Results 1 to 3 of 3

Math Help - circle

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    circle

    In the figure , AB and CD are chords of a circle with centre O . They intersect at right angle at E . THe perpendiculars from teh centre O meet AB at H and CD at K . Given that EA=27 cm , EB=13 cm , and ED=9cm , prove that CD=30 cm . Hence calculate OH and the radius of the circle .
    Attached Thumbnails Attached Thumbnails circle-bbb.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    In the figure , AB and CD are chords of a circle with centre O . They intersect at right angle at E . THe perpendiculars from teh centre O meet AB at H and CD at K . Given that EA=27 cm , EB=13 cm , and ED=9cm , prove that CD=30 cm . Hence calculate OH and the radius of the circle .
    There's a theorem - see for instance here - that tells you that in the diagram you have here:
    EB \times EA = ED \times EC
    (It derives from the fact that the triangles ECB and EAD are similar. Can you see why? Hint: use angles in the same segment.)

    So, if we plug in the numbers given:
    13\times27=9\times EC

    \Rightarrow EC = 3\times 13 = 39

    \Rightarrow CD = 30 cm
    Now (although it doesn't look like it in your diagram) K is the mid-point of CD (and H the mid-point of AB). So CK = \tfrac12 CD = 15 cm. Also OHED is a rectangle. So OH = KE = 39-15=24 cm.

    AH = \tfrac12 AB = 7

    \Rightarrow OA^2 = 7^2 + 24^2 = 25^2

    \Rightarrow radius of circle = 25 cm.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    381
    Quote Originally Posted by Grandad View Post
    Hello thereddevilsThere's a theorem - see for instance here - that tells you that in the diagram you have here:
    EB \times EA = ED \times EC
    (It derives from the fact that the triangles ECB and EAD are similar. Can you see why? Hint: use angles in the same segment.)

    So, if we plug in the numbers given:
    13\times27=9\times EC

    \Rightarrow EC = 3\times 13 = 39

    \Rightarrow CD = 30 cm
    Now (although it doesn't look like it in your diagram) K is the mid-point of CD (and H the mid-point of AB). So CK = \tfrac12 CD = 15 cm. Also OHED is a rectangle. So OH = KE = 39-15=24 cm.

    AH = \tfrac12 AB = 7

    \Rightarrow OA^2 = 7^2 + 24^2 = 25^2

    \Rightarrow radius of circle = 25 cm.

    Grandad

    Awesome.. Thanks !!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle, tangent line, and a point not on the circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 31st 2011, 01:40 PM
  2. Replies: 6
    Last Post: July 8th 2010, 05:39 PM
  3. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  4. Replies: 2
    Last Post: February 6th 2010, 08:31 AM
  5. Replies: 0
    Last Post: October 12th 2008, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum