Figure 1 shows the triangle ABC, with AB = 8 cm, AC = 11 cm and ∠ BAC = 0.7 radians - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

To find the length of BC, we can use the cosine rule:

$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$

Substituting the values:

$BC^2 = 8^2 + 11^2 - 2 \cdot 8 \cdot 11 \cdot \cos(0.7)$

Calculating:

$BC \approx 7.098 \text{ cm}.$

Now, the perimeter of region R is:

$\text{Perimeter} = (8 + 11) + BC = 19 + 7.098 = 26.098 \text{ cm},$ which rounded to 3 significant figures is approximately 26.1 cm.

To find the area of region R, we separate it into two parts: the area of triangle ABC and the area of sector ADB.

First, we calculate the area of triangle ABC:

$\text{Area}_{\triangle ABC} = \frac{1}{2} ab \sin(\angle BAC) = \frac{1}{2} \cdot 8 \cdot 11 \cdot \sin(0.7).$

Calculating this gives approximately 28.3 cm².

Next, we calculate the area of the sector ADB:

$\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 8^2 \cdot 0.7 = 22.4 \text{ cm}^2.$

Thus, the area of region R can be found by subtracting:

$\text{Area}_{R} = \text{Area}_{\triangle ABC} - \text{Area}_{\text{sector}} = 28.3 - 22.4 = 5.9 \text{ cm}^2.$