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In the given figure, $\Delta ABC$ is a right triangle in which $\angle B = 90^\circ$, $AB = 4$ cm and $BC = 3$ cm. Find the radius of the circle inscribed in the triangle ABC.
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$AC = \sqrt{3^2 + 4^2} = 5$ cm (1/2 Mark)
Let $BE = BD = x$ cm (1/2 Mark)
$AD = 4 - x = AF$, $CE = 3 - x = CF$ (1/2 Mark)
$AF + CF = AC \implies 4 - x + 3 - x = 5$ (1/2 Mark)
$\therefore x = 1$ (1 Mark)
$BD = BE = 1$ and $\angle B = 90^\circ$
Hence radius of circle $= x = 1$ cm (1/2 Mark)
**ALTERNATE SOLUTION:**
$AC = \sqrt{3^2 + 4^2} = 5$ cm (1/2 Mark)
Let $r$ be the radius of the circle
$ar(\Delta ABC) = \frac{1}{2} \times 4 \times 3 = 6$ cm$^2$ (1/2 Mark)
Also, $ar(\Delta ABC) = (\frac{1}{2} \times r \times 4) + (\frac{1}{2} \times r \times 3) + (\frac{1}{2} \times r \times 5)$ (1 Mark)
$\implies 6r = 6 \implies r = 1$ (1 Mark)
Hence the radius of the circle is $1$ cm.
Let $BE = BD = x$ cm (1/2 Mark)
$AD = 4 - x = AF$, $CE = 3 - x = CF$ (1/2 Mark)
$AF + CF = AC \implies 4 - x + 3 - x = 5$ (1/2 Mark)
$\therefore x = 1$ (1 Mark)
$BD = BE = 1$ and $\angle B = 90^\circ$
Hence radius of circle $= x = 1$ cm (1/2 Mark)
**ALTERNATE SOLUTION:**
$AC = \sqrt{3^2 + 4^2} = 5$ cm (1/2 Mark)
Let $r$ be the radius of the circle
$ar(\Delta ABC) = \frac{1}{2} \times 4 \times 3 = 6$ cm$^2$ (1/2 Mark)
Also, $ar(\Delta ABC) = (\frac{1}{2} \times r \times 4) + (\frac{1}{2} \times r \times 3) + (\frac{1}{2} \times r \times 5)$ (1 Mark)
$\implies 6r = 6 \implies r = 1$ (1 Mark)
Hence the radius of the circle is $1$ cm.