Results 1 to 4 of 4

Math Help - Help me solve this exercise, please? Parallelogram Area

  1. #1
    Newbie buujiji's Avatar
    Joined
    Apr 2014
    From
    Tbilisi
    Posts
    6

    Help me solve this exercise, please? Parallelogram Area

    Hello ^^


    Please help me solve this. This might appear a little too easy for those of you who are pros but still .. It's 24th ex. and everything that was before and after this turned out quite easy for me. But for some reason I got stuck with this one :/


    Since I doubt I can translate the contest of the exercise decently, I scanned what I wrote. I hope you can understand the symbols at least? I mean, isn't maths the international language and all I did start and I did try but kjaskjgkagf


    Therefore, I hope you can help me^^~~~





    P.S. for Area (A), in my country we use symbol (S). And in case it's not obvious (because damn I'm not good at drafting) enough, it's a parallelogram
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Help me solve this exercise, please? Parallelogram Area

    Quote Originally Posted by buujiji View Post
    Hello ^^


    Please help me solve this. ...





    P.S. for Area (A), in my country we use symbol (S). And in case it's not obvious (because damn I'm not good at drafting) enough, it's a parallelogram
    Hello,

    I've modified your sketch a little bit:
    Help me solve this exercise, please? Parallelogram Area-anwinkelparallelgrm.png

    As you can see the exterior angle at the right lower vertex has the same size as \alpha

    Therefore.

    h_2 = b \cdot \sin(\alpha) ~\implies~ 4 = b \cdot \frac{13}{17}

    Determine b and consequently the value of the area.

    You should come out with A = 68
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie buujiji's Avatar
    Joined
    Apr 2014
    From
    Tbilisi
    Posts
    6

    Re: Help me solve this exercise, please? Parallelogram Area

    oh my god! thank you so much! I couldn't see that with the angle. Thank you so much for actually taking time to answer! Thanks a lot! This was a great help!
    Hope you'll help me again someday if I need it! Thank you again ^^

    P.S. I wont double-post anymore, I promise ^^
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    695

    Re: Help me solve this exercise, please? Parallelogram Area

    Hello, buujiji!

    A slightly different approach . . .



    \text{Parallelgram }PQRS\!:\;PS = a,\:PQ = b
    \text{Altitude from }Q\text{ to }RS\!:\:h_1 = QA = 13
    \text{Altitude from }Q\text{ to }PS\!:\:h_2 = QB = 4
    \alpha = \angle AQB,\;\sin\alpha = \tfrac{13}{17}

    Code:
                  : - - - - a - - - - :
                Q *-------------------* R
                 /. α . θ'         θ /
                / .       .         /
               /θ'.       13  .    /
            b /   .4              .
             /    .              / A
            /     .             /
           / θ    .            /
        P *-------.-----------* S
                  B
          : - - - - a - - - - :
    \text{Find the area of parallelogram }PQRS.

    Let \theta \,=\,\angle P \,=\,\angle R.
    Then \angle BQP \,=\, \angle AQR \,=\, 90^o-\theta \,=\,\theta'


    \angle PQR is the supplement of \angle R.

    . . \begin{array}{c}\theta' + \alpha + \theta' \;=\;180^o - \theta \\ \\ (90^o-\theta) + \alpha + (90^o - \theta) \;=\;180^o - \theta \\ \\ \alpha \;=\;\theta \end{array}

    In right triangle QAR\!:\:\sin\theta \,=\,\frac{13}{a} \quad\Rightarrow\quad a \,=\,\frac{13}{\sin\theta}
    . . a \:=\:\frac{13}{\sin\alpha} \:=\:\frac{13}{\frac{13}{17}} \:=\:17

    Therefore: . \text{Area} \:=\:a\cdot h_2 \:=\:17\cdot 4 \:=\:68
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of Parallelogram
    Posted in the Geometry Forum
    Replies: 4
    Last Post: April 16th 2009, 01:58 PM
  2. Area of a Parallelogram
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 6th 2008, 05:32 PM
  3. area of parallelogram
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: September 6th 2008, 01:50 PM
  4. area of parallelogram
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 21st 2008, 03:00 PM
  5. Area of Parallelogram
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 21st 2008, 12:32 PM

Search Tags


/mathhelpforum @mathhelpforum