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Math Help - Trigonometry - modelling and problem solving

  1. #1
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    Trigonometry - modelling and problem solving

    I am having trouble finding the length of a sloping edge in the following problem:

    A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the legth of a sloping edge and the area of a triangular face.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Trigonometry - modelling and problem solving

    Quote Originally Posted by elmidge View Post
    I am having trouble finding the length of a sloping edge in the following problem:

    A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the length of a sloping edge and the area of a triangular face.
    Suppose the right pyramid is like the one in the following image:

    Trigonometry - modelling and problem solving-right_pyramid.png

    Right pyramid ABCDE has a square base BCDE. This square base BCDE has all equal sides where each side has length 5 \text{ cm}. We also know that the height of the right pyramid is 6\text{ cm} which is AO

    Now in sqare BCDE, \triangle BED is a right triangle with \angle BED a right angle.
    So by Pythogorean theorem:

     BD^2 = BE^2 + ED^2

    But BE = ED = 5 \text{ cm}

    So BD = 5\sqrt{2}

    BO = \frac{1}{2}BD = \frac{5}{\sqrt{2}} \text{ cm}

    We know that \text{height} = AO = 6\text{ cm}

    Now \triangle AOB is a right triangle with \angle AOB a right angle.
    So:

    AB^2 = AO^2 + BO^2
    \therefore AB = \frac{\sqrt{97}}{\sqrt{2}}......\text{[because AO = } 6 \text{ cm  and  BO }= \frac{5}{\sqrt{2}}\text{ cm]}

    That's the length of the slopping edge which is \frac{\sqrt{97}}{\sqrt{2}}\text{ cm}

    So the area of:
    \begin{align*}\triangle ABC =& \frac{1}{2} \times BC \times AO\\=& 15\text{ cm}^2...\text{[because BC} = 5 \text{ cm and AO} = 6 \text{ cm]}\end{align*}

    So the area of the triangular face is 15\text{ cm}^2
    Thanks from HallsofIvy and elmidge
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