Figure 2 shows $ABC$, a sector of a circle of radius 6 cm with centre $A$ - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 4

To find the area of the sector, we use the formula:

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

Substituting in the values gives us:

$\text{Area} = \frac{1}{2} \times 6^2 \times 0.95 = 17.1 \text{ cm}^2.$

We need to express the triangle relations. Since angle $BAC = 0.95$ radians,

Using the sine rule:

$\frac{AD}{\sin(0.95)} = \frac{6}{\sin(\theta)}$

Finding $AD$ involves using the side opposite the angle.

After calculating, we find:

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

The perimeter $P$ of region $R$ can be found by adding the lengths of lines $CD$, $DB$, and arc $BC$:

Using values:

$P = AD + AB + BC = 5.16 + 6 + 5.7 = 16.86 \text{ cm}.$

To find the area of the region $R$, we need the area of triangle $ABD$ and the area of sector $ABC$:

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

Then, area of region $R$ = Area of sector $ABC$ - Area of triangle $ABD$:

$\text{Area of } R = 17.1 - 12.6 = 4.5 \text{ cm}^2.$