$$8^{2}+15^{2}=64+225=289=17^{2}$$
$\therefore \triangle L M N$ is a right-angled triangle
$$A=\frac{8 \times 15}{2}=60$$

Reference: 1989 HG9
Let $O$ be the centre and the radius of the circle be $r$, which touches the triangle at $C, D$ and $E$.
$O C \perp L M, O D \perp M N, O E \perp L N$ (tangent $\perp$ radius)
$O D M C$ is a rectangle (which consists of 3 right angles)
$$O C=r=O D \text { (radii) }$$
$\Rightarrow O C M D$ is a square.
$$\begin{array}{l} C M=M D=r \text { (opp. sides, rectangle) } \\ L C=15-r, N D=8-r \\ L E=L C=15-r, N E=N D=8-r \text { (tangent from ext. pt.) } \\ L E+N E=L N \\ \Rightarrow 15-r+8-r=17 \\ \Rightarrow r=3 \end{array}$$