Page 1 of 3 123 LastLast
Results 1 to 15 of 39

Math Help - A problem on triangles (area, ratios)

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    40

    Question A problem on triangles (area, ratios)

    Let Triangle ABC be a right triangle such that <A = 90*, AB = AC and let M be the midpoint 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.

    Find the ratio between the 2 triangles ABH and AHM
    Find the ratio of BP : PC

    So far the -BH^2 + HM^2 = 3 is the only equation I was able to formulate :/ Help please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    May 2011
    Posts
    40

    Answer for part 1?

    Can i just say that

    "Since the internal angles are equal among the 2 triangles, then the triangles are deemed similar. If so...

    Area of BHA = BA = 2AM = 2
    Area of AHM AM AM

    Since 2AM = BA
    "
    Is this a justified answer?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    917
    Thanks
    27

    problem of triangles

    Hi gundanium.
    Assume a 6x6 rt triangle.Compare the area ofAHM to ABM and then ABH to ABM. From these two compare the areas of AHM to ABH.


    bjh
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2011
    Posts
    40
    What do you mean compare? Cause I'm having trouble finding the relations of the other points/segments to point H. Other than that H forms a right angle for the 2 triangles BHA and AHM
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    917
    Thanks
    27
    Hi gundanium,
    Triangle ABH is similar to AHM . AM = 1/2 ofAB. Area of AHM =1/4 of ABH. Numbers would certainly help for the second part of problem which is not easy.If you try it and repost I will help
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2011
    Posts
    40
    First off. How are you able to determine that the triangles are similar? I can only see one similarity which is that they are right triangles from point H. But it ends there.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Actually, you have all three angles in the triangles being the same.

    Let angle BAH be x. Angle ABH then is 90 - x.

    The angle HAM is 90-x and the angle AMH is x.

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2011
    Posts
    40
    Thank you! I did not see that!

    I FINALLY SOLVED THE FIRST PART. YES! It involved a lot of finding the relationships between the sides of different triangles to one another. My approach was to find the relationship of Triangle BAM with AHM to get AH. Then from there related Triangle BAH to AHM then. Put them into A=bxh/2 and came out 4:1. Is that the easiest/most practical way to solve it? Seems long.

    Moving on to part II. I can't find any parameter/relation of point P to anything...
    Last edited by mr fantastic; May 6th 2011 at 04:34 AM. Reason: Merged posts.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    917
    Thanks
    27
    Hi gundanium,


    Part2 first stage,
    Let AB =6 then BM =3rad5
    Let BH=x and MH = 3rad5 -x

    x^2 +AH^2 =36
    (3rad5-x)^2 +AH^2 =9
    Solve forx and 3rad5-x and then for AH

    If you get this then stage 3 will be next



    bjh
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    May 2011
    Posts
    40
    I'm sorry but... and this might be a very stupid question but... what is 3rad5? how did you get that? I'm not so much familiar when using radians.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Well, as soon as you prove that triangles ABH and AHM are similar, you say that side AM corresponds to side AH and AM:AH = 1:2

    The area is then simply the square, which is 1:4 for triangle ABH:AHM.

    2. Oh well, I used some trigonometry for that (sine rule), and it gets quite long, but I get 1:2.25 = 4 : 9 (for PC : AP)

    EDIT: To answer this, 3rad5 is 3\sqrt5
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    May 2011
    Posts
    40
    Quote Originally Posted by Unknown008 View Post
    Well, as soon as you prove that triangles ABH and AHM are similar, you say that side AM corresponds to side AH and AM:AH = 1:2

    The area is then simply the square, which is 1:4 for triangle ABH:AHM.

    2. Oh well, I used some trigonometry for that (sine rule), and it gets quite long, but I get 1:2.25 = 4 : 9 (for PC : AP)

    EDIT: To answer this, 3rad5 is 3\sqrt5

    AM:AH = 1:2? Really? Cause I came up with Sqrt(5)/2
    And "the area is simply the square"? How did squaring a side ratio equivalent to area ratio?

    OH 3rad5 HAHA NOOB Thanks man
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    May 2011
    Posts
    40
    Quote Originally Posted by bjhopper View Post
    Hi gundanium,


    Part2 first stage,
    Let AB =6 then BM =3rad5
    Let BH=x and MH = 3rad5 -x

    x^2 +AH^2 =36
    (3rad5-x)^2 +AH^2 =9
    Solve forx and 3rad5-x and then for AH

    If you get this then stage 3 will be next



    bjh

    Yes, i have MH and BH already from my solution when solving the first part. But I don't see how point P will come in.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    Oops, sorry, I actually meant AM : AB = 1 : 2, my bad

    Also, the 'conversion' from length to area is the square of the ratio.

    That is, the ratio of length is 1:2, the ratio of area will then be 1^2:2^2 = 1:4
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    May 2011
    Posts
    40
    REALLY? WOW. I HAD NO IDEA. Is there a derivation for this? I had no idea that to be true. Nor did that exist
    Follow Math Help Forum on Facebook and Google+

Page 1 of 3 123 LastLast

Similar Math Help Forum Discussions

  1. Finding similar triangles using ratios
    Posted in the Geometry Forum
    Replies: 19
    Last Post: November 14th 2011, 11:17 AM
  2. Replies: 5
    Last Post: May 29th 2011, 04:10 PM
  3. Area of Triangles Problem
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 14th 2010, 06:09 AM
  4. Ratios of areas of triangles..
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 15th 2007, 09:04 PM
  5. Trig Ratios Right Triangles
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 28th 2007, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum