\begin{tabular}{l} Length of court \( =30,5 \mathrm{~m} \) \\ Width of court \( =15,25 \mathrm{~m} \) \\ Radius of semicircle \( =4,9 \mathrm{~m} \) \\ Area of centre circle \( =0,64 \mathrm{~m}^{2} \) \\ 2.2 .3 Calculate the length of lines to be painted if all the lines need to be painted \\ \hline\end{tabular}

We note that the playing‐area is marked as a rectangle with additional curved markings. In this problem the painted lines are taken to be: 1. The boundary of the rectangle (with length and width given). 2. The two semicircular arcs (each having radius \(4.9\,m\)). Because two semicircles make a full circle, their combined length (arc–length) equals the circumference of a circle with radius \(4.9\,m\). 3. The circumference of the centre circle. (Its area is given as \(0.64\,m^2\), so from that we can compute its circumference.) We now calculate each part step by step. --- **1. Perimeter of the Rectangle** The rectangle has length \[ L=30.5\,m \] and width \[ W=15.25\,m. \] Its perimeter is \[ P_{\text{rect}}=2(L+W)=2(30.5+15.25). \] Calculating the sum inside the parentheses: \[ 30.5+15.25=45.75\,m, \] so \[ P_{\text{rect}}=2\times45.75=91.5\,m. \] --- **2. Combined Arc Length of the Two Semicircles** Each semicircle has radius \[ r=4.9\,m, \] and the arc length of a full circle is \[ \text{Circumference}=2\pi r. \] Since two semicircles make a full circle, the combined length is \[ L_{\text{semi}}=2\pi(4.9)=9.8\pi\,m. \] --- **3. Circumference of the Centre Circle** The centre circle has area \[ A=0.64\,m^2. \] For a circle, \[ A=\pi r^2, \] so the radius of the centre circle is \[ r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{0.64}{\pi}}. \] Its circumference is therefore \[ C=2\pi r=2\pi\sqrt{\frac{0.64}{\pi}}. \] We can simplify this as follows: \[ \sqrt{\frac{0.64}{\pi}}=\frac{\sqrt{0.64}}{\sqrt{\pi}}=\frac{0.8}{\sqrt{\pi}}, \] so \[ C=2\pi\cdot\frac{0.8}{\sqrt{\pi}}=\frac{1.6\pi}{\sqrt{\pi}}=1.6\sqrt{\pi}\,m. \] --- **4. Total Length of Painted Lines** Adding the three contributions we get \[ L_{\text{total}}=P_{\text{rect}}+L_{\text{semi}}+C. \] Substituting the values computed: \[ L_{\text{total}}=91.5+9.8\pi+1.6\sqrt{\pi}\,m. \] We may also give an approximate numerical value. With \(\pi\approx3.14\) and \(\sqrt{\pi}\approx1.772\) we have: \[ 9.8\pi\approx9.8\times3.14\approx30.77\,m, \] \[ 1.6\sqrt{\