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Fifth Graders Workbook

Posted by trickofmind on May 3, 2010 – 3:29 am

You are given a Rectangle with a length of 20 inches.
There is a circle dead center on the rectangle, and its
diameter happens to be sqrt(50) inches.
The circle is noted to have two drawn diameters,
connecting the points at which the edge of the
rectangle and the circle intersect that are opposite of
each other. The diameters are at right angles to each other.
What is the area of the rectangle that is not covered by the circle?

– Jordan McMichael


This post is under “SharedPuzzle, Tom” and has 6 respond so far.
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6 Responds so far- Add one»

  1. 1. Ragknot Said:

    102.1514918 square inches.

    The rectangle is 141.4213562
    The circle is 39.26990817

  2. 2. Jappa Said:

    hmmmmm
    i dont see where youre coming from ragnot
    I got the rectangles area to be 100(given sqrt(50)=5*sqrt(2), divided by two and multiplied by sqrt 2 to get the hypotenuse of that triangle, equaling 5, 5*20=100), the circles area to be 12.5*pi
    and the remaining area of the rectangle not covered by the circle(keep in mind some of the circle does not cover the rectangle) to be half the circle and two isoceles right triangles formed with the right angle at the center of the circle and the hypotenuse is on the edge of the rectangle
    so 100 – (12.5*pi)/2 – ((2.5*sqrt(2))^2)*2 =
    100 – 19.634954 – 25 = 55.3650459

  3. 3. Chris Said:

    The radius of the disc is sqrt(50)/2 = r ≈ 3.5355 and r² = 12.5. The
    width of the rectangle is sqrt(r² + r²) = 5. The total area of the
    rectangle is 5*20 = 100. The area of the disc overlapping the
    rectangle is πr²/2 – 2(r²/2) = (π/2 – 1)r² ≈ 7.13495 ; the first part
    is due to the to pie slices and the second part is due to the two
    triangles. So the uncovered area is is approx 92.865 in².

  4. 4. Chris Said:

    Aaargh, I subtracted when I should have added:
    The radius of the disc is sqrt(50)/2 = r ≈ 3.5355 and r² = 12.5. The
    width of the rectangle is sqrt(r² + r²) = 5. The total area of the
    rectangle is 5*20 = 100. The area of the disc overlapping the
    rectangle is πr²/2 + 2(r²/2) = (π/2 + 1)r² ≈ 32.13495 ; the first part
    is due to the to pie slices and the second part is due to the two
    triangles. So the uncovered area is is approx 67.865 in².

  5. 5. Chris Z Said:

    Answer is 100.
    The two drawn diameters are at right angles.
    A triangle can be drawn with two of the radii at right angles and the hypotenuse equal to the width of the rectangle. The radius = 1/2 sqrt(50).
    a^2 + b^2 = c^2
    then (sqrt50/2)^2 + (sqrt50/2)^2 = c^2
    then 50/4 + 50/4 = c^2
    then 25 = c^2
    so c =5 = width of rectangle
    length is 20
    so area is 100.

  6. 6. Chris Said:

    Hi Chris Z. You have found the area of the entire rectangle. The question asks for the uncovered area.

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