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Math Help - 6 geometry questions

  1. #1
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    Post 6 geometry questions

    I have what I think are the correct answers to all of these but the last two. Can someone help a poor mom out please?
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  2. #2
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    Quote Originally Posted by sanee66 View Post
    I have what I think are the correct answers to all of these but the last two. Can someone help a poor mom out please?
    Hello,

    I've attached a sketch which will show you how to do the construction problem.

    To the last problem:

    Calculate the centres of the circles by completing the square:

    x^2+y^2 - 8y = 12 ~ \Longrightarrow ~ x^2 + y^2 - 8y +16 = 12 + 16 ~\Longrightarrow~x^2+(y-4)^2=28

    Thus you get the coordinates of the first centre as: C_1(0, 4)

    x^2-4x+y^2=16~\Longrightarrow~x^2-4x+4+y^2=16+4~\Longrightarrow~(x-2)^2+y^2=20

    Thus you get the coordinates of the second centre as: C_2(2, 0)

    Now use Pythagorean theorem to calculate the distance:

    d = \sqrt{4^2+2^2}=\sqrt{20} = 2\cdot \sqrt{5} \approx 4.47
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  3. #3
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    Hello, Sanee!

    I have what I think are the correct answers to all of these but the last two.
    Your four answers are correct.


    In the fifth problem: .OQ is a radius (12) perpendicular to the tangent QR.

    . . Since OS \parallel QR, then: . OS \perp\ OQ . . . and OS = 12.


    From S draw perpendicular ST to QR.

    . . Then quadrilateral OQTS is a square: . QT= 12


    \Delta RTS is an isosceles right triangle: . ST \:=\:TR = 12

    . . Therefore: . QR \:=\:QT + TR \:=\:12 + 12 \:=\:24

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  4. #4
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    area of a parallegram in a rectangle

    Please take a look and see if I have the correct answer. Thanks for all you have done.
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  5. #5
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    Hello, sanee!

    Sorry, your answer is incorrect . . .


    For the area of a parallelogram, you need its base (GD = 11)
    . . and its height (DB)

    \Delta DBF is an isosceles right triangle, so DB = FB.

    Let x = DB = FB
    . . Then we have: . x^2 + x^2 \:=\:(8\sqrt{2})^2\quad\Rightarrow\quad 2x^2 \:=\:128\quad\Rightarrow\quad x \:=\:8

    Now you can find the area . . .

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  6. #6
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    Thank you Soroban for looking at this!
    So the answer would then be 88? Base of 11 times the height of 8?
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  7. #7
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    Hello, Sanee!

    Yes . . . you've got it!

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