In $△ A B C, m ∠ B = 90$ . $A$ circle touches all the sides of $△ A B C$ . If $A B = 5, B C = 12$ , find the radius of the circle.

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Solution

Since $Δ A B C$ is Right Angled
$A C^{2} = A B^{2} + B C^{2}$
$⟹ A C = 13$
Now Consider The $Δ A B C$ to be on Coordinate axis
with $A B$ on $x-axis$ and $B C$ on $y-axis$
And the side $A C$ is given by $\frac{x}{5} + \frac{y}{12} = 1 ⟹ 12 x + 5 y - 60 = 0$

$Incenter$ is given by $(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$

where $a = 12, b = 13, c = 5$

$(x_{1}, y_{1}) = (5, 0); (x_{2}, y_{2}) = (0, 0); (x_{3}, y_{3}) = (0, 12)$
By substituting,
$Incenter = (2, 2)$
So the Radius is $2$

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In △ABC,m∠B=90. A circle touches all the sides of △ABC. If AB=5,BC=12, find the radius of the circle.

In $△ A B C, m ∠ B = 90$ . $A$ circle touches all the sides of $△ A B C$ . If $A B = 5, B C = 12$ , find the radius of the circle.