Figure 1 shows ABC, a sector of a circle with centre A and radius 7 cm - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

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

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

Plugging in the values:

• $r = 7 \text{ cm}$
• $\theta = 0.8 \text{ radians}$

We find:

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

To find the perimeter of region R, we will need to calculate the lengths of segments CD, DB, and the arc BC. We already calculated the length of the arc BC as 5.6 cm.

Let:

• $BD = \sqrt{(7^2) + (3.5^2) - (2 \times 7 \times 3.5 \times \cos(0.8))}$ Because D is the midpoint of AC, we can assume $AC=2 imes AD$.

Calculating the lengths: $BD = \sqrt{7^2 + 3.5^2 - (2 \times 7 \times 3.5 \times \cos(0.8))} = 5.21 \text{ cm}$

Now, we can find the perimeter: $\text{Perimeter} = CD + DB + \text{arc BC} = (3.5 + 5.6 + 5.21) = 14.3 \text{ cm}$ (to 3 significant figures: 14.3 cm)

To find the area of region R, we subtract the area of triangle ABD from the area of the sector ABC. First, we need to calculate the area of triangle ABD using the formula:

$\text{Area} = \frac{1}{2} ab \sin C$

where:

• $a$ and $b$ are the sides of the triangle (7 cm), and
• $C$ is the angle (0.8 radians).

Thus, the area of triangle ABD is:

$\text{Area}_{ABD} = \frac{1}{2} \times 7 \times 7 \times \sin(0.8) = 7 \times \sin(0.8) \approx 7 \times 0.717 = 5.019 \text{ cm}^2$

Now, subtracting this from the area of sector ABC:

$\text{Area}_R = \text{Area}_{ABC} - \text{Area}_{ABD} \approx 19.6 - 5.019 = 14.581 \text{ cm}^2$ (to 3 significant figures: 14.6 cm²)