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Math Help - Lateral Area...

  1. #1
    Junior Member whytechocolate01's Avatar
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    Post Lateral Area...

    Find the lateral area of a regular triangular pyramid with base edge 3 in. and slant height 6 in.

    Here's how I worked it out. Please tell me if this is correct. If it isn't please explain why and how to get the right answer. Thank you!

    V=1/3*s^2*h
    s=3
    h=6
    V=1/3*3^2*6
    = 1/3*9*6
    V=18
    Attached Thumbnails Attached Thumbnails Lateral Area...-slanted-triangle.jpg  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by whytechocolate01 View Post
    Find the lateral area of a regular triangular pyramid with base edge 3 in. and slant height 6 in.

    Here's how I worked it out. Please tell me if this is correct. If it isn't please explain why and how to get the right answer. Thank you!

    V=1/3*s^2*h
    s=3
    h=6
    V=1/3*3^2*6
    = 1/3*9*6
    V=18
    This is not answering the question. You have found a volume (not that
    of the pyramid as it happens), but the question asks for the lateral area,
    which is the surface area of the solid less the area of the base.

    The area of a single face is b*s/2, so the area of three of these is:

    A = 3*b*s/2 = 3*3*6/2=27 sq inches.

    where b is the lebgth of one of the base edges (they are all the same)
    and s is the slant height.

    RonL
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