Figure 1 shows $ABC$, a sector of a circle with centre $A$ and radius 7 cm - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

The area of the sector can be found using the formula:

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

Substituting the known values:

$A = \frac{1}{2} \cdot 7^2 \cdot 0.8 = \frac{1}{2} \cdot 49 \cdot 0.8 = 19.6 \text{ cm}^2$

To find the perimeter of region $R$, we need to calculate:

1. The length of $DC$ (which is the same as $AD$ since $D$ is the midpoint).
2. The length of arc $BC$ (calculated previously: $5.6$ cm).

We first find $BD$:

Using the formula for $BD$:

$BD = \sqrt{(AB)^2 + (AD)^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle ABC)}$

Assuming $AD = 3.5$ cm, we calculate:

$BD = \sqrt{(7)^2 + (3.5)^2 - 2 \cdot 7 \cdot 3.5 \cdot \cos(0.8)}$

Once we compute $BD$, the perimeter $P$ of region $R$ is given by:

$P = DC + BD + BC = (3.5 + 5.6 + BD)$

Finally, we round our result to 3 significant figures.

To find the area of region $R$, we can subtract the area of triangle $ABD$ from the area of the sector $ABC$.

The area of triangle $ABD$ can be calculated using:

$A_{ABD} = \frac{1}{2} \cdot base \cdot height$

Where base = $AD$ and height can be found using:

$height = AB \cdot \sin(0.8)$

Thus,

$A_{ABD} = \frac{1}{2} \cdot 3.5 \cdot (7 \cdot \sin(0.8))$

Now, the area of region $R$ is

$Area_R = Area_{ABC} - A_{ABD}$

Finally, round the area to 3 significant figures.