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Math Help - Sat

  1. #1
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    Sat

    hi,

    my name is sara ...

    I,m 15

    and i has a sat test today

    please

    help me in this question

    >>>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<


    in the figure at the right.
    the legs of right tringle ACB are diameters of the two semicircles. If AB=4 ,What is the sum of the areas of the semicircles?



    (a) 3.14 (b) 2 x 3.14 (c) 4 x 3.14 (d) 8 x 3.14 (c) 16 x 3.14


    please help me.....
    Last edited by MathGuru; May 5th 2006 at 02:14 PM.
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  2. #2
    Site Founder MathGuru's Avatar
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    Assume it is an isosceles triangle and therefore D^2 + D^2 = 4^2

    solve for D and you will find one leg of the triangle or the diameter of the semicircle.

    The area of the semi circle will be

    1/2*(Pi*(D/2)^2) and since you have two of them you will find the sum of the area is:

    Pi*(D/2)^2
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  3. #3
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    so, What is the answer??


    (( sorrry i am not good in english because english is not my mother language))
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  4. #4
    Site Founder MathGuru's Avatar
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    D^2 + D^2 = 4^2

    2(D^2) = 16

    D^2 = 8

    D = Sqrt(8)

    can you do the rest?
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  5. #5
    Site Founder MathGuru's Avatar
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    Here is the rest:

    Given:
    Area= Pi*(D/2)^2 and D=sqrt(8)



    Area= Pi*(sqrt(8)/2)^2

    Area= Pi*(8/4)

    Area= 2*Pi

    so B is the answer
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  6. #6
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    Thank you

    i hope i get a good mark in SAt text
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  7. #7
    Site Founder MathGuru's Avatar
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    Your welcome,

    always try to understand what you are doing rather than just focus on the answer. Please tell your friends in Qatar about this website, I hope you found it useful.

    Cheers,
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  8. #8
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    how do you know that i am from Qatar?
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by MathGuru
    Assume it is an isosceles triangle and therefore D^2 + D^2 = 4^2

    solve for D and you will find one leg of the triangle or the diameter of the semicircle.

    The area of the semi circle will be

    1/2*(Pi*(D/2)^2) and since you have two of them you will find the sum of the area is:

    Pi*(D/2)^2
    No assumptions are needed the areas of the semi-circles obey a semi-circle
    version of Pythagoras's theorem.

    The area of the semi-circle on the hypotenuse of a right triangle is equal
    to the sum of the areas of the semi-circles on the other two sides.

    Hence the sum of the areas of the two semi-circles shown is equal to
    the area of a semi-circle on the hypotenuse, which is:

    \pi.2^2/2=2.\pi

    RonL
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  10. #10
    Site Founder MathGuru's Avatar
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    Thanks Ron for the correction.

    Sara I checked your IP address which show that you are in Qatar.
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