Results 1 to 7 of 7

Math Help - ratio of figures

  1. #1
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177

    ratio of figures

    i already proved the triangles.i need help with finding ratio of BXC to rectangle ABCD. it is given that CM=3MB

    THANKS
    Attached Thumbnails Attached Thumbnails ratio of figures-img039.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by helloying View Post
    i already proved the triangles.i need help with finding ratio of BXC to rectangle ABCD. it is given that CM=3MB

    THANKS
    the ratio of the areas ?? Perhaps you mean CM=3MD ?

    Pls clarify . Thanks .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177
    Quote Originally Posted by mathaddict View Post
    the ratio of the areas ?? Perhaps you mean CM=3MD ?

    Pls clarify . Thanks .
    sorry i left out the areas.i need to find the ratio of area of that triangle to the ractangle.CM=3MD is given from the question.
    Last edited by helloying; October 26th 2009 at 09:49 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2009
    Posts
    147
    Thanks
    1
    Produce a straight line where Y lies on AB, Z lies on DC, and YXZ//AD//BC.
    CM = 3DM, hence
    CM = 0.75*DC = 0.75*AB

    hence
    AB / CM
    = AB / 0.75AB
    = 4/3
    = AX / CX
    = BX / MX

    triangle AYX similiar to CZX (AAA), therefore
    AY / CZ = AX / CZ = 4/3
    Therefore CZ = BY = (3/7)*AB

    area of triangle BXC = BC*BY/2
    = BC*(3/7*AB)*1/2
    = (BC)(AB)(3/14)
    area of rectangle ABCD = (BC)(AB)
    Hence, the ratio concerned
    = 3:14
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177
    Quote Originally Posted by ukorov View Post
    Produce a straight line where Y lies on AB, Z lies on DC, and YXZ//AD//BC.
    CM = 3DM, hence
    CM = 0.75*DC = 0.75*AB

    hence
    AB / CM
    = AB / 0.75AB
    = 4/3
    = AX / CX
    = BX / MX

    triangle AYX similiar to CZX (AAA), therefore
    AY / CZ = AX / CZ = 4/3
    Therefore CZ = BY = (3/7)*AB

    area of triangle BXC = BC*BY/2
    = BC*(3/7*AB)*1/2
    = (BC)(AB)(3/14)
    area of rectangle ABCD = (BC)(AB)
    Hence, the ratio concerned
    = 3:14
    i think this AY / CZ = AX / CZ = 4/3 should be AY / CZ = AX / CX = 4/3

    And i dont understand this step
    Therefore CZ = BY = (3/7)*AB

    By the way, this qn is only 1 mark.perhaps there a shorter method?

    thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    Posts
    147
    Thanks
    1
    Quote Originally Posted by helloying View Post
    i think this AY / CZ = AX / CZ = 4/3 should be AY / CZ = AX / CX = 4/3

    And i dont understand this step
    Therefore CZ = BY = (3/7)*AB

    By the way, this qn is only 1 mark.perhaps there a shorter method?

    thanks
    ax /cx <--- oh yeah I typed it wrong

    regarding to BY = (3/7)*AB :
    first have a look at AY:BY again. it is 4:3. it means that WHEN AY is 4, BY is 3.....when AY is 20, BY is 15, etc.
    it tells us that WHEN the whole length of AB is EQUALLY divided into 7 PARTS, then AY constitutes 4 parts, and BY consitutes 3.
    therefore AY:BY = 4:3, AY:AB = 4:7, BY:AB = 3:7
    These numbers do NOT have units (cm, mm, etc). it is all about ratios.
    if you want to prove with real lengths, consider AY = 20cm, then BY = 15cm, and hence AB = 35cm. you will get the same ratios as aforesaid.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    147
    Thanks
    1
    to shorten the calculation, it is not really necessary to have point Z and line XZ but you need Y and line BY which is equivalent to the height of triangle required to calculate its area. simply use the ratio of AX:CX to work out the ratios of AY:BY and AB:BY, and eventually present BY in terms of AB.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ratio of areas of similar figures
    Posted in the Geometry Forum
    Replies: 7
    Last Post: January 20th 2011, 06:43 AM
  2. Ratio for curve and point-ratio form
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 11th 2010, 01:07 AM
  3. significant figures
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 16th 2009, 11:20 AM
  4. 3 dimensional figures
    Posted in the Geometry Forum
    Replies: 0
    Last Post: October 12th 2009, 07:45 PM
  5. Significant Figures
    Posted in the Geometry Forum
    Replies: 1
    Last Post: September 29th 2007, 10:34 AM

Search Tags


/mathhelpforum @mathhelpforum