Results 1 to 7 of 7

Thread: Triangle ratio problem

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    20

    Triangle ratio problem

    Let $\displaystyle \bigtriangleup ABC$ be a right angled triangle such that $\displaystyle \angle A =90$$\displaystyle , AB =AC$ and let $\displaystyle M$ 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 $\displaystyle \bigtriangleup ABH : \bigtriangleup AHM$.

    (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 :

    $\displaystyle AB : AM = 2:1$

    $\displaystyle \bigtriangleup ABH : \bigtriangleup AHM$

    $\displaystyle = 2^2 : 1^2 = 4 : 1$

    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, $\displaystyle \bigtriangleup AHM = \bigtriangleup HCM$ --- (1)

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

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

    therefore from (3) and question (a)

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

    $\displaystyle \frac{PC}{BP} = \frac{\bigtriangleup HPC}{\bigtriangleup HBP}$ 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by Patrick_John View Post
    ...
    Now the answer to (a) is :

    $\displaystyle AB : AM = 2:1$

    $\displaystyle \bigtriangleup ABH : \bigtriangleup AHM$

    $\displaystyle = 2^2 : 1^2 = 4 : 1$

    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
    $\displaystyle A_1 = \frac{1}{2} \cdot b_1 \cdot h_1$ is the area of the first triangle and you know that
    $\displaystyle b_2 = 2 \cdot b_1 \text{ and } h_2 = 2 \cdot h_1$

    then you get:

    $\displaystyle A_2 = \frac{1}{2} \cdot b_2 \cdot h_2 = \frac{1}{2} \cdot 2 \cdot b_1 \cdot 2 \cdot h_1 = 2 \cdot 2 \cdot \frac{1}{2} \cdot b_1 \cdot h_1 = 2^2 \cdot A_1$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    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 $\displaystyle |AB| = |AC| = 5 ~cm$ then $\displaystyle |BC| = \sqrt{5^2+5^2} = 7 \cdot \sqrt{2}$

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

    M is in the middle of AC, $\displaystyle \bigtriangleup AHM = \bigtriangleup HCM$ --- (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, $\displaystyle \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)
    The same method is used here too.

    Maybe you can complete the problem from here on ...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    I will note $\displaystyle S_{ABC}$ the area of the triangle $\displaystyle ABC$.
    Now:
    For (1): $\displaystyle \triangle AHM$ and $\displaystyle \triangle HCM$ have the same height $\displaystyle h_1$ from $\displaystyle H$.
    $\displaystyle \displaystyle S_{AHM}=\frac{AM\cdot h_1}{2}=\frac{MC\cdot h_1}{2}=S_{HCM}$.

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

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

    For (4): $\displaystyle \displaystyle S_{HPC}=\frac{PC\cdot h_4}{2},S_{HBP}=\frac{BP\cdot h_4}{2}\Rightarrow$
    $\displaystyle \displaystyle \frac{PC}{BP}=\frac{S_{HPC}}{S_{HBP}}=\frac{2S_{AH M}}{S_{ABH}}=\frac{2S_{AHM}}{4S_{AHM}}=\frac{1}{2}$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2007
    Posts
    20
    This is still giving some problems,

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

    $\displaystyle S_{ABH}$ is equal to $\displaystyle 4S_{AHM}$?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Patrick_John View Post
    This is still giving some problems,

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

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

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

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

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

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

    So when the legs of a triangle double, the area quadruples. That is the reason why
    $\displaystyle S_{ABH}$ is equal to $\displaystyle 4S_{AHM}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2007
    Posts
    20
    OK! Now I understand, thnks a lot everyone!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Triangle Ratio Problem (I think)
    Posted in the Geometry Forum
    Replies: 4
    Last Post: May 4th 2011, 06:32 PM
  2. ratio of circle to triangle
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Dec 14th 2010, 06:09 AM
  3. Triangle Ratio Problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Apr 14th 2008, 07:33 PM
  4. Ratio of a Triangle
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Mar 16th 2008, 10:28 AM
  5. triangle ratio
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Dec 10th 2006, 01:52 PM

Search Tags


/mathhelpforum @mathhelpforum