Results 1 to 2 of 2

Math Help - QUICK Yo, I need me some help!!!!

  1. #1
    Cflow
    Guest

    30 60 90 and isosceles triangles

    If a 30-60-90 degree right triangle and an isosolies right triangle have the same perimeter, which one will have the bigger area, and why?

    I need to come up with the perimeters for both triangles as well, can ya'll help me?

    Thank you bunches!
    Love,
    Carly
    Last edited by MathGuru; May 26th 2005 at 12:15 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    30-60-90 triangle:
    The leg opposite the 30-degree angle is the short leg.
    The leg opposite the 60-degree angle is the long leg.

    The hypotenuse is twice the short leg.
    If x = short leg, then hypotenuse = 2x.
    The longer leg then is x*sqrt(3).

    Perimeter, P1,
    = x +x*sqrt(3) +2x
    = x +x(1.732) +2x
    = x(4.732)

    Isosceles right triangle:
    The two legs are equal in length.
    If y = one leg, then hypotenuse = y*sqrt(2).

    Perimeter, P2,
    = y +y +y*sqrt(2)
    = y +y +y(1.4142)
    = y(3.4142)

    ----------
    "... have the same perimeter, which one will have the bigger area, and why?"

    If P1 = P2, then,
    x(4.732) = y(3.4142)
    y = x(4.732)/(3.4142)
    y = x(1.386) ...........****

    Let us get the areas of both right triagles in terms of x.

    Area of right triangle = (1/2)(base)(altitude) = (1/2)(leg)(other leg)

    30-60-90:
    A1 = (1/2)(x)(x*sqrt(3))
    A1 = (0.866)(x^2)

    isosceles:
    A2 = (1/2)(y)(y)
    A2 = (0.5)(y^2)
    Substitute x(1.386) for y,
    A2 = (0.5)(1.386x)^2
    A2 = (0.96)(x^2)

    So, since (A2 = 0.96x^2) is greater than (A1 = 0.866x^2), then,
    Therefore, the isosceles right triangle has a bigger area. .....answer.

    ---------
    "I need to come up with the perimeters for both triangles as well,..."

    Since both perimeters are equal, then if you assign any value to x, (or y), you can get the perimeters.

    If x = 1,
    P1 = P2 = x(4.732) = 1(4.732) = 4.732

    If x = 2,
    P1 = P2 = x(4.732) = 2(4.732) = 9.464

    If x = 8,
    P1 = P2 = x(4.732) = 8(4.732) = 37.856

    Based upon the given information, there is no unique value for the perimeter.
    Last edited by ticbol; May 13th 2005 at 02:28 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Just a quick one...
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 5th 2009, 05:54 PM
  2. Help me quick
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 16th 2009, 02:09 PM
  3. need some quick help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 18th 2008, 02:56 PM
  4. How many squares? quick quick quick!
    Posted in the Math Challenge Problems Forum
    Replies: 8
    Last Post: December 10th 2007, 08:11 PM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum