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Math Help - [SOLVED] Equation of an arc/pythagorus

  1. #1
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    [SOLVED] Equation of an arc/pythagorus

    on number 13 Im confused. I thought a-b would be pythagorus a^2+b^2=c^2 so sqrt (2r^2) .... It gives sqrt(2r)?
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  2. #2
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    I'm assuming you're referring to exercise 13(a)i, the straight-line distance between A and B.

    Since the triangle (inside the quarter-circle) is isosceles right, with equal-length legs "r", the hypotenuse is, as you suggest, given by:

    . . . . . \sqrt{r^2\, +\, r^2}\, =\, \sqrt{2r^2}\, =\, \sqrt{2}\,r

    I would guess that \sqrt{2r} is a typo.
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  3. #3
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    thanks, i just have one more question

    15 a/b

    for a
    it says the answer is a = r^2xθ all over 2
    I know a area of a triangle is BxH all over 2
    and a circle is A = Pi  r^2 but I dont understand how to get the answer.

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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    thanks, i just have one more question

    15 a/b

    for a
    it says the answer is a = r^2xθ all over 2
    I know a area of a triangle is BxH all over 2
    and a circle is A = Pi times r squared but I dont understand how to get the answer.

    A = \left(\frac{\theta}{2\pi}\right)\; \pi r^2 The fraction is the ratio of the angle of the sector to that of the entire circle.
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  5. #5
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    the ratio of the angle theta with respect to one full revolution is the same as the sector area with respect to the area of a circle. Put mathematically:

    theta/2pi = SectorArea/ pi*r^2

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  6. #6
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    ok I pretty much get it now ie 90%
    Ill look at it latter with fresher eyes to make sure.

    Thank you.
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