Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

The area of the sector BAC can be calculated using the formula:

$\text{Area} = \frac{1}{2} r^2 \theta$

Substituting the known values:

$\text{Area} = \frac{1}{2} \times (2.12)^2 \times 0.65 = 1.46 \text{ m}^2$

The size of ( \angle CAD ) is given as:

$\alpha = \frac{\pi}{2} - 0.65 = 0.92 \text{ radians}$

The area of the cross-section ABCD can be found by adding the area of the sector to the area of triangle ACD:

1. The area of triangle ACD is given by:

$\Delta ACD = \frac{1}{2} \times (2.12) \times (1.86) \times \sin(\alpha)$ Here, substituting ( \alpha = 0.92 ) radians:

$\Delta ACD = \frac{1}{2} \times (2.12) \times (1.86) \times \sin(0.92) \approx 1.57 \text{ m}^2$

2. Therefore, the total area:

$\text{Total Area} = 1.46 + 1.57 = 3.03 \text{ m}^2$