# Find out the lenght of the circle radius inscribed in the triangle which has the sides lenghts of 12, 14, 15 cm

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We know that:

area of the triangle= radius of the circle* perimeter of the triangle/2

OR A= R*P/2

The perimeter= 12+14+15= 41

The p/2= 20.5

We can obtain the area of the triangle through the rule:

a= sqrt(p(p-12)(p-14)(p-15)

= sqrt(20.5)(8.5)(6.5)(5.5)

= 79 (approx.)

Then :

A=R+p/2

79= R*20.5

==> R= 79/20.5= 3.9 (approx.)

Let O be the centre of the circle inscribed in the triangle ABC and let the cirlcle touch the sided BC,CA and AB at D,E and F. Then OD = OE = OF = r , the radius of the inscribed circle. So OD , OE and OF are pependiculars to BC,CA and AB as the tangent at D , E, F aare pependicular to the radius DO, EO and FO respectively.

Therefore the area of triangles OBC+ OCA+OAB = area of ABC Or

(1/2) a*OD+(1/2)b*OE+(1/2)c*OF = Area ABC = sqrt{(s)(s-a)s-b)(s-c)}, where a, b,c are the sides of the sides opposite to A,B, C and s = (a+b+c)/2.

(1/2) r (a+b+c) = sqrt[ s(s-a)(s-b)(s-c)}. Solving for r, we get:

r = (1/s)sqrt{s(s-a)(s-b)(s-c)}. Given in this example, a= 12 cm,b=14cm and c =15 cm. So s = (12+14+15)/2 = 20.5cm

r = (1/20.5)sqrt{ 20.5*(20.5-12)(20.5-14)(20.5-16)}

= (1/20.5)sqrt{20.5*8.5*6.5*4.5}

= 3.4825 cm is the radius of the circle.

Because of the fact that we know the lengths of the sides of the triangle , we could calculate the radius of the inscribed circle, using the following formula:

R=S/p, where R-radius, S-aria of the triangle and p-half-perimeter of the triangle.

p=(13+14+15)/2=21

S= sqrt [21(21-13)(21-14)(21-15)]=84

**R=84/21=4**