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Helen is creating a mosaic pattern by placing square tiles next to each other along a straight line - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 3

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Helen is creating a mosaic pattern by placing square tiles next to each other along a straight line. The area of each tile is half the area of the previous tile, an... show full transcript

Worked Solution & Example Answer:Helen is creating a mosaic pattern by placing square tiles next to each other along a straight line - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 3

Step 1

Find, in terms of w, the length of the sides of the second largest tile.

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Answer

The area of the first tile is given by the side length squared, which can be expressed as: A1=w2A_1 = w^2 The area of the second tile is half that of the first tile: A2=12A1=12w2A_2 = \frac{1}{2} A_1 = \frac{1}{2} w^2 Since the area of a square is equal to the square of its side length, we set: A2=s22A_2 = s_2^2 Thus: s22=12w2s_2^2 = \frac{1}{2} w^2 Taking the square root gives: s2=w12s_2 = w \cdot \frac{1}{\sqrt{2}}

Step 2

Show that, no matter how many tiles are in the pattern, the total length of the series of tiles will be less than 3.5w.

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Answer

In this case, the lengths of the tiles form a geometric sequence where:

  • First term: a=wa = w
  • Common ratio: r=12r = \frac{1}{\sqrt{2}} The formula for the sum SnS_n of the first n terms of a geometric series is: Sn=a1rn1rS_n = a \frac{1 - r^n}{1 - r} As nn approaches infinity, since r<1r < 1, we can find the sum: S=w112S_\infty = \frac{w}{1 - \frac{1}{\sqrt{2}}} To simplify: S=w212=w221S_\infty = \frac{w}{\frac{\sqrt{2} - 1}{\sqrt{2}}} = \frac{w \sqrt{2}}{\sqrt{2} - 1} We can approximate 2\sqrt{2} as approximately 1.414, leading to: S<3.5wS_\infty < 3.5w

Step 3

Explain how you could refine the model used in part (b) to account for the 3 millimetre gap.

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Answer

To refine the model accounting for a 3 mm gap between tiles, we would need to add the additional length for gaps between each tile. If there are n tiles, there will be n - 1 gaps. Hence, the total additional length required is: 3(n1) mm3(n - 1) \text{ mm} Thus, the new total length of the tiles including gaps becomes: Snew=Sn+3(n1) mmS_{new} = S_n + 3(n - 1) \text{ mm} The total length will no longer have an upper limit since adding gaps will contribute a consistently increasing amount of length.

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