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Math Help - Similar triangles desperate help

  1. #1
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    Similar triangles desperate help

    I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
    Thank you very much!
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  2. #2
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    Hello f1f2f3
    Quote Originally Posted by 354732
    I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
    Thank you very much!
    Sorry I haven't long to give you a complete answer, but since you said you are desperate, here's some help for the first two.

    1) y = \sqrt{xz}

    \Rightarrow y^2 = xz

    \Rightarrow \frac{y}{x}= \frac{z}{y}

    Can you complete?


    2) \triangle's ABM, DEA are similar (Can you see why?)

    AM = \sqrt{a^2 + (\tfrac12b)^2} and \frac{DE}{AD}=\frac{AB}{AM}

    Can you complete?


    Grandad
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  3. #3
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    Hello again f1f2f3
    Quote Originally Posted by f1f2f3 View Post
    I have a few similar triangles problems. I'm really not sure how to do them. Could you please help me, or at least get me started on how to solve it.
    Thank you very much!
    Number 3 is a bit more tricky, I think.

    The easiest way is to use the formula \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}, noting that \tan (\angle EBF) = \tfrac13, and \tan\angle EBG = \tfrac12. Then

    \tan(\angle EBF + \angle EBG)=\frac{\tfrac13+\tfrac12}{1-\tfrac13\cdot \tfrac12}=\frac{\tfrac56}{\tfrac56}=1

    \Rightarrow \angle EBF + \angle EBG =45^o

    But if you must use similar triangles...

    Since \angle CBH = 45^o, if we can prove that \angle EBF = \angle GBH, then we're there. For then we should have \angle EBF + \angle EBG = \angle CBH.

    So, in the diagram attached, J is the foot of the perpendicular from H to BG. We aim to prove that \triangle's BJH, BEF are similar, and hence that \angle EBF = \angle JBH.

    Let's assume that the length of the side of any of the squares is 1 unit. So BG = \sqrt5 (Pythag.)

    Now \triangle's GJH, GAB are similar (Can you supply the reason?)

    \Rightarrow \frac{HJ}{HG}=\frac{BA}{BG}

    \Rightarrow \frac{HJ}{1}=\frac{1}{\sqrt5}

    and AG = 2AB \Rightarrow JG = 2JH = \frac{2}{\sqrt5}

    \Rightarrow BJ = BG-JG = \sqrt5-\frac{2}{\sqrt5}=\frac{5-2}{\sqrt5}=\frac{3}{\sqrt5}=3 HJ

    But BE = 3EF

    Therefore, since \angle BJH = \angle BEF =90^o, \triangle's BJH, BEF are similar, and hence \angle EBF = \angle JBH.

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello again f1f2f3Number 3 is a bit more tricky, I think.

    The easiest way is to use the formula \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}, noting that \tan (\angle EBF) = \tfrac13, and \tan\angle EBG = \tfrac12. Then

    \tan(\angle EBF + \angle EBG)=\frac{\tfrac13+\tfrac12}{1-\tfrac13\cdot \tfrac12}=\frac{\tfrac56}{\tfrac56}=1

    \Rightarrow \angle EBF + \angle EBG =45^o

    But if you must use similar triangles...

    Since \angle CBH = 45^o, if we can prove that \angle EBF = \angle GBH, then we're there. For then we should have \angle EBF + \angle EBG = \angle CBH.

    So, in the diagram attached, J is the foot of the perpendicular from H to BG. We aim to prove that \triangle's BJH, BEF are similar, and hence that \angle EBF = \angle JBH.

    Let's assume that the length of the side of any of the squares is 1 unit. So BG = \sqrt5 (Pythag.)

    Now \triangle's GJH, GAB are similar (Can you supply the reason?)

    \Rightarrow \frac{HJ}{HG}=\frac{BA}{BG}

    \Rightarrow \frac{HJ}{1}=\frac{1}{\sqrt5}

    and AG = 2AB \Rightarrow JG = 2JH = \frac{2}{\sqrt5}

    \Rightarrow BJ = BG-JG = \sqrt5-\frac{2}{\sqrt5}=\frac{5-2}{\sqrt5}=\frac{3}{\sqrt5}=3 HJ

    But BE = 3EF

    Therefore, since \angle BJH = \angle BEF =90^o, \triangle's BJH, BEF are similar, and hence \angle EBF = \angle JBH.

    Grandad
    Sorry, I'm still not sure how triangles GJH and GAB are similar.
    And also for the first question, I'm still not sure how the two triangles are similar. Could you please explain a little more for me?
    Thank you so much!
    Last edited by f1f2f3; August 27th 2009 at 05:55 AM.
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  5. #5
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    Hello f1f2f3
    Quote Originally Posted by f1f2f3 View Post
    Sorry, I'm still not sure how triangles GJH and GAB are similar. Could you please explain? Thank you!
    \angle G is common.

    \angle A = \angle HJG = 90^o

    So the triangles are similar (Equal angles).

    Grandad
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