### In the given figure, $ABC$ is a right-angled triangle with $AB = 7 \space cm$ and $AC = 9 \space cm$. A circle with center $O$ has been inscribed inside the triangle. Calculate the value of $r$, the radius of the inscribed circle. O A B C F D E 7 cm 9 cm

$2.3 \space cm$

Step by Step Explanation:
1. Let us join $O$ to $A, B,$ and $C$ and draw $OD \perp AB$, $OE \perp BC$ and $OF \perp CA$.

We see that $OD, OE,$ and $OF$ are the radius of the circle with center $O$.
$\implies OD = OE = OF = r \space cm$

Also, $\triangle ABC$ is a right-angled triangle. \begin{aligned} \implies \text { Area of } \triangle ABC & = \dfrac { 1 }{ 2 } \times AB \times AC \\ & = \dfrac { 1 }{ 2 } \times 7 \space cm \times 9 \space cm \\ & = 31.5 \space cm^2 \end{aligned}
2. Let us now find the area of $\triangle ABC$ in terms of $r$. \begin{aligned} \text{Area of } \triangle ABC & = \text{Area of } \triangle OAB + \text{Area of } \triangle OBC + \text{Area of } \triangle OCA \\ & = \dfrac { 1 }{ 2 } \times AB \times OD + \dfrac { 1 }{ 2 } \times BC \times OE + \dfrac { 1 }{ 2 } \times CA \times OF \\ & = \dfrac { 1 }{ 2 } \times AB \times r + \dfrac { 1 }{ 2 } \times BC \times r + \dfrac { 1 }{ 2 } \times CA \times r \\ & = \dfrac { 1 }{ 2 } (AB + BC + CA) \times r \\ & = \dfrac { 1 }{ 2 } \text{(Perimeter of } \triangle ABC) \times r \end{aligned}
3. Comparing the area of $\triangle ABC$ obtained in step 1 and step 2, we have \begin{aligned} & \text{ Area of } \triangle ABC = 31.5 \space cm^2 = \dfrac { 1 }{ 2 } \times (AB + BC + CA) \times r \\ \implies & 31.5 \space cm^2 = \dfrac { 1 }{ 2 } \times (7 + BC + 9 ) \times r && \ldots \text{(i)} \end{aligned}
4. Applying Pythagoras theorem in $\triangle ABC$, we have \begin{aligned} & BC^2 = AB^2 + AC^2 \\ \implies & BC = \sqrt{(7)^2 + (9)^2} = 11.4 \space cm \end{aligned}
5. Now, substituting the value of $BC$ in $eq \space \text{(i)}$, we have \begin{aligned} & 31.5 = \dfrac { 1 } { 2 } \times (7 + 11.4 + 9) \times r \\ \implies & 31.5 \times 2 = 27.4 \times r \\ \implies & r = \dfrac { 31.5 × 2 } { 27.4 } = 2.3 \space cm \end{aligned}
6. Hence, the radius of the inscribed circle is $2.3 \space cm$.