Results 1 to 4 of 4

Math Help - find the area

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    42

    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!
    Attached Thumbnails Attached Thumbnails find the area-12121212121212.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    42
    no i can't
    is the solution available ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 5th 2010, 08:48 PM
  2. Find the area
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 27th 2009, 05:38 AM
  3. Find the area
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 21st 2008, 07:37 PM
  4. Find the area
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 19th 2008, 06:24 PM
  5. find area of shaded area (yellow)
    Posted in the Geometry Forum
    Replies: 8
    Last Post: September 4th 2007, 08:34 AM

Search Tags


/mathhelpforum @mathhelpforum