1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Express the area of an equilateral triangle as a function

  1. Mar 2, 2005 #1
    If A=x^2 and 2y-3x=12, express A as a function of y.
    i got A= 4/9y^2 - 2/3y + 20
    is that right?


    Express the area of an equilateral triangle as a function of its side.
  2. jcsd
  3. Mar 2, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    How did you arrive at your equation for A(y) ? Can you show your intermediary steps? (I found a different equation).

    For equilateral triangle, I'll steer you in the right direction. What can you tell me about he properties of an equilateral triangle?
    What are their angles equal to? Do you know the general formula for the area of a triangle?
  4. Mar 3, 2005 #3
    1) A=x^2 and 2y-3x=12;x=2/3y-4
    A=(2/3y-4)^2 - (2/3y-4)
    =4/9y^2 + 16 - 2/3y + 4
    =4/9y^2-2/3y +20
    Where is my mistake?

    2) 3 equal sides...all 60 degrees angle...B x H/2
  5. Mar 3, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    1) okay [tex] A=x^2 [/tex] and [tex] 2y-3x=12 [/tex] is given
    you solved the right expression for x,
    [tex] x= \frac{2}{3}y-4 [/tex] which looks fine

    Next I see you plugging in your value for x into an equation:
    [tex] A = x^2-x [/tex] Why?

    Where did you get that equation from? :confused:

    If you follow my meaning, next be careful when you expand [tex] (\frac{2}{3}y-4)^2 [/tex]
    it is the same as this operation [tex] (\frac{2}{3}y-4)(\frac{2}{3}y-4) [/tex]

    2) equilateral [tex]\Delta [/tex], 3 equal sides, all angles 60 deg, Area = 1/2 base*ht .. good

    Let each side of your [tex]\Delta [/tex] be length s
    If you draw your [tex]\Delta [/tex] with a vertex at the top, drop a vertical
    from that vertex to the bottom of [tex]\Delta [/tex] (so it bisects it).
    You now have two equal right [tex]\Delta [/tex]'s, inside your equilateral [tex]\Delta [/tex]. And the vertical line is now your height.
    Can you see where I'm leading?

    What are the angles inside your two right [tex]\Delta [/tex]s?
    Is there something special about them such that you can determine the
    sides of each right triangle in terms of s ?
    With that information you should be able to find the area of the equilateral
    [tex]\Delta [/tex] as a function of its side s. :cool:
    Last edited: Mar 3, 2005
  6. Mar 3, 2005 #5
    As a quick tip, if you want to check if you have performed manipulations like this correctly, put some numbers in to make sure that both sides of the equation are still the same.
  7. Mar 6, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    Did you see what I mean about this equation?
    [tex] A = x^2-x [/tex]?

    You have x=2/3y-4 and then A=(2/3y-4)^2 - (2/3y-4).
    But in the beginning you said that [tex] A = x^2 [/tex].

    How did you make out with the triangle part of your question?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook