Determine the lenghth of the radius of inscribed circle in a triangle whose lengths of sides are15,13,14.

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hala718 eNotes educator| Certified Educator

We know that :

Area of the triangle = radius of the circle * (1/2)perimeter of the triangle

==> a = r*p

==> r = p/a

But since we are giver the lebgth of the triangle sides, we can obtain a and P

P = (15 + 13 + 14)/2 = 42 /2 = 21

a= sqrt[p(p-a)(p-b)(p-c) ]  

 = sqrt[21*(21-15)(21-14)(21-13)]

 = sqrt[21*6* 7*8]

= sqrt7056.= 84

==> r= a/p = 84/ 21= 4

Then the radius is r = 4

 

a =

giorgiana1976 | Student

Because the circle is inscribed in a triangle whose lengths of it's sides are given, we'll apply the formula:

r = A/p, where r is the radius of the inscribed circle, A is the area of the triangle and p is the half-perimeter of the triangle.

We'll calculate the half-perimeter of the triangle:

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

Because we know the lengths of the sides of the triangle, we'll calculate it's area, applying Heron's formula:

A = sqrt [p(p-13)(p-14)(p-15)]

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

Now, we have all the elements to calculate the radius:

r = 84/21

r = 4

neela | Student

We know by sine rule that a/sinA = 2R . Or R = a/2sinA, where r is the radius of circumcircle.

a = 15, b= 13 and  c = 14. s = (15+13+14)/2 =21

Threfore sinA/2 = sqrt{(s-b)(s-c)/bc} = sqrt{{21-13)(21-14)/bc} = sqrt(8*7)/13*14} = sqrt (4/13)

cosA/2 = sqrt{s(s-a)/bc} = sqrt(21*6/13*14} = sqrt (9/13)

Therefore sinA = 2sinA/2*cosA/2 = 2sqrt {(4/13)(9/13) = 2*6/13 = 12/13.

Therefore R = a/2sinA =15 /2(12/13) = 15*13/12*2 = 65/8

 

 

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