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 is given by the formula:

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

Substituting in the values, we have:

• r = 2.12 m
• ( \theta = 0.65 ) radians.

Thus,

$A = \frac{1}{2} \times (2.12)^2 \times 0.65 = 1.459\, m²$

Rounding to 2 decimal places, the area of the sector BAC is 1.46 m².

Given that the size of ∠BAC is 0.65 radians, and since the angles in a triangle add up to π radians, we can find ∠CAD by:

$\angle CAD = \frac{\pi}{2} - 0.65 \approx 0.92\, \text{radians}$

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

The total area of the cross-section ABCD can be found by adding the area of the sector BAC and the area of triangle CAD. The area of triangle CAD is calculated using:

$Area_{CAD} = \frac{1}{2} \times 2.12 \times 1.86 \times \sin(0.92)$

Calculating this gives:

$Area_{CAD} \approx \frac{1}{2} \times 2.12 \times 1.86 \times 0.81 \approx 1.57\, m²$

Adding the area of the sector BAC:

$Area_{Total} = 1.46 + 1.57 = 3.03\, m²$

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