A triangle has sides with lengths of 4, 9, and 7. What is the radius of the triangles inscribed circle?

Question

A triangle has sides with lengths of 4, 9, and 7.  What is the radius of the triangles inscribed circle?

Autumn Armstrong · Accepted Answer

#Radius=(3sqrt5)/5#

Radius of circle inscribed in a triangle#=A/s#

Where,

#A#=Area of triangle,

#s#=Semi-perimeter of triangle#=(a+b+c)/2# Note #a,b,c# are sides of the triangle

So,#s=(4+9+7)/2=20/2=10#

We can find the area of triangle using Heron's formula: Heron's formula: #Area = sqrt(s(s-a)(s-b)(s-c)) #

#rarrArea=sqrt(10(10-4)(10-9)(10-7))#

#Area=sqrt(10(6)(1)(3))#

#Area=sqrt(10(18))#

#Area=sqrt180=6sqrt5#

#Radius=A/s=(6sqrt5)/10=(3sqrt5)/5#

Allison Allen · Answer

undefined The radius of the triangle's inscribed circle, also known as the inradius, can be calculated using the formula:

$ r = \frac{2 \times \text{Area of the triangle}}{\text{Perimeter of the triangle}} $

Given that the sides of the triangle have lengths of 4, 9, and 7, the semiperimeter (half of the perimeter) is $ s = \frac{4+9+7}{2} = 10 $.

Using Heron's formula, we can calculate the area of the triangle:
$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $
where $a$, $b$, and $c$ are the lengths of the sides of the triangle.

$ \text{Area} = \sqrt{10(10-4)(10-9)(10-7)} = \sqrt{10 \times 6 \times 1 \times 3} = \sqrt{180} = 6\sqrt{5} $

Substituting the area and semiperimeter into the formula for the inradius:
$ r = \frac{2 \times 6\sqrt{5}}{10} = \frac{12\sqrt{5}}{10} = \frac{6\sqrt{5}}{5} $

So, the radius of the triangle's inscribed circle is $ \frac{6\sqrt{5}}{5} $.