Discussion Forum

The lengths of the sides  of a triangle are13,14 and 15. If R and r repectively  denote circumradius and inradius of that  triangle , then  8R+r=

Answer:

Explanation:

Let a=13, b=14 and c=15, Then

 $S= \frac{a+b+c}{2}=\frac{13+14+15}{2}=21$

And area of triangle

          $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$

                              = $\sqrt{21(21-13)(21-14)(21-15)}$

            $\triangle=\sqrt{21\times8\times7\times6}=7\times3\times4=84$

Now, as we know   $R=\frac{abc}{4 \triangle}  $ and $r=\frac{\triangle}{s}$

 $\therefore$       $R= \frac{ 13 \times 14 \times 15}{4 \times 84}$ and

  $r=\frac{84}{21}$

$\Rightarrow$   $R=\frac{65}{8}$  and r=4

 So, 8R+r=65+4=69