Results 1 to 4 of 4

Math Help - Find SURFACE AREA of a pyramid (hexagonal-base)?

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    7

    Question Find SURFACE AREA of a pyramid (hexagonal-base)?

    How to find the surface area of a pyramid which has a regular hexagonal base of edge 6 cm and a height of 8 cm?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Surface area

    The area of the base should be simple, so the problem is the six triangular sides. Each triangle has area \tfrac{1}{2}bl, where b = 6 cm is the length of a side of the base, and l is the height of the triangle. Consider the right triangle formed by a line from the center of the base to the center of one of the sides of the base, by the center axis of the pyramid, and the height l of the side in question. Since b=6 cm, the distance from center of base to the side of the base is 6 \cdot \tfrac{\sqrt{3}}{2} = 3\sqrt{3}. Thus by the pythagorean theorem, l^2 = (8 cm)^2 + (3\sqrt{3} cm)^2 = 64 cm^2 + 27 cm^2 = 91 cm^2, and l = \sqrt{91} \approx 9.54 cm. From here, you can find the area of one of the triangles. Then just multiply by 6 to get the entire lateral area, and add the area of the base for the entire surface area.


    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Quote Originally Posted by kapitanvicki View Post
    How to find the surface area of a pyramid which has a regular hexagonal base of edge 6 cm and a height of 8 cm?
    For the base (img1):
    You can divide it up into 6 equilateral triangles by drawing diagonals from opposite points.

    The area of each of the equilateral triangles is

    A = \frac{1}{2}\left(6 \cdot \sqrt{6^2 -3^2}\right)=3\sqrt{27}=9\sqrt{3}

    So the Total Base Area is

    6 \cdot 9\sqrt{3} = 54\sqrt{3}



    Now for the slanting plane areas (img2):

    From before, we learnt that the length of b is \sqrt{6^2-3^2}=3\sqrt{3}

    Also, h = 8, as given.

    Therefore, a can be found by Pythagoras:

    a = \sqrt{(3\sqrt{3})^2+8^2}=\sqrt{91}

    a is the height of the triangular plane, so we can now work out its area:
    \frac{1}{2}\cdot 6 \cdot \sqrt{91}=3\sqrt{91}

    Multiplying by 6 gives us the area of all of them = 18\sqrt{91}

    So the total surface area is 54\sqrt{3}+18\sqrt{91}
    Attached Thumbnails Attached Thumbnails Find SURFACE AREA of a pyramid (hexagonal-base)?-img1.jpg   Find SURFACE AREA of a pyramid (hexagonal-base)?-img2.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2007
    Posts
    7

    Smile

    Thank you, divideby0! excellent answer (dont have much of an aptitude for maths) hehe...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface area of a pyramid
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 30th 2011, 06:08 PM
  2. hexagonal pyramid height
    Posted in the Geometry Forum
    Replies: 2
    Last Post: February 17th 2010, 08:52 AM
  3. Hexagonal Spherical Pyramid
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 28th 2009, 06:30 AM
  4. Replies: 1
    Last Post: February 5th 2009, 04:17 PM
  5. Replies: 2
    Last Post: May 30th 2008, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum