The radius of the incircle of the triangle with sides 25 cm, 51 cm, and 52
cm is 26 cm.
The area of the triangle is given as follows;
A = √(s·(s - a)·(s - b)·(s - c))
Where;
a = 25 cm
b = 51 cm
c = 52 cm
s = The semi perimeter of the triangle = (25 + 51 + 52) ÷ 2 = 64
A = √(64·(64 - 25)·(64 - 51)·(64 - 52)) = 624
The area of the triangle is the area of ΔABC + ΔAOC + ΔBOC + ΔAOB
Which gives;
[tex]A = \mathbf{\dfrac{a \cdot r}{2} + \dfrac{b \cdot r}{2} + \dfrac{c \cdot r}{2}}[/tex]
Therefore;
[tex]A = r \cdot \left(\dfrac{a + b + c }{2} \right) = \mathbf{r \cdot s}[/tex]
[tex]r = \dfrac{A}{s}[/tex]
Therefore;
[tex]r = \dfrac{624}{64} = 26[/tex]
The radius of the incircle of the triangle is 26 cm
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