Results 1 to 14 of 14

Math Help - Perimeter of a triangle

  1. #1
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468

    Perimeter of a triangle

    AB=3.2
    BC=8.4
    Area=10cm^2

    Calculate the perimeter.

    Method not answer. Calculator allowed.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Mukilab View Post
    AB=3.2
    BC=8.4
    Area=10cm^2

    Calculate the perimeter.

    Method not answer. Calculator allowed.
    What have you tried? Where are you stuck?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by mr fantastic View Post
    What have you tried? Where are you stuck?
    I don't know the formula at all for this. I cannot even attempt to try.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Hi there,

    Let me first say there is probably a few ways to do this other than the way I am going to suggest. Something with Herron's formula may be easier.

    My solution takes a few steps so be patient.

    First of all, in triangle with sides a,b,c and with angles A,B,C the perimeter needs all sides to be found and summed.

    You have information about the area and sides a= 8.4 and side c = 3.2

    Now use Area = \frac{1}{2}\times a \times c \times \sin(B) to solve for B

    Here is the sum, use inverse \sin to complete.

    10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)

    After you have your angle value for B you should employ the \cos rule to find b

    It is b = \sqrt{a^2+c^2-2ac\times \cos(B)}

    Now go ahead a change the world!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by pickslides View Post
    Hi there,

    Let me first say there is probably a few ways to do this other than the way I am going to suggest. Something with Herron's formula may be easier.

    My solution takes a few steps so be patient.

    First of all, in triangle with sides a,b,c and with angles A,B,C the perimeter needs all sides to be found and summed.

    You have information about the area and sides a= 8.4 and side c = 3.2

    Now use Area = \frac{1}{2}\times a \times c \times \sin(B) to solve for B

    Here is the sum, use inverse \sin to complete.

    10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)

    After you have your angle value for B you should employ the \cos rule to find b

    It is b = \sqrt{a^2+c^2-2ac\times \cos(B)}

    Now go ahead a change the world!
    I don't know how to use reverse sin for lines, only for angles and google is not helping me. Although thank you for trying
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Mukilab View Post
    I don't know how to use reverse sin for lines,

    use \sin^{-1} on the side length, it will work.
    Last edited by mr fantastic; January 8th 2010 at 12:23 PM. Reason: Fixed latex
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    Quote Originally Posted by pickslides View Post
    use \sin^{-1} on the side length, it will work.
    What do you mean side length? I know how to use sin^{-1} with angles. I tried getting the whole thing in (apart from the area) into that and there was a math error as spoken by my calculator.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Mukilab View Post
    What do you mean side length? I know how to use sin^{-1} with angles. I tried getting the whole thing in (apart from the area) into that and there was a math error as spoken by my calculator.

    Show me your workings I will try to correct them.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468
    sin^{-1}(1/2 x 3.2 x 8.4)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Mukilab View Post
    sin^{-1}(1/2 x 3.2 x 8.4)
    This is how it goes

    10 = \frac{1}{2}\times 8.4 \times 3.2 \times \sin(B)

    \frac{10}{\frac{1}{2}\times 8.4 \times 3.2} = \sin(B)

    \sin(B) = \frac{10}{\frac{1}{2}\times 8.4 \times 3.2}

    Now take the inverse.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Hi Mukilab,

    The side measurements you are given are

    3.2cm and 8.4cm.

    If you knew the 3rd side length, you could add all 3 lengths together
    to calculate the perimeter.

    The clue you are given to find the 3rd side is.. triangle area = 10 sq cm.

    You must understand how the area of a triangle is calculated in order to
    go in reverse.
    Inverse Sine is not confusing when you have mastered Sine.
    Inverse addition is subtraction, inverse multiplication is division,
    inverse square is square root and so on.

    Triangle area is half the base times the perpendicular height.
    When you are not given the perpendicular height directly, then the

    triangle\ area=0.5abSinC gives the same result.

    You should study why this is so....
    if "a" is the base, then bSinC is the perpendicular height.

    I want to add something here to make this more complete.

    The formula area=0.5abSinC

    means that if you have the 3 dimensions of a corner then you can calculate the area.

    0.5abSinC=0.5acSinB=0.5bcSinA all give triangle area.

