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The shape shown in Figure 1 is a pattern for a pendant - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 2

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The shape shown in Figure 1 is a pattern for a pendant. It consists of a sector OAB of a circle centre O, of radius 6 cm, and angle AOB = \( \frac{\pi}{3} \). The ci... show full transcript

Worked Solution & Example Answer:The shape shown in Figure 1 is a pattern for a pendant - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 2

Step 1

(a) the area of the sector OAB

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Answer

To find the area of the sector OAB, we use the formula for the area of a sector:

Area=12r2θ\text{Area} = \frac{1}{2} r^2 \theta

where ( r = 6 ) cm and ( \theta = \frac{\pi}{3} ) radians.

Thus,

Area=12×62×π3=12×36×π3=12π3=12π cm2\text{Area} = \frac{1}{2} \times 6^2 \times \frac{\pi}{3} = \frac{1}{2} \times 36 \times \frac{\pi}{3} = \frac{12\pi}{3} = 12\pi \text{ cm}^2

The area is approximately ( 12\pi \approx 37.7 ) cm² or exactly ( 12\pi \text{ cm}^2 ).

Step 2

(b) the radius of the circle C

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Answer

For the circle C that touches the straight edges OA and OB, we need to use trigonometric properties. From the geometry of the sector,

Let ( r ) be the radius of circle C. The height from O to the tangent where circle C touches OA and OB can be described as:

\Rightarrow r = \frac{6 - r}{\sqrt{3}}$$ Solving for \( r \): Multiplying both sides by \( \sqrt{3} (6 - r) \): $$r\sqrt{3} + r = 6\sqrt{3}$$ Thus, $$r(1 + \sqrt{3}) = 6\sqrt{3}\n r = \frac{6\sqrt{3}}{1 + \sqrt{3}} = 2\text{ cm}$$.

Step 3

(c) Find the area of the shaded region

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Answer

To find the area of the shaded region, we first calculate the area of the sector OAB, which is ( 12\pi \text{ cm}^2 ), and then subtract the area of circle C:

The area of circle C is:

Area=πr2=π×22=4π cm2\text{Area} = \pi r^2 = \pi \times 2^2 = 4\pi \text{ cm}^2

Thus, the area of the shaded region is:

Shaded Area=Area of Sector OABArea of Circle C=12π4π=8π cm2\text{Shaded Area} = \text{Area of Sector OAB} - \text{Area of Circle C} = 12\pi - 4\pi = 8\pi \text{ cm}^2

The shaded area is approximately ( 8\pi \approx 25.1 \text{ cm}^2 ) or exactly ( 8\pi \text{ cm}^2 ).

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