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Figure 1 shows a sketch of a design for a scraper blade - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 2

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Figure 1 shows a sketch of a design for a scraper blade. The blade AOBCDA consists of an isosceles triangle COD joined along its equal sides to sectors OBC and ODA o... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of a design for a scraper blade - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 2

Step 1

a) Show that the angle COD is 0.906 radians, correct to 3 significant figures.

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Answer

In triangle COD, we can use the cosine rule to find angle COD. By the cosine rule:

extcos(C)=a2+b2c22ab ext{cos}(C) = \frac{a^2 + b^2 - c^2}{2ab}

Here, we let a = 8 cm, b = 8 cm, and c = 7 cm. Thus,

cos(C)=82+8272288=64+6449128=791280.6171875\text{cos}(C) = \frac{8^2 + 8^2 - 7^2}{2 \cdot 8 \cdot 8} = \frac{64 + 64 - 49}{128} = \frac{79}{128} \approx 0.6171875

Now, to find angle COD:

COD=2×arccos(0.6171875)0.906extradians\angle COD = 2 \times \text{arccos}(0.6171875)\approx 0.906 ext{ radians}

Thus, we have shown that the angle COD is approximately 0.906 radians, correct to 3 significant figures.

Step 2

b) Find the perimeter of AOBCDA, giving your answer to 3 significant figures.

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Answer

To find the perimeter of AOBCDA, we will calculate the lengths of the triangle edges and the arc lengths:

  1. The length of AB is given as 16 cm.
  2. Lengths of sides OA and OB are both 8 cm (as they are radii of the circle).
  3. The length of DC is given as 7 cm.

Putting it all together for the perimeter:

extPerimeter=AB+OA+OB+DC=16+8+8+7=39extcm ext{Perimeter} = AB + OA + OB + DC = 16 + 8 + 8 + 7 = 39 ext{ cm}

Thus, the perimeter of AOBCDA is 39.0 cm, correct to 3 significant figures.

Step 3

c) Find the area of AOBCDA, giving your answer to 3 significant figures.

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Answer

To find the area, we will calculate the area of the triangle COD and the areas of the sectors OBC and ODA:

  1. The area of triangle COD can be found using the formula:
extArea=12bh ext{Area}_{\triangle} = \frac{1}{2} \cdot b \cdot h

Here, if the height is calculated using sine:

Area=1287sin(0.906)12870.78928.0extcm2\text{Area} = \frac{1}{2} \cdot 8 \cdot 7 \cdot \sin(0.906) \approx \frac{1}{2} \cdot 8 \cdot 7 \cdot 0.789 \approx 28.0 ext{ cm}^2
  1. The area of each sector is given by:
extAreasector=12r2θ=12820.906ext(foronesector)=28.928extcm2 ext{Area}_{sector} = \frac{1}{2} \cdot r^2 \cdot \theta = \frac{1}{2} \cdot 8^2 \cdot 0.906 ext{ (for one sector)} \\= 28.928 ext{ cm}^2

Thus, for two sectors:

  1. Adding the area of the triangle:

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