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\includegraphics[max width=\textwidth, alt={}, center]{d1eb99a1-04e5-43bc-87b4-d0f7c962135c-4_353_579_248_744} The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a trapezium \(A B C D\) with \(A B\) and \(C D\) perpendicular to \(A D\). The lengths of \(A B\) and \(A D\) are each 5 cm and the length of \(C D\) is \(( a + 5 ) \mathrm { cm }\).

1. Show the distance of the centre of mass of the prism from \(A D\) is $$\frac { a ^ { 2 } + 15 a + 75 } { 3 ( a + 10 ) } \mathrm { cm } .$$ The prism is placed with the face containing \(A B\) in contact with a horizontal surface.
2. Find the greatest value of \(a\) for which the prism does not topple. The prism is now placed on an inclined plane which makes an angle \(\theta ^ { \circ }\) with the horizontal. \(A B\) lies along a line of greatest slope with \(B\) higher than \(A\).
3. Using the value for \(a\) found in part (ii), and assuming the prism does not slip down the plane, find the greatest value of \(\theta\) for which the prism does not topple.