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Math Help - find the area

  1. #1
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    Thumbs up find the area

    In this regular pyramid find the area of
    i think that we have to solve this by using similar triangles and than continue with other things , and we have to solve it in terms of a and angle and .

    I'm having problems with this.
    Please help!
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  2. #2
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    Area

    Hello beq!x
    Quote Originally Posted by beq!x View Post
    In this regular pyramid find the area of
    i think that we have to solve this by using similar triangles and than continue with other things , and we have to solve it in terms of a and angle and .

    I'm having problems with this.
    Please help!
    You haven't told us much about the diagram, but I'll make the following assumptions:

    • the sides of the base are a units long
    • JK is parallel to AH
    • M and N are the mid-points of the sides BG and AH

    You need to know the formula:

    • Area of a trapezium = \tfrac12(a+b)h

    where the lengths of the parallel sides are a and b, and the height is h.

    So in BGKJ, the area we need is \tfrac12(BG + JK)LM.

    Now BG = a (I'm assuming!), and you can find LM using the Sine Rule on \triangle NML:

    \frac{LM}{\sin\alpha}= \frac{a}{\sin(180 - (\alpha+\beta))}

    \Rightarrow LM = \frac{a\sin\alpha}{\sin(\alpha+\beta)}

    To find JK, you need to use the fact that triangles IJK and IAH are similar, together with the lengths of IL and IN. First IN: drop a perpendicular from I to the base, and then from the right-angled triangle thus formed:

    IN = \tfrac12a\tan\alpha

    From \triangle LMN,\, NL = \frac{a\sin\beta}{\sin(\alpha+\beta)}

    \Rightarrow IL = \tfrac12a\tan\alpha-\frac{a\sin\beta}{\sin(\alpha+\beta)}

    So from similar triangles \frac{JK}{a}=\frac{IL}{IN} = \frac{\tfrac12a\tan\alpha-\dfrac{a\sin\beta}{\sin(\alpha+\beta)}}{\tfrac12a\  tan\alpha}

    Can you complete it from here?

    Grandad
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  3. #3
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    no i can't
    is the solution available ?
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  4. #4
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    Hello beq!x

    A correction to my first posting, and then the solution. I meant \sec\alpha rather than \tan\alpha. I.e.

    IN = \tfrac12a\sec\alpha

    \Rightarrow IL = \tfrac12a\sec\alpha-\frac{a\sin\beta}{\sin(\alpha+\beta)}

    \Rightarrow\frac{JK}{a}= \frac{IL}{IN}=\frac{\tfrac12a\sec\alpha-\dfrac{a\sin\beta}{\sin(\alpha+\beta)}}{\tfrac12a\  sec\alpha}

    \Rightarrow JK = \frac{a\sec\alpha-\dfrac{2a\sin\beta}{\sin(\alpha+\beta)}}{\sec\alph  a}

    = a - \frac{2a\sin\beta\cos\alpha}{\sin(\alpha+\beta)}

    So the area = \tfrac12\Big(2a -\frac{2a\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\B  ig)\frac{a\sin\alpha}{\sin(\alpha+\beta)}

    = \Big(1 -\frac{\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big  )\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}

    = \Big(\frac{\sin(\alpha+\beta)-\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big)\frac  {a^2\sin\alpha}{\sin(\alpha+\beta)}

    = \Big(\frac{\sin\alpha\cos\beta}{\sin(\alpha+\beta)  }\Big)\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}

    = \frac{a^2\sin^2\alpha\cos\beta}{\sin^2(\alpha+\bet  a)}

    Grandad
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