Area of a Triangle

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Heron's Formula

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1 of 2 videos by turksvids

Key Questions

  • Answer:

    Please see below.


    How do you find the area of a triangle using the sine function?

    The question doesn't specify anything you may already know about the triangle, and that information is critical in deciding which procedure is necessary to arrive at a set of conditions that call for using the sine function for the calculation.

    However, I shall look at one possible set of facts that would be solved easily by using the sine function.

    Suppose you are given a triangle with the lengths of two adjacent sides and the size of the angle between them, as in:
    side a, side b, and
    angle C, between side a and side b.

    And, side a is the "base" of the triangle.

    Then, you could use this for the calculation, where the letters represent the numbers you have been given:
    # "Area of the triangle" = 1/2 * base * "altitude" #
    # "Area of the triangle" = 1/2 * a * b * sin(C) #

    If you draw a simple diagram to illustrate the above information, you might observe that the quantity
    # b * sin(C) # is the altitude of the triangle.

    Example diagram:

    enter image source here

  • Heron's formula allows you to evaluate the area of a triangle knowing the length of its three sides.
    The area #A# of a triangle with sides of lengths #a, b# and #c# is given by:


    Where #sp# is the semiperimeter:


    For example; consider the triangle:
    enter image source here
    The area of this triangle is #A=(base×height)/2#
    So: #A=(4×3)/2=6#
    Using Heron's formula:

    The demonstration of Heron's formula can be found in textbooks of geometry or maths or in many websites. If you need it have a look at:

  • Answer:

    Heron's Formula is almost always the wrong formula to use; try Archimedes' Theorem for a triangle with area #A# and sides #a,b,c#:

    #16A^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2#

    #quad = ( a^2+b^2+c^2)^2 - 2 (a^4+b^4+c^4) #

    #quad = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) #

    # quad = 16s(s-a)(s-b)(s-c) # where #s=1/2(a+b+c)#

    This last is thinly veiled Heron.


    Hero of Alexandria wrote in the first century AD. Why we continue to torture students with his result when there are much nicer modern equivalents I have no idea.

    Heron's formula for the area #A# of a triangle with sides #a,b,c# is

    # A = sqrt{s(s-a)(s-b)(s-c)} # where #s=1/2(a+b+c)# is the semiperimeter.

    There's no doubt this formula is awesome. But it's awkward to use because of the fraction and, if we start from coordinates, the four square roots.

    Let's just do the math. We square and eliminate #s# which mostly serves to hide a #16# and an important factorization. You might want to try it yourself first.

    #A^2 = 1/2(a+b+c)(1/2(a+b+c)-a)(1/2(a+b+c)-b)(1/2(a+b+c)-c) #

    #A^2 = 1/2(a+b+c)(1/2(-a+b+c))(1/2(a-b+c))(1/2(a+b-c)) #

    #16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) #

    That's already much better than Heron's form. We save the fraction to the end and there's no more wondering about the meaning of the semiperimeter.

    The degenerate case is telling. When one of those factors with a minus sign is zero, that's when two sides add up to exactly the other side. Those are distances between three collinear points, the degenerate triangle, and we get zero area. Makes sense.

    The #a+b+c# factor is interesting. What it tells us is this formula still works if we use displacements, signed lengths, instead of all positive.

    The formula is still awkward to use given coordinates. Let's multiply it out; you might want to try it yourself;

    #16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) #

    #= (-a^2-ab-ac + ab+b^2+bc + ac+bc+c^2) ( a^2-ab+ac + ab-b^2+bc -ac+bc-c^2)#

    #= (-a^2+b^2+c^2 +2bc)(a^2 - b^2-c^2 + 2bc) #

    #= (-a^2+b^2+c^2 +2bc)(a^2 - b^2-c^2 + 2bc) #

    #16A^2 = 2( a^2 b^2 + a^2 c^2 + b^2 c^2)-(a^4 +b^4+ c^4)#

    That form depends only on the squares of the lengths. It's clearly fully symmetrical. We can go beyond Heron now and say if the squared lengths are rational, so is the squared area.

    But we can do better if we note

    #( a^2+b^2+c^2)^2 = (a^4+b^4+c^4)+2(a^2b^2+a^2c^2+b^2c^2)#


    # 16A^2 = ( a^2+b^2+c^2)^2 - 2 (a^4+b^4+c^4) #

    That's the prettiest form.

    There's an asymmetric looking form that's usually the most useful. We note

    #( a^2+b^2 - c^2)^2 = (a^4+b^4+c^4)-2(-a^2b^2+a^2c^2+b^2c^2)#

    Adding this to

    #16A^2 = 2( a^2 b^2 + a^2 c^2 + b^2 c^2)-(a^4 +b^4+ c^4)#

    #16A^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2#

    That's the most useful form. There are really three ways to write it, swapping sides.

    Collectively these are called Archimedes' Theorem, from NJ Wildberger's Rational Trigonometry.

    When given 2D coordinates, often the Shoelace Formula is the quickest path to the area, but I'll save that for other posts.