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Math Help - Perimeter of triangle

  1. #1
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    Perimeter of triangle

    Perimeter of triangle-capture.jpg
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  2. #2
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    Re: Perimeter of triangle

    I don't think you've been given enough information...
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    Re: Perimeter of triangle

    How ?,i have the answer but i want to know the way to solve it .
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    Re: Perimeter of triangle

    Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.
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    Re: Perimeter of triangle

    Get AM by Pythagoras (=8)
    AngleACM=angleB (both=90-A)
    sinB=6/CB But from triangle ACM sinACM=8/10 So sinB=8/10 So 8/10=6/CB So CB=15/2
    In triangle BCM cosB=BM/CB So 6/10=BM/15/2 BM=9/2
    Perimeter=AC+CB+BM+MA=10+7.5+4.5+8=30
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    Re: Perimeter of triangle

    Really helpful ,thanks
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    Re: Perimeter of triangle

    Quote Originally Posted by Mhmh96 View Post
    Click image for larger version. 

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    |AM|=\sqrt{10^2-6^2}=8

    Since \Delta AMC \sim \Delta BCM it follows :

    |BM| : 6 = 6 : 8

    |BM|=\frac{9}{2} \Rightarrow |AB|=\frac{25}{2}

    Hence :

    |BC|=\sqrt{\left(\frac{25}{2}\right)^2-10^2}=\frac{15}{2}

    therefore perimeter is : 30
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    Re: Perimeter of triangle

    Thank you !
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    Re: Perimeter of triangle

    Can you explain this part of your solution ?
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    Re: Perimeter of triangle

    Quote Originally Posted by Mhmh96 View Post
    Click image for larger version. 

Name:	Capture.JPG 
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ID:	23521
    Here is a slightly different approach:

    1. Determine |\overline{AM}| = 8 by Pythagorean theorem.

    2. Use Pythagorean theorem in triangle ABC:

    (8+\overline{BM})^2-10^2 = (\overline{BC})^2 and in the small right triangle:
    6^2+ (\overline{BM})^2 = (\overline{BC})^2

    3. Subtract both equations columnwise and solve for \overline{BM}
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  11. #11
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    Re: Perimeter of triangle

    Quote Originally Posted by Mhmh96 View Post
    Can you explain this part of your solution ?
    This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.
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    Re: Perimeter of triangle

    Quote Originally Posted by emakarov View Post
    This is the right triangle altitude theorem. Its proof involves the fact that triangles AMC and CMB are similar.
    Of course you are right but you also can use 2 right triangles to prove the altitude theorem:

    (Using the labels of the attached diagram):

    CM^2+BM^2=BC^2

    CM^2+AM^2=AC^2

    Add both equations columnwise:

    2CM^2 + AM^2 + BM^2 = BC^2+AC^2 = AB^2

    Since AB = AM + BM the right side of the last equation becomes:

    2CM^2 + AM^2 + BM^2 = (AM + BM)^2 = AM^2 + 2AM \cdot BM + BM^2

    Collect like terms, divide through by 2 and you'll get the equation of the altitude theorem.
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