The shape BCD shown in Figure 3 is a design for a logo - Edexcel - A-Level Maths Pure - Question 9 - 2009 - Paper 2

To find ( \angle DAC ), we note that ( \angle BAC + \angle DAC = \pi ) radians (since the angles are supplementary in a straight line).

Thus,

$\angle DAC = \pi - 2.2 \approx 3.14 - 2.2 = 0.94 \text{ radians}$

Rounding to 3 significant figures gives

$\angle DAC \approx 0.940\text{ radians}$

The complete area of the logo design consists of the area of the sector BAC and two triangles, DAC and DAB.

The area of triangle DAC can be calculated using:

$\text{Area}_{DAC} = \frac{1}{2} \cdot AD \cdot DC \cdot \sin(\angle DAC)$

Here, ( AD = 4 ) cm and ( DC = 6 ) cm:

$\text{Area}_{DAC} = \frac{1}{2} \cdot 4 \cdot 6 \cdot \sin(0.94) \approx 11.6 \text{ cm}^2$

Adding the two triangle areas and the sector area gives:

Total area = ( 39.6 + 11.6 + \text{Area of triangle DAB} ). We can assume symmetry, so:

$\text{Area of triangle DAB} \approx 11.6 \text{ cm}^2$

Thus,

Total area = ( 39.6 + 11.6 + 11.6 \approx 62.8 \text{ cm}^2$$

Rounding to the nearest cm² gives:

Total area = 63 cm².