Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths Pure - Question 10 - 2006 - Paper 2

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

$A = \frac{1}{2} r^2 \theta$ Where:

• $r = 2.12 \ m$
• $\theta = 0.65 \ rad$

Substituting the values:

$A = \frac{1}{2} (2.12)^2 \times 0.65 = 1.4586 \ m²$

Thus, rounding to 2 decimal places, the area of sector BAC is approximately 1.46 m².

We know that

$\text{Since } \theta = \frac{\pi}{2} - \alpha$

Thus, $\alpha = \frac{\pi}{2} - 0.65 \approx 0.92 \ radians$

Therefore, the size of ∠CAD is approximately 0.92 radians.

To find the total area of the cross-section ABCD, we sum the areas of sector BAC and triangle ACD.

1. Area of Sector BAC: We found this to be 1.46 m².

2. Area of Triangle ACD: The area can be calculated using:

   = \frac{1}{2} (2.12)(1.86) \sin(\alpha)$$

Substituting the value of\ \alpha:\n$\alpha ≈ 0.92$:

$A_{ACD} ≈ \frac{1}{2} (2.12)(1.86) \times 0.81 ≈ 1.57 \ m²$

Summing both areas:

$\text{Total Area} = 1.46 + 1.57 = 3.03 \ m²$

Thus, the area of the cross-section ABCD is approximately 3.03 m².