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Math Help - Please i need help finding the cross sectional area trapezium!...

  1. #1
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    Question Please i need help finding the cross sectional area trapezium!...

    1(b) A cutting is to be made through rough ground. The width of the bottom has to be a constant 4m and the sides of the cutting must be at a slope of 1 in 1. The depth s of the cutting at 6 m intervals are :-

    (a) 1.8m (b) 1.4m (c) 1.4m (d) 1.8m (e) 2.0m (f) 2.6m
    (g) 3.2m (h) 3.6m (i) 2.8m (j) 2.6m (k) 2.4m (l) 2.8m

    (m) 3.8m

    (ii) Determine the cross sectional area for each cross section.
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  2. #2
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    Re: Please i need help finding the cross sectional area trapezium!...

    Use the expression for the area of a trapezoid, which is equal to 1/2 [ sum of the parallel sides x distance between them ]
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  3. #3
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    Re: Please i need help finding the cross sectional area trapezium!...

    Hello, ashicus!

    \text{1. A channel is to be dug through level ground.}
    \text{The width of the bottom has to be a constant 4m,}
    \text{and the sides of the channel must be at a slope of 1-to-1.}
    \text{The depth }h\text{ of the channel, at 6 m intervals, are:}

    . . \begin{array}{c}\text{(a) 1.8m} \quad \text{(b) 1.4m} \quad \text{(c) 1.4m } \quad  \text{(d) 1.8m} \quad  \text{(e) 2.0m} \quad \text{(f) 2.6m } \quad \text{(g) 3.2m} \\ \text{(h) 3.6m } \quad \text{(i) 2.8m} \quad \text{(j) 2.6m} \quad \text{(k) 2.4m} \quad \text{(l) 2.8m} \quad \text{(m) 3.8m} \end{array}

    \text{Determine the cross-sectional area for each cross section.}


    The cross section is an isosceles trapezoid.
    Code:
          : - h - : - - 4 - - : - h - :
          * * * * * * * * * * * * * * *
            *     |           |     *
              *   |h          |   *
                * |           | * 45o
                  * * * * * * * - - -
                  : - - 4 - - :
    The area of a trapezoid is: . A \;=\;\tfrac{h}{2}(b_1+b_2)
    . . where h is the height, and b_1,\,b_2 are the lengths of the two bases.

    We have height h and bases 4 and 2h+4.

    Therefore: . A \;=\;\tfrac{h}{2}(4+2h+4) \quad\Rightarrow\quad \boxed{A \;=\;h(h+4)}


    Now plug in the thirteen values for h.
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