Question

Let abc be a triangle with sides a= 13,b=14,c=15.then radius of the inscribed circle is:

Answer 1

r=∆/s
∆=√[(s)(s-a)(s-b)(s-c)]
s=(a+b+c)/2=42/2=21
∆=√[21×8×7×6] = 7×3×4 = 84

r=84/21=4

Answer 2

Therefore the radius of the inscribed circle is $4$

Step-by-step explanation:

Given;

For a triangle,

$a=13$ , $b=14$ and $c=15$

By Hero's formula;

Area of triangle $abc=\sqrt{s(s-a)(s-b)(s-c)}$  where $s=\frac{1}{2} (a+b+c)$

Then,

  $s=\frac{1}{2} (13+14+15)$

⇒$s=21$

Now, Area of triangle $abc=\sqrt{21(21-13)(21-14)(21-15)}$$=84$

From figure;

Area of $\bigtriangleup abc$$=$Area of $\bigtriangleup aob$$+$ Area of $\bigtriangleup boc$+Area of $\bigtriangleup aoc$

⇒Area of $\bigtriangleup abc=(\frac{1}{2} \times or \times ab)+(\frac{1}{2} \times op \times bc)+(\frac{1}{2} \times oq \times ac)$

⇒Area of $\bigtriangleup abc=(\frac{1}{2} \times r \times 13)+(\frac{1}{2} \times r \times 14)+(\frac{1}{2} \times r \times 15)$  (where $r$ is the radius of the circle)

⇒Area of $\bigtriangleup abc=\frac{13r}{2}+\frac{14r}{2} +\frac{15r}{2}$

⇒$84=21r$

∴ $r=4$

So the radius of the inscribed circle is $4$