\includegraphics{figure_3}
Fig. 3 shows the cross-sections of two drawer handles. Shape \(X\) is a rectangle \(ABCD\) joined to a semicircle with \(BC\) as diameter. The length \(AB = d\) cm and \(BC = 2d\) cm. Shape \(Y\) is a sector \(OPQ\) of a circle with centre \(O\) and radius \(2d\) cm. Angle \(POQ\) is \(\theta\) radians. Given that the areas of the shapes \(X\) and \(Y\) are equal,
- prove that \(\theta = 1 + \frac{1}{4}\pi\). [5]
Using this value of \(\theta\), and given that \(d = 3\), find in terms of \(\pi\),
- the perimeter of shape \(X\), [2]
- the perimeter of shape \(Y\). [3]
- Hence find the difference, in mm, between the perimeters of shapes \(X\) and \(Y\). [2]