    Remember, side a is opposite angle A, b is opposite B, c is opposite C.

    You need to study and practice with that.

    If that's very clear, you can continue....

    Your problem is asking you to work this in reverse.
    You are given the area and two sides.

    What dimensions calculate the area?

    two sides and the angle that those two sides make! ok?

    So 0.5(3.2)(8.4)SinX=10, where X is the angle between those sides.

    SinX=\frac{10}{(0.5)(3.2)(8.4)}

    Therefore X is the inverse Sine of that fraction.

    That's putting it simply, but unfortunately, there are 2 possible answers.
    Your calculator gives you the one that is less than 90 degrees.
    The other one is 180 degrees minus the one your calculator gave you.

    Just work with the one your calculator gave you.

    there are more steps after you've got the angle.
    You should just go step by step.

    So try to calculate the angle between the two lines first.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, Mukilab!

    We don't need inverse trig functions . . . just Pythagorus.


    AB\,=\,3.2,\;\;BC\,=\,8.4,\;\;\text{Area}\:=\:10\t  ext{ cm}^2

    Calculate the perimeter.
    Code:
                      A
                      o
                    *  \
         c = 3.2  *     \
                *        \ b
              *           \
            * θ            \
        B *  *  *  *  *  *  * C
               a = 8.4

    \text{Area: }\:A \:=\:\tfrac{1}{2}ac\,\sin B

    We have: . \text{Area} \:=\:\tfrac{1}{2}(3.2)(8.4)\sin B \:=\:10

    . . \sin B \:=\:\frac{10}{13.44} \quad \Rightarrow\quad \sin B \:=\:\frac{125}{168} .[1]


    Since \sin^2\!B + \cos^2\!B \:=\:1\text{, then: }\:\cos B \:=\:\sqrt{1-\sin^2\!B}

    Substitute [1]: . \cos B \:=\:\sqrt{1 - \left(\frac{125}{168}\right)^2} \quad\Rightarrow\quad \cos B \:=\:\frac{\sqrt{12599}}{168}


    Law of Cosines: . b^2 \;=\;a^2+c^2-2ac\cos B

    We have: . b^2 \;=\;8,4^2 + 3.2^2 - 2(8.4)(3.2)\left(\frac{\sqrt{12599}}{168}\right)

    . . . . . . . . . =\;70.56 + 10.24 - 35.91848549 \;=\;44.88151451


    Hence: . b \;=\;\sqrt{44.88151451} \;=\;6.699366724 \;\approx\;6.7\text{ cm}


    Got it?

    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Or...

    10=0.5(3.2)(8.4)SinB=3.2(4.2)SinB

    SinB=\frac{(\frac{10}{4.2})}{3.2}

    which is \frac{opposite}{hypotenuse}

    for the right-angled triangle on the left with perpendicular height \frac{10}{4.2}

    Then split the base "a" into 2 parts x and y.

    x^2+(\frac{10}{4.2})^2=3.2^2

    x^2=3.2^2-(\frac{10}{4.2})^2

    x=\sqrt{3.2^2-(\frac{10}{4.2})^2}

    Then y=8.4-x

    Finally y^2+h^2=b^2

    b=\sqrt{y^2+h^2}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Dec 2009
    Posts
    14
    You can try Heron's formula for the area of a triangle :
    E=sqrt[t(t-a)(t-b)(t-c)], sqrt:square root, t=1/2 of perimeter or (a+b+c)/2.
    I tried the method, but i found out that it leads to a 4th degree equation(if you take y=b^2 the equation becomes quadratic) with no solutions!!! Check the facts and figures of your problem again. Maybe i had made a mistake during my calculations.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Perimeter of right triangle?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 9th 2012, 07:56 AM
  2. triangle perimeter
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 18th 2010, 06:25 AM
  3. perimeter of a triangle?? huh??
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 26th 2008, 06:55 PM
  4. Perimeter of a Triangle
    Posted in the Geometry Forum
    Replies: 4
    Last Post: July 23rd 2007, 03:50 PM
  5. perimeter of a triangle
    Posted in the Geometry Forum
    Replies: 5
    Last Post: June 13th 2007, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum