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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

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

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Answer

To find angle COD in triangle COD, we can use the cosine rule:

Using the formula:

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

where:

  • a = 8 cm (side OC)
  • b = 8 cm (side OD)
  • c = 7 cm (side CD)

Calculating:

cos(COD)=82+82722×8×8=64+6449128=791280.6171875\text{cos}(COD) = \frac{8^2 + 8^2 - 7^2}{2 \times 8 \times 8} = \frac{64 + 64 - 49}{128} = \frac{79}{128} \approx 0.6171875

Then, taking the arccosine:

COD=cos1(0.6171875)0.906extradians(3significantfigures)\angle COD = \cos^{-1}(0.6171875) \approx 0.906 ext{ radians (3 significant figures)}

Step 2

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

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Answer

To find the perimeter of AOBCDA, we consider each side:

  • AB = 16 cm
  • DC = 7 cm
  • OA = OB = 8 cm (radius of the circles)

Thus:

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

To 3 significant figures, the perimeter is 39.0 cm.

Step 3

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

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Answer

To find the area of AOBCDA, we need the area of triangle COD and the two circular sectors OBC and ODA:

  1. Area of Triangle COD: Using the formula:
Areatriangle=12×b×h\text{Area}_{triangle} = \frac{1}{2} \times b \times h

Where base = CD = 7 cm and height can be calculated using

Areatriangle=12×7×8×sin(0.906)25.2extcm2\text{Area}_{triangle} = \frac{1}{2} \times 7 \times 8 \times \sin(0.906) \\ \approx 25.2 ext{ cm}^2
  1. Area of the sectors OBC and ODA: The angle for each sector is 0.906 radians, so the area for each sector is:
Areasector=r22×θ=822×0.906=28.8extcm2\text{Area}_{sector} = \frac{r^2}{2} \times \theta = \frac{8^2}{2} \times 0.906 \\ = 28.8 ext{ cm}^2

Total area of two sectors:

  1. Total Area AOBCDA:
Total Area=Areatriangle+Areasectors25.2+57.6=82.8extcm2\text{Total Area} = \text{Area}_{triangle} + \text{Area}_{sectors} \approx 25.2 + 57.6 = 82.8 ext{ cm}^2

Final area to 3 significant figures is 82.8 cm².

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