C. 30
• D. 54

Answer 1

Answer: (D)

It is given in a$\Delta ABC$
$r_1=\frac{\Delta }{s-a}=36,r_2=\frac{\Delta }{s-b}=18$ and $r_3=\frac{\Delta }{s-c}=12$
$\Rightarrow \frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{36}+\frac{1}{18}+\frac{1}{12}=\frac{1+2+3}{36}=\frac{1}{6}$
$\Rightarrow \frac{s-a+s-b+s-c}{\Delta }=\frac{1}{6}$
$\Rightarrow \frac{s}{\Delta }=\frac{1}{6}$
$\Rightarrow \Delta =6s$
So, $r_1=\frac{\Delta }{s-a}=\frac{6s}{s-a}=36$
$\Rightarrow s=6s-6a$
$\Rightarrow 6a=5s$
$r_2=\frac{\Delta }{s-b}=\frac{6s}{s-b}=18$
$\Rightarrow s=3s-3b$
$\Rightarrow 3b=2s$
and $r_3=\frac{\Delta }{s-c}=\frac{6s}{s-c}=12$
$\Rightarrow s=2s-2c$
$\Rightarrow 2c=s$
So, ${\Delta }^2=36s^2$
$\Rightarrow s(s-a)(s-b)(s-c)=36s^2$
$\Rightarrow \left(s-\frac{5s}{6}\right)\left(s-\frac{2s}{3}\right)\left(s-\frac{s}{2}\right)=36s$
$\Rightarrow s^3\left(\frac{1}{6}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=36s$
$\Rightarrow s^2=(36{)}^2$
$\Rightarrow s=36$
$a=30$ and $b=24$
So, $a+b=30+24=54$