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 6 - 2010 - Paper 3

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

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

where $r = 9 \text{ cm}$ and $\theta = 0.7 \text{ radians}$. Substituting the values gives:

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

Since AC is perpendicular to OA, we can use trigonometric relationships. We have:

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

This leads to:

$AC = 9 \tan(0.7)$.

Calculating this:

$AC \approx 9 \times 0.755 \approx 6.795$.

Rounding to 2 decimal places, we get:

$AC \approx 6.80 \text{ cm}$.

The area of triangle AOC can be determined using the formula for the area of a triangle:

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

where the base is AC and the height is the arc AB.

From previous results, we know:

$Area_{AOC} = \frac{1}{2} \times 6.80 \times 6.3$.

Calculating this gives:

$Area_{AOC} \approx \frac{1}{2} \times 6.80 \times 6.3 \approx 21.42 \text{ cm}^2$.

Thus, the area of region H is approximately:

$Area_H = \frac{1}{2} \times 9 \times 28.35 - Area_{AOC}$,

which would result in:

$Area_H \approx 28.35 - 21.42 = 6.93 \text{ cm}^2$.

Finally, rounding to 2 decimal places gives:

$Area_H \approx 6.93 \text{ cm}^2$.