Answer:
4.33 unit ( approx )
Step-by-step explanation:
The radius of the circle inscribed in triangle is,
[tex]r=\frac{A}{S}[/tex]
Where,
A = Area of the triangle,
S = Semi perimeter of the triangle,
Given,
In triangle ABC,
AB = 15, AC = 41, BC = 52
[tex]S=\frac{AB+AC+BC}{2}=\frac{15+41+52}{2}=\frac{108}{2}=54[/tex]
By the Heron's formula,
Area of the triangle ABC,
[tex]A=\sqrt{S(S-15)(S-41)(S-52)}[/tex]
[tex]=\sqrt{54(54-15)(54-41)(54-52)}[/tex]
[tex]=\sqrt{54\times 39\times 13\times 2}[/tex]
[tex]=\sqrt{54756}[/tex]
[tex]=234\text{ square unit}[/tex]
Hence, the radius of the circle inscribed in triangle ABC,
[tex]r=\frac{234}{54}\approx 4.33\text{ unit}[/tex]
