### Given a right-angled $\Delta ABC$. The lengths of the sides containing the right angle are $7 \space cm$ and $24 \space cm$. A circle is inscribed in $\Delta ABC$. Find the radius of the circle.

$3 \space cm$

Step by Step Explanation:
1. In $\Delta ABC$, we have \begin{aligned} \angle B = 90^\circ, AB = 7 \space cm \space \text{ and } BC = 24 \space cm \end{aligned}
2. A circle is inscribed in $\Delta ABC$. Let $O$ be its centre and $M$, $N$ and $P$ be the points where it touches the sides $AB$, $BC$ and $CA$ respectively. \begin{aligned} \text { Then, } OM \perp AB, ON \perp BC, OP \perp CA. \end{aligned}
3. Let $r \space cm$ be the radius of the circle.

Then, $OM = ON = OP = r\space cm$.

$\text{ Now, } AB^2 + BC^2 = CA^2 \space\space [\text{ By pythagoras' theorem }]$

$\implies (7)^2 cm^2 + (24)^2 cm^2 = CA^2 \\ \implies CA = 25 \space cm$.
4. Now, \begin{aligned} &ar(\Delta ABC) = ar(\Delta AOB) + ar(\Delta BOC) + ar(\Delta COA) \\ \implies& \dfrac { 1 } { 2 } \times AB \times BC = \bigg( \dfrac { 1 } { 2 } \times AB \times OM \bigg) + \bigg( \dfrac { 1 } { 2 } \times BC \times ON \bigg) + \bigg( \dfrac { 1 } { 2 } \times CA \times OP \bigg) \\ \implies& \dfrac { 1 } { 2 } \times 7 \times 24 = \bigg( \dfrac { 1 } { 2 } \times 7 \times r \bigg) + \bigg( \dfrac { 1 } { 2 } \times 24 \times r \bigg) + \bigg( \dfrac { 1 } { 2 } \times 25 \times r \bigg) \\ \implies& 168 = 7 r + 24 r + 25 r = 56 r \\ \implies& r = \dfrac { 168 } { 56 } = 3 \space cm. \end{aligned} Hence, the radius of the circle is $3 \space cm$.