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OAC is a sector of a circle, centre O, radius 10 m - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 2

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OAC is a sector of a circle, centre O, radius 10 m. BA is the tangent to the circle at point A. BC is the tangent to the circle at point C. Angle AOC = 120° Calc... show full transcript

Worked Solution & Example Answer:OAC is a sector of a circle, centre O, radius 10 m - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 2

Step 1

Find the length of BC

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Answer

To find the length of BC, we note that triangle OAC is formed, where angle AOC is 120° and the radius OA = OC = 10 m. We can use the Law of Cosines:

BC2=OA2+OC2−2(OA)(OC)⋅cos⁡(120°)BC^2 = OA^2 + OC^2 - 2(OA)(OC) \cdot \cos(120°)

Substituting the values:

BC2=102+102−2(10)(10)(−0.5)=100+100+100=300BC^2 = 10^2 + 10^2 - 2(10)(10)(-0.5) = 100 + 100 + 100 = 300

So,

BC=300=103≈17.32 mBC = \sqrt{300} = 10\sqrt{3} \approx 17.32 \, m

Step 2

Find the area of triangle AOC

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Answer

The area of triangle AOC can be calculated using the formula:

Area=12⋅base⋅height\text{Area} = \frac{1}{2} \cdot base \cdot height

Using vertex O as the tip of the triangle, we use the sine of angle AOC:

Area=12⋅OA⋅OC⋅sin⁡(120°)\text{Area} = \frac{1}{2} \cdot OA \cdot OC \cdot \sin(120°)

Substituting the values:

Area=12⋅10⋅10⋅32=253≈43.3 m2\text{Area} = \frac{1}{2} \cdot 10 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3} \approx 43.3 \, m^2

Step 3

Find the area of the sector OAC

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Answer

The area of the sector OAC is found using:

Area of Sector=θ360⋅πr2\text{Area of Sector} = \frac{\theta}{360} \cdot \pi r^2

Where (\theta = 120°) and (r = 10 m):

Area of Sector=120360⋅π(10)2=13⋅100π=100π3≈104.72 m2\text{Area of Sector} = \frac{120}{360} \cdot \pi (10)^2 = \frac{1}{3} \cdot 100\pi = \frac{100\pi}{3} \approx 104.72 \, m^2

Step 4

Calculate the area of the shaded region

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Answer

The area of the shaded region is obtained by subtracting the area of triangle AOC from the area of sector OAC:

Area of Shaded Region=Area of Sector−Area of Triangle\text{Area of Shaded Region} = \text{Area of Sector} - \text{Area of Triangle}

Substituting the values we calculated earlier:

Area of Shaded Region≈104.72−43.3≈61.42 m2\text{Area of Shaded Region} \approx 104.72 - 43.3 \approx 61.42 \, m^2

Finally, rounding to three significant figures, we get:

Area of Shaded Region≈61.4 m2\text{Area of Shaded Region} \approx 61.4 \, m^2

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