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\includegraphics[max width=\textwidth, alt={}, center]{334b4bdf-6d9c-4208-9032-572eb7c5f9ee-2_549_579_1505_781} A uniform lamina is made by joining a rectangle \(A B C D\), in which \(A B = C D = 0.56 \mathrm {~m}\) and \(B C = A D = 2 \mathrm {~m}\), and a square \(E F G A\) of side 1.2 m . The vertex \(E\) of the square lies on the edge \(A D\) of the rectangle (see diagram). The centre of mass of the lamina is a distance \(h \mathrm {~m}\) from \(B C\) and a distance \(v \mathrm {~m}\) from \(B A G\).

1. Find the value of \(h\) and show that \(v = h\). The lamina is freely suspended at the point \(B\) and hangs in equilibrium.
2. State the angle which the edge \(B C\) makes with the horizontal. Instead, the lamina is now freely suspended at the point \(F\) and hangs in equilibrium.
3. Calculate the angle between \(F G\) and the vertical.