Results 1 to 2 of 2

Math Help - Fun with the right trianlge.

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9

    Fun with the right trianlge.

    The other day I saw something interesting,
    Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by ThePerfectHacker
    The other day I saw something interesting,
    Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.
    Hello,

    I'm a little bit puzzled about your question: What are the segments of the hypotenuse and what exactly do you take as divisor? The radius, the perimeter or the area of the inscribed circle?

    I've attached a drawing of a right triangle so you can check which names I use for the variables.

    Heron of Alexandra had found that in a right triangle is \rho=\frac{a \cdot b}{a+b+c}


    The three bisectors of a triangle divide it into three smaller triangles. The height in every of these smaller triangles is always ρ. That's true with every triangle.

    So you get:
     A=\frac{1}{2} \cdot \rho \cdot a+\frac{1}{2} \cdot \rho \cdot b+\frac{1}{2} \cdot \rho \cdot c=\frac{1}{2} \cdot \rho \cdot (a+b+c)

    Heron set \frac{1}{2} \cdot (a+b+c)=s and you get: A=\rho \cdot s

    With a right triangle you've additional conditions:
    A=\frac{1}{2} \cdot c \cdot h=\frac{1}{2} \cdot a \cdot b=\frac{1}{2} \cdot \rho \cdot (a+b+c)

    and therefore \rho=\frac{a\cdot b}{a+b+c}

    With a^2=p \cdot c and b^2=q \cdot c you can change p \cdot q=\frac{a^2}{c} \cdot \frac{b^2}{c}=\frac{a^2 \cdot b^2}{c^2}

    Now you want to divide this term by a part of the inscribed circle. But I cann't help you any further because I don't know which part you want to take. But maybe you can use my remarks to get a little bit ahead.

    Bye
    Attached Thumbnails Attached Thumbnails Fun with the right trianlge.-heron_rwd.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove for trianlge ABC
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 25th 2010, 12:44 AM
  2. centriod of trianlge
    Posted in the Geometry Forum
    Replies: 1
    Last Post: January 28th 2010, 02:39 AM
  3. How to draw an isosceles trianlge
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 23rd 2006, 12:22 AM

Search Tags


/mathhelpforum @mathhelpforum