Question

ABC is a right-angle triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are \[6cm\]and \[8cm\], then the radius of the circle is:
A. \[3cm\]
B. \[2cm\]
C. \[4cm\]
D. \[8cm\]

Answer 1

Hint: As, we are given an right angled triangle with the lengths of two sides, we first find the length of hypotenuses using the Pythagoras theorem. As, we have a right angled triangle we can easily calculate its area. Now also as we know the sides we calculate the semi-perimeter of triangle, then we use the formula of radius of incentre to calculate the radius of circle which is inscribed in the given triangle.

Complete step by step answer:

As given that the lengths of the two sides containing the right angle are \[6cm\]and \[8cm\]

Diagram:

Hence, then calculating third side using Pythagoras theorem,
\[B{C^2} = A{B^2} + A{C^2}\]
On substituting values we get,
\[ \Rightarrow \]\[B{C^2} = {8^2} + {6^2}\]
Hence, expanding the above term
\[ \Rightarrow \]\[B{C^2} = 64 + 36\]
On addition, we get,
\[ \Rightarrow \]\[B{C^2} = 100\]
On taking square root on both side, we get,
\[ \Rightarrow \]\[BC = 10cm\]
Hence, calculating the area of triangle using \[\Delta ABC = \dfrac{1}{2} \times AB \times AC\]
Hence, putting the value of sides,
\[ \Rightarrow \]\[\Delta ABC = \dfrac{1}{2} \times 8 \times 6\]
On calculating, we get,
\[ \Rightarrow \]\[\Delta ABC = 24c{m^2}\]
Now, calculate the semi perimeter of triangle as \[s = \dfrac{{a + b + c}}{2}\]
Putting the values of sides as,
\[ \Rightarrow \]\[s = \dfrac{{8 + 6 + 10}}{2}\]
Hence, on calculation, we get,
\[ \Rightarrow \]\[s = 12cm\]
Hence, put the values in the formula of incentre as \[I = \dfrac{{area\,of\,\Delta BAC}}{{semi - perimeter\,of\,\Delta BAC}}\]
\[ \Rightarrow \]\[I = \dfrac{{24}}{{12}}\]
Hence, on dividing , we get,
\[ \Rightarrow \]\[I = 2cm\]
Hence, the radius of incentre is\[2cm\].

Note: The incenter is the centre of the incircle for a polygon or in sphere for a polyhedron (when they exist). The corresponding radius of the incircle or in sphere is known as the inradius. The incenter can be constructed as the intersection of angle bisectors. In geometry, the incenter of a triangle is a triangle centre, a point defined for any triangle in a way that is independent of the triangle's placement or scale. In this case the incentre is the centre of this circle and is equally distant from all sides.