Results 1 to 3 of 3

Math Help - Proof: Area of an isosceles triangle

  1. #1
    Member
    Joined
    Aug 2012
    From
    Utah
    Posts
    88

    Proof: Area of an isosceles triangle

    Area of an isosceles triangle = c/4*sqrt(4a^2 - c^2).

    Can someone give proof for this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,407
    Thanks
    1294

    Re: Proof: Area of an isosceles triangle

    Are "a" and "c" the equal sides of the triangle or are they two sides with different lengths?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2014
    From
    USA
    Posts
    3
    Thanks
    1

    Re: Proof: Area of an isosceles triangle

    Hi hisajesh,

    An isosceles triangle has two equal sides - so let's call the length of these sides "a". Also, let's call the other side of the isosceles triangle "c".
    Something neat about isosceles triangles, if you create a segment (call it "b") that bisects the angle between the two equivalent sides to "c" (try to visualize this before moving on), then the segment will also bisect the length of "c" into two equal parts.

    This will split your isosceles triangle into two congruent right triangles; the hypotenuse of these right triangles is "a", one leg is "c/2", and the last leg is our new segment "b".
    Let's try to express "b" in terms of "a" and "c". Since we are dealing with a right triangle, we have a^2 = b^2 + (c/2)^2 (*important* remember that "a" is our hypotenuse - not "c"!!!)
    This means b = sqrt (a^2 - (c/2)^2)

    Let's try to visualize our isosceles triangle again - we have two equivalent sides ("a") and a third side between them ("c"). Then we created a line segment from the angle between the two equivalent sides to the middle of "side c", and we called this "b".
    The area of a triangle is:
    A = (1/2)*base*height = (1/2) * c * b = (1/2) * c * sqrt(a^2 - (c/2)^2) = (1/2) * c * sqrt( 1 * (a^2 - (c/2)^2)) = (1/2) * c * sqrt( (4/4) * (a^2 - (c/2)^2))
    = (1/2) * c * sqrt( (1/4) * (4*a^2 - 4*(c/2)^2)) = (1/2) * c * sqrt( (1/4)) * sqrt (4*a^2 - 4 * (c/2)^2)
    = (1/2) * c * 1/2 * sqrt(4*a^2 - 4*c^2/4) = (1/4) * c * sqrt(4*a^2 - c^2)
    = c/4 * sqrt(4*a^2 - c^2)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of isosceles right triangle
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 15th 2009, 08:57 PM
  2. Area of an Isosceles Triangle
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 29th 2008, 06:31 PM
  3. Possible isosceles triangle/area?
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 6th 2008, 05:54 PM
  4. Replies: 27
    Last Post: April 27th 2008, 10:36 AM
  5. Area of an Isosceles Triangle
    Posted in the Geometry Forum
    Replies: 2
    Last Post: October 21st 2006, 11:30 AM

Search Tags


/mathhelpforum @mathhelpforum