Results 1 to 4 of 4

Math Help - Area of a Equil. Triangle given 1 measure

  1. #1
    Junior Member
    Joined
    Nov 2011
    Posts
    28

    Area of a Equil. Triangle given 1 measure

    Here is the question: Gyazo - 8d2dd59242fb7fa392dbdc6fa84e79a5.png (question 25)
    Here is my work: Gyazo - 9fcbe6021e7df14bd5c03df38dd82d3a.png

    I could've used A=.5*b*h (regular area formula for triangle) but since I've been using A=.5*apothem*perimeter, I figured I'd continue using it. My final answer was 32in squared.

    Used 30-60-90 special right triangle rules etc. If something isn't clear in my scrap, let me know.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,559
    Thanks
    1002

    Re: Area of a Equil. Triangle given 1 measure

    Quote Originally Posted by Iceycold View Post
    Here is the question: Gyazo - 8d2dd59242fb7fa392dbdc6fa84e79a5.png (question 25)
    Here is my work: Gyazo - 9fcbe6021e7df14bd5c03df38dd82d3a.png

    I could've used A=.5*b*h (regular area formula for triangle) but since I've been using A=.5*apothem*perimeter, I figured I'd continue using it. My final answer was 32in squared.

    Used 30-60-90 special right triangle rules etc. If something isn't clear in my scrap, let me know.
    This isn't right.

    You've got a 30/60/90 degree triangle with the side opposite the 30 deg angle, $h=4$.

    so by the properties of this triangle the hypotenuse is 8, and the other, $b=4\sqrt{3}$.

    The area of a right triangle is $\large \frac{b h}{2}$ and you have 6 of these triangles in your original triangle.

    See if you can come up with the correct answer now.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648

    Re: Area of a Equil. Triangle given 1 measure

    Hello, Iceycold!

    Find the area.

    Code:
                  *
                 *|*
                * | *
               *  |4 *
              *   |   *
             *    |    * 2x
            *     *     *
           *      |      *
          *       |2      *
         *        |     60 *
        * * * * * * * * * * *
                       x

    The "center" (centroid) of a triangle
    . . divides a median in the ratio 2:1.

    We have a 30-60-90 right triangle
    . . with legs x and 6, and hypotenuse 2x.

    Hence: . x^2 + 6^2 \:=\:(2x)^2 \quad\Rightarrow\quad x^2 + 36 \:=\:4x^2

    . . . 3x^2 \:=\:36 \quad\Rightarrow\quad x^2 \:=\:12 \quad\Rightarrow\quad x \:=\:2\sqrt{3}


    We have a triangle with base b = 4\sqrt{3} and height h = 6.

    It area is: . A \;=\;\tfrac{1}{2}(4\sqrt{3})(6) \;=\;12\sqrt{3}\text{ in}^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2011
    Posts
    28

    Re: Area of a Equil. Triangle given 1 measure

    Quote Originally Posted by romsek View Post
    This isn't right.

    You've got a 30/60/90 degree triangle with the side opposite the 30 deg angle, $h=4$.

    so by the properties of this triangle the hypotenuse is 8, and the other, $b=4\sqrt{3}$.

    The area of a right triangle is $\large \frac{b h}{2}$ and you have 6 of these triangles in your original triangle.

    See if you can come up with the correct answer now.
    I believe this is what you're talking about: Gyazo - 6bde3478b337984dd9c270f12dc82a3b.png

    If the height is 4, wouldn't the short leg (side opposite of 30deg) 4/sqrt3? Then *2 for the hypotenuse?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the angle measure of a triangle
    Posted in the Geometry Forum
    Replies: 13
    Last Post: August 3rd 2011, 10:02 AM
  2. Replies: 1
    Last Post: October 28th 2008, 07:02 PM
  3. Replies: 7
    Last Post: July 19th 2008, 06:53 AM
  4. Replies: 27
    Last Post: April 27th 2008, 10:36 AM
  5. Replies: 6
    Last Post: August 10th 2005, 08:29 AM

Search Tags


/mathhelpforum @mathhelpforum