Results 1 to 9 of 9

Math Help - Equilateral Triangles..Finding Sides..Areas...

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    55

    Equilateral Triangles..Finding Sides..Areas...

    Hello. Here is the question I have to solve. Its attached. Now, what I came up with was the tick marks are each 6 (whatever unit) (doesn't specify). However I am very confused, on how to solve this.

    Thanks
    Attached Thumbnails Attached Thumbnails Equilateral Triangles..Finding Sides..Areas...-photo.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    55

    Smile

    I thought I was supposed to use the pythagorean therorum. What did you do?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    Quote Originally Posted by zachcumer View Post
    I thought I was supposed to use the pythagorean therorum. What did you do?
    The tick marks indicate they are of the same length, so each side is 12.
    Because each side is split in the middle we know that each \frac{1}{2} is 6 units.
    We also know that the 3 corners of the triangle each have 60^0 because all side lengths are the same, meaning it's equilateral.
    So for our seperate triangles we have 30^0 angles, right angles and 6 unit lengths, which is enough for trigonometry.

    Though if you're looking for a solution using Pythagorean theorem then I'll take another look.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2008
    Posts
    55
    cool. thanks. any help I could get would be great.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    Bah I keep disproving myself, I got x=2.1962...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by zachcumer View Post
    Hello. Here is the question I have to solve. Its attached. Now, what I came up with was the tick marks are each 6 (whatever unit) (doesn't specify). However I am very confused, on how to solve this.

    Thanks
    (See photo2.jpg)

    Solve for L using pythagoras. I find L = 6 \sqrt{3}

    ----

    (See photo3.jpg)

    That red line is half the length of the green line.
    Therefore the red line has length: 2 \sqrt{3}

    ----

    (See photo4.jpg)

    We once again have a nice right angled triangle to work with.
    We have two sides so solve for the third(green) one.
    The green side is 4 \sqrt{3}

    ----

    (See photo5.jpg)

    A right angled triangle again, solving for the green side, which is finally x.

    x = 2 \sqrt{3}



    =============

    Man i was stupid. This is an equiliteral triangle. All these lines inside the triangle have length  6 \sqrt{3} !! anyway you have the value for x now...
    Attached Thumbnails Attached Thumbnails Equilateral Triangles..Finding Sides..Areas...-photo2.jpg   Equilateral Triangles..Finding Sides..Areas...-photo3.jpg   Equilateral Triangles..Finding Sides..Areas...-photo4.jpg   Equilateral Triangles..Finding Sides..Areas...-photo5.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    Quote Originally Posted by janvdl View Post
    (See photo2.jpg)


    x = 2 \sqrt{3}



    =============

    Man i was stupid. This is an equiliteral triangle. All these lines inside the triangle have length  6 \sqrt{3} !! anyway you have the value for x now...
    Ah good, well if x = 2 \sqrt{3} then:

    y^2=(2 \sqrt{3})^2+36 \Rightarrow y^2= 12+36 \Rightarrow y= 4\sqrt{3} = 6.928
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Nyoxis View Post
    Ah good, well if x = 2 \sqrt{3} then:

    y^2=(2 \sqrt{3})^2+36 \Rightarrow y^2= 12+36 \Rightarrow y= 4\sqrt{3} = 6.928
    Why such a long approach??

    y = 2x according to that rule... Now what was it called again? Anyway, it does exist
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equilateral Triangles.
    Posted in the Geometry Forum
    Replies: 4
    Last Post: January 6th 2011, 05:48 AM
  2. Replies: 3
    Last Post: October 29th 2009, 07:53 AM
  3. equilateral triangles
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 21st 2009, 02:03 AM
  4. Finding a triangles angles from its area + 2 sides?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 9th 2009, 02:45 PM
  5. Replies: 7
    Last Post: March 10th 2008, 06:02 AM

Search Tags


/mathhelpforum @mathhelpforum