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Math Help - Geometrical Relations

  1. #1
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    Talking Geometrical Relations

    Please any one can help me in these two calculations.


    5.1) A sheet of thick plastic is used to make a traffic cone. What area of plastic must be cut from the sheet to make a cone of vertical height 40cm and diameter 30cm, as shown below?



    5.2

    Referring to the trapezoidal prism show above, find the following, showing formulae and calculations.





    6.1) Area of the trapezoid (i.e. the front view)



    6.2) Volume of the prism



    6.3) Total surface area of the prism.




    2.
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  2. #2
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    Hello, Sane!

    Here's the first one . . . it's very messy!


    1) A sheet of thick plastic is used to make a traffic cone.
    What area of plastic must be cut from the sheet to make
    a cone of vertical height 40cm and diameter 30cm?
    Code:
                *
               /|\
              / | \
             /  |  \ R
            /   |40 \
           /    |    \
          *-----+-----*
                   15
    We have a right triangle with legs 15 and 40.
    . . The hypotenuse is: . R \:=\:\sqrt{15^2 + 40^2} \:=\:\sqrt{1875} \:=\:5\sqrt{73}


    The cone will be cut from a circle of radius R.
    Code:
                  * * *
              *           *
            *               *
           *                 *
    
          *         O         *
          *         *         *
          *       *:::*   R   *
                *:::::::*
           *  *:::::::::::*  *
           A*:::::::::::::::*B
              *:::::::::::*
                  * * *
    The area of the entire circle is: . A \:=\:\pi(5\sqrt{73})^2 \:=\:1825\pi cm

    The cone takes up some fraction of the circle . . . What fraction?


    The arc AB is some fraction of the circumference of the circle.
    . . And that is the fraction we seek.

    The circle has circumference: . C_1 \:=\:2\pi(5\sqrt{73}) \:=\:10\sqrt{73}\pi cm

    Arc AB is the circumference of the base-circle of the cone.
    . . The base-circle has: . C_2 \:=\:\pi D \:=\:30\pi cm

    There is our fraction . . . \frac{C_2}{C_1}\;=\;\frac{30\pi}{10\sqrt{73}} \;=\;\frac{3}{\sqrt{73}}


    The area of the cone is: . \frac{3}{\sqrt{73}} \times 1825\pi \;=\;75\sqrt{73}\pi cm

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  3. #3
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    Hello, Sane!

    Do you know any area formulas?


    Referring to the trapezoidal prism shown above,
    find the following, showing formulae and calculations.

    6.1) Area of the trapezoid (i.e. the front view)
    Area of trapezoid: . A \;=\;\frac{h}{2}(b_1 + b_2)

    We have: . h \,= \,4,\;b_1\,=\,4,\;b_2\,=\,10

    You finish it . . .



    6.2) Volume of the prism
    \text{Volume of prism } \:= \:\text{ area of face} \times \text{length}

    In part (6.1) we found that: A = 28

    You finish it . . .



    6.3) Total surface area of the prism
    This takes a bit more Thinking . . . ready?
    Code:
                *---------*
               /|         |\
              / |         | \
             /  |        4|  \ 5
            /   |         |   \
           /    |         |    \
          *-----+---------+-----*
                             3

    On the right, we have a right triangle with sides 3 and 4.
    . . Hence, the hypotenuse is 5.


    There are two 5-by-10 rectangles on the left and right sides.
    . . Their total area is: . 2 \times (5\cdot10) \:=\:100 m

    There are two trapezoidal face, each with area 28 m.
    . . Their total area is: . 2 \times 28 \:=\: 56 m

    The top is a 4-by-10 rectangle; its area is: . 4 \times 10\:=\:40 m

    The bottom is a 10-by-10 rectangle; its area is: . 10 \times 10 \:=\:100 m


    The total surface area is: . 100 + 56 + 40 + 100 \:=\:\boxed{296\text{ m}^2}

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