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Math Help - Really hard questions...

  1. #1
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    Really hard questions...

    1. XY and YZ are diameters of two tangent circles and XY = 10 cm. Find YZ if the area of hte shaded region equals the area of hte unshaded region.

    The answer is: 24.14, but how and why?

    2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

    Notes: These drawings are not EXACT as the real one because I used paint, however they really close.

    The answer is 4 cm and 8 cm, but how and why...?

    These questions are really hard. Nobody in my class were able to answer them.
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  2. #2
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    Suggested soln

    ok.. here we go...
    let's say the dist YZ is x cm, then the diameter of the larger circle is (10+x) cm. And the area is pi [(10+x)/2]^2. Similarly, the area of the smaller circle can be found ie pi(x/2)^2. We can then find an expression for the shaded area and equate that to the area of the smaller circle. You should get a wuadratic equation which can be reduced to
    x^2 - 20x -100 = 0.
    Solving this would give you the answer as required.

    As for the second question, I think you should work on it alonmg similar lines as per question 1.

    Hope it helps. Let me know if u need further clarifications.

    musMD, Singapore
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  3. #3
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    Quote Originally Posted by AlphaRock View Post
    [B]...

    2. Find the diameters of the two smaller circles if the shaded area is 4/9 of the area of the larger circle which has a diameter of 12 cm.

    ...
    to #2:

    I've modified your sketch a little bit.

    1. Let x denote the length of the diameter of the upper circle then (12-x) is the length of the diameter of the lower circle.

    2. The shaded region is the difference of the large circle and the two small circles:

    A=\pi \cdot \left(\frac{12}2 \right)^2 - \pi \cdot \left(\frac{x}2 \right)^2 - \pi \cdot \left(\frac{12-x}2 \right)^2 = \frac49 \cdot \pi \cdot \left(\frac{12}2 \right)^2

    3. Expand the brackets, collect like terms. You'll get a quadratic equation:

    -\frac12x^2+6x-16=0

    Solve for x.
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