Figure 1 shows the sector OAB of a circle with centre O, radius 9 cm and angle 0.7 radians - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 3

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

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

Substituting in the radius and angle:

$A = \frac{1}{2} \times 9^2 \times 0.7 = \frac{1}{2} \times 81 \times 0.7 = 28.35 \text{ cm}^2.$

Given that AC is perpendicular to OA and we know the radius and angle, we can apply trigonometric relationships. Here, using the tangent function:

$\tan(0.7) = \frac{AC}{9}$

Solving for AC gives:

$AC = 9 \times \tan(0.7) \approx 9 \times 0.757 = 6.81 \text{ cm}.$

The area of region H can be found using the formula for the area of a triangle:

$\text{Area} = \frac{1}{2} \times base \times height$.

Using the segment AC as the base and the height can be derived from the sine relation:

$\text{Area of } H = \frac{1}{2} \times AC \times height = \frac{1}{2} \times 6.81 \times 9 \times \sin(0.7) \approx 5.76 \text{ cm}^2.$