This looks pretty long actually. Both parts a) and b) require you to be able to identify all the overlapping shapes present in the question.
For a), the hatched area is obtained by noting that:
- \(ABDC\) is really just the triangle \(\triangle ABC\).
- \(ABEC\) is a sector of a circle, whose centre is \(A\).
Observe that the area you require is the area of the sector, less the area of the triangle. (Side note: In 2u maths, they frequently call this the area of a minor segment.) So you need to compute the areas of the two shapes listed above first:
- Area of \(ABDC\): Using \( \frac12bh\), we obtain \( \frac12\times 2\times 2 = 2\) cm.
- Area of \(ABEC\): Using \( \frac{\theta}{360} \times \pi r^2\), since our angle is a right angle, we obtain \( \frac{90}{360}\times \pi \times 2^2 = \pi\).
Hence the area of the required region, i.e. the hatched area \(BDCE\), is \( \pi - 2\) cm2. If you use your calculator, you'll find that indeed the answer is approximately 1.14 cm2.
Then for part b), you need to first realise that \(BDCF\) is a semi-circle. This is because \(D\) is the centre of the circle through \(B\), \(C\) and \(F\), and hence we conclude that \(BC\) is a diameter of this circle.
Once you realise that \(BDCF\) is a semi-circle, you need to realise. that the shaded area \(BECF\) is now just the area of this semicircle, less the hatched area from the previous part. This should help you get to the answer.
The key point: Try not to focus on the picture as a whole - it's very easy to get lost this way. Build up what you require bit by bit. Compound area problems always require you to see if one area can be formed by somehow manipulating other areas. In particular, other areas, which you know how to compute (e.g. area of a sector).