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Math Help - triangular blocks problem

  1. #1
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    triangular blocks problem

    the lowest row has 198 triangular blocks.For each of the other rows,the number of triangular blocks are 4 more than in the above row.Joe builds a triangular blocks in this way.The number of triangular blocks in the top row is 2.

    a)Calculate the number of rows in this arrangement

    b)If the height and the base of each triangular block is 6 cm and 8 cm respectively,calculate the area,in cm,of the triangle.
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  2. #2
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    Hello mastermin346
    Quote Originally Posted by mastermin346 View Post
    the lowest row has 198 triangular blocks.For each of the other rows,the number of triangular blocks are 4 more than in the above row.Joe builds a triangular blocks in this way.The number of triangular blocks in the top row is 2.

    a)Calculate the number of rows in this arrangement

    b)If the height and the base of each triangular block is 6 cm and 8 cm respectively,calculate the area,in cm,of the triangle.
    (a) Let's suppose there are n rows.

    In row 1 there are 198 blocks.

    In row 2 there are 198 - 4 \times 1 blocks.

    In row 3 there are 198 - 4\times 2 blocks.

    ...

    In row n there are 198 -4\times(n-1) blocks.

    So if there are 2 blocks in the top row:
    198-4(n-1)=2

    \Rightarrow 198 -4n +4 =2

    \Rightarrow 4n = 200

    \Rightarrow n = 50
    So there are 50 rows.

    (b) The area of each block = \tfrac12 \text{ base x height} = \tfrac12\; 6\times 8 = 24 \text{ cm}^2

    So we need to find the number of blocks altogether and multiply by 24.

    That total number, S, is:
    S=\underbrace{198 + 194 + 190 + ... + 6 + 2}_{50 \text{  items}}
    If we write the sum backwards:
    S=\underbrace{2+6+...+194+198}_{50 \text{  items}}
    and add the two together:
    2S=\underbrace{200 + 200 + ... + 200 + 200}_{50 \text{  items}}
    =50\times200

    =10\,000
    \Rightarrow S = 5\,000
    So the area = 24\times 5\,000 = 120\,000 \text{ cm}^2.

    Grandad
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