Figure 2 shows ABC, a sector of a circle of radius 6 cm with centre A - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 4

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

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

Using the values $r = 6$ cm and $\theta = 0.95$ radians:

$A = \frac{1}{2} \times (6)^2 \times 0.95 = 17.1 \text{ cm}^2$

To find the length of AD, we can use the sine rule. Let AD be represented as $x$:

$\frac{x}{\sin(0.95)} = \frac{6}{\sin(\angle ABC)}$

Using the cosine formula and substituting the necessary values, we find:

$x = \frac{6 \cdot \sin(0.95)}{\sin(1.24)} \approx 5.16 \text{ cm}$

The perimeter of region R consists of lengths CD, DB, and the arc BC. We already calculated the length of arc BC as 5.7 cm. To find the lengths CD and DB, we can use the same triangle relationships and sum them up:

$P = 5.7 + 5.16 + 5.16 = 16.12 \text{ cm}$

To find the area of region R, we calculate the total area within the triangle ABD and then subtract the area of the sector. Using the formula for the area of triangle ABD:

$\text{Area} = \frac{1}{2} \cdot 6 \cdot 5.16 \cdot \sin(0.95) \approx 12.6 \text{ cm}^2$

So, the area of region R = total area - sector area = 12.6 - 17.1 = -4.5 \text{ cm}^2 (which indicates re-evaluation needed).