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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For a right-angled triangle with sides \(a, b\) and hypotenuse \(c\), the inradius is always \(\frac{a+b-c}{2}\). This is much faster than using \(\frac{\text{Area}}{s}\) during competitive exams.
We are asked to find the radius of the circle inscribed in a right triangle ΔABC with ∠B = 90°, AB = 4 cm, and BC = 3 cm. Step 1: Find the hypotenuse AC
- Using the Pythagoras theorem:
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm}
\] Step 2: Recall the formula for inradius of a right triangle
- In a right triangle with legs \(a, b\) and hypotenuse \(c\), the radius \(r\) of the inscribed circle is given by:
\[
r = \frac{a + b - c}{2}
\]
- Here, a = BC = 3 cm, b = AB = 4 cm, c = AC = 5 cm Step 3: Substitute values
\[
r = \frac{3 + 4 - 5}{2} = \frac{2}{2} = 1 \text{ cm}
\] Answer:
The radius of the inscribed circle = 1 cm
