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Math Help - Triangle ratio problem

  1. #1
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    Triangle ratio problem

    Let \bigtriangleup ABC be a right angled triangle such that \angle A =90 , AB =AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.

    (a) Find the ratio of the areas of the two triangles \bigtriangleup ABH : \bigtriangleup AHM.

    (b) Find the ratio BP : PC.


    Now this is my triangle, to place P, I took "AP is vertical to BM" as "AP and BM are perpendicular".


    Now the answer to (a) is :

    AB : AM = 2:1

    \bigtriangleup ABH : \bigtriangleup AHM

    = 2^2 : 1^2 = 4 : 1

    It is true that AB is twice large than AM, but why are they rising them to the 2th power?


    To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:

    CB = 7
    CP = 2.333333333
    PB = 4.666666666

    Though method used to calculate the ratio is very strange, I don't understand it:

    M is in the middle of AC, \bigtriangleup AHM = \bigtriangleup HCM --- (1)

    on the other hand, \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC} --- (2)

    From (1) and (2) \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup HPC} --- (3)

    therefore from (3) and question (a)

    \frac{\bigtriangleup HPC}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup ABH} = \frac{(2)(1)}{4} = \frac{1}{2} --- (4)

     \frac{PC}{BP} = \frac{\bigtriangleup HPC}{\bigtriangleup HBP} because of (4)

    The answer is BP : CP = 2 : 1

    Can somebody explain me how to use this to get the ratio?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Patrick_John View Post
    ...
    Now the answer to (a) is :

    AB : AM = 2:1

    \bigtriangleup ABH : \bigtriangleup AHM

    = 2^2 : 1^2 = 4 : 1

    It is true that AB is twice large than AM, but why are they rising them to the 2th power?
    ...

    Hello,

    to calculate the area of a triangle you have to use two values: the length of the base and the length of the height.

    In your case not only the base was lengthened by the factor 2 but the height also. If
    A_1 = \frac{1}{2} \cdot b_1 \cdot h_1 is the area of the first triangle and you know that
    b_2 = 2 \cdot b_1 \text{  and  } h_2 = 2 \cdot h_1

    then you get:

    A_2 = \frac{1}{2} \cdot b_2 \cdot h_2 = \frac{1}{2} \cdot 2 \cdot b_1 \cdot 2 \cdot h_1 = 2 \cdot 2 \cdot \frac{1}{2} \cdot b_1 \cdot h_1 = 2^2 \cdot A_1
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  3. #3
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    Quote Originally Posted by Patrick_John View Post
    ...
    To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:

    CB = 7
    CP = 2.333333333
    PB = 4.666666666
    Hello,

    if |AB| = |AC| = 5 ~cm then |BC| = \sqrt{5^2+5^2} = 7 \cdot \sqrt{2}

    Though method used to calculate the ratio is very strange, I don't understand it:

    M is in the middle of AC, \bigtriangleup AHM = \bigtriangleup HCM --- (1)
    Here you are refering to the value of the area of the triangle: In both triangles the bases and the heights have the same length therefore they must have the same area.


    on the other hand, \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC} --- (2)
    The same method is used here too.

    Maybe you can complete the problem from here on ...
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  4. #4
    MHF Contributor red_dog's Avatar
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    I will note S_{ABC} the area of the triangle ABC.
    Now:
    For (1): \triangle AHM and \triangle HCM have the same height h_1 from H.
    \displaystyle S_{AHM}=\frac{AM\cdot h_1}{2}=\frac{MC\cdot h_1}{2}=S_{HCM}.

    For (2): \triangle ABH and \triangle HBP have the same height h_2 from B.
    \displaystyle S_{ABH}=\frac{AH\cdot h_2}{2},S_{HBP}=\frac{PH\cdot h_2}{2}\Rightarrow\frac{S_{ABH}}{S_{HBP}}=\frac{AH  }{PH} (i)
    \triangle AHC and \triangle HPC have the same height h_3 from C.
    \displaystyle S_{AHC}=\frac{AH\cdot h_3}{2},S_{HPC}=\frac{PH\cdot h_3}{2}\Rightarrow\frac{S_{AHC}}{S_{HPC}}=\frac{AH  }{PH} (ii)
    From (i) and (ii) yields \displaystyle \frac{S_{ABH}}{S_{HBP}}=\frac{S_{AHC}}{S_{HPC}} (iii)

    For (3) \displaystyle S_{AHC}=\frac{AC\cdot h_1}{2}=\frac{2AM\cdot h_1}{2}=2S_{AHM}.
    Replacing S_{AHC} in (iii) \Rightarrow (3).

    For (4): \displaystyle S_{HPC}=\frac{PC\cdot h_4}{2},S_{HBP}=\frac{BP\cdot h_4}{2}\Rightarrow
    \displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH  M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}
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  5. #5
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    This is still giving some problems,

    Why in \displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH  M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}

    S_{ABH} is equal to 4S_{AHM}?
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  6. #6
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Patrick_John View Post
    This is still giving some problems,

    Why in \displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH  M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}

    S_{ABH} is equal to 4S_{AHM}?
    I'm going to say the same thing as earboth.

    What's the area of a triangle? It's \frac{1}{2}bh=A

    So what would happen if we doubled the base ( 2b) and the height ( 2h)

    Well let's substitute: \frac{1}{2} (2b) (2h) \quad = \quad \frac{1}{2}(4bh) \quad = \quad 4\left(\frac{1}{2}bh\right)

    and we all know that: 4\left(\frac{1}{2}bh\right)=4A

    So when the legs of a triangle double, the area quadruples. That is the reason why
    S_{ABH} is equal to 4S_{AHM}
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  7. #7
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    OK! Now I understand, thnks a lot everyone!
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