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Math Help - quarter cirlces

  1. #1
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    quarter cirlces

    HI
    I would be most grateful if someone can help me in the below question

    Thanks

    The shaded figure below is formed by a semicircle and quarter circles. It lies within the rectangle PQRS in which RU = 1cm, UQ = 2cm and PT = 3cm.
    Find the area of the shaded region (Give your answer correct to 2 decimal places)
    (ans 6.57 square cm
    Attached Thumbnails Attached Thumbnails quarter cirlces-777.bmp  
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I suggest you find the area of the 3 cm radius sector, then subtract this area from the 3 by 3 square to get the unshaded area.
    Do the same for the 2 cm radius sector.

    Then, find the area of the semicircle with radius 1 cm, subtract that from the square with sides 2 by 2 to get the unshaded area.

    now, you only have to find the last unshaded area, at the top right corner. Find the area of the segment with 1 cm radius, subtract that from the 1 by 1 square.

    Add all the unshaded areas together, then subtract that from the total area of the rectangle with area 4 by 3.
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  3. #3
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    Quote Originally Posted by kingman View Post
    HI
    I would be most grateful if someone can help me in the below question

    Thanks

    The shaded figure below is formed by a semicircle and quarter circles. It lies within the rectangle PQRS in which RU = 1cm, UQ = 2cm and PT = 3cm.
    Find the area of the shaded region (Give your answer correct to 2 decimal places)
    (ans 6.57 square cm
    1. You have a rectangle with A = 3 \cdot 4 = 12

    2. "Cut off" the 4 areas A1 to A4. What's left is the area in question:

    A1 = 3^2-\frac14 \cdot \pi \cdot 3^2

    A2 = 1^2-\frac14 \cdot \pi \cdot 1^2

    A3 = 4^2-\frac12 \cdot \pi \cdot 1^2

    A4 = 2^2-\frac14 \cdot \pi \cdot 2^2

    3. Subtract these areas from A.

    EDIT: ... too late - as usual
    Attached Thumbnails Attached Thumbnails quarter cirlces-krsteile_inrechteck.png  
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Nah, that's okay, your post has a pretty drawing, so... it's kind of better than mine. If I had to draw that, I'd have taken more time
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  5. #5
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    Thanks

    Thanks very much for the solution but I really wonder whether there is any shortcut by looking at the symmetry of the diagtram. Also I wonder how you can replicate the diagram so nice!.
    thanks
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  6. #6
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    Hi earboth,

    A3 = 2^2 - 1/2*π*1^2
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  7. #7
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    Hi earboth,

    A3 = 2^2 - 1/2*π*1^2
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  8. #8
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Hi earboth,

    A3 = 2^2 - 1/2*π*1^2
    ... of course you are right - but as you can see my brain is calculating faster than I can type
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  9. #9
    MHF Contributor Unknown008's Avatar
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    Ok, since you're mentioning this, I think we can use, not symmetry, but similarity.

    I've made a drawing this time ( though not as good as yours earboth)

    The red and green parts together represent one segment, the pale blue another segment, and the green, a third segment.

    The area of the largest segment is given by:

    Area = \frac14 \pi r^2  - \frac12 (base)(height)= \frac14 \pi (3)^2 - \frac{(3\times3)}{2}= \frac94 \pi - \frac92 = 9(\frac{\pi - 2}{4})

    Through similarity, we know then that segments, from largest to smallest are in the ratio:

    9 : 4 : 1

    This means, the area of the smaller segments are \pi - 2 and \frac{\pi-2}{4} respectively.

    Now, we have all three segments. Take the largest area, add the second largest area and remove the smallest area. We only need to add the area of the smallest circle (yellow + green), which we get by:

    Area = \pi (1)^2 = \pi

    The total shaded area then is:

    Total\ Area = (\frac{\pi - 2}{4})(9+4-1) +   \pi= 4\pi - 6

    quarter cirlces-test.png
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  10. #10
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    Thanks very very much.,
    You have really provided a very neat and creative solution
    Big thanks
    Kingman
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  11. #11
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    dear Sir,
    I would be glad if you showus what software you used to draw such nice diagram.
    thanks
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  12. #12
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    Quote Originally Posted by kingman View Post
    dear Sir,
    I would be glad if you showus what software you used to draw such nice diagram.
    thanks
    Uh... is it directed to me?

    If so, I first used Open Office Drawing to draw the geometric lines, and used GIMP image editor to add colors...
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  13. #13
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    Quote Originally Posted by kingman View Post
    dear Sir,
    I would be glad if you showus what software you used to draw such nice diagram.
    thanks
    ...only in case I was meant, have a look here: euklid dynageo homepage

    I don't know if there exists an English version of the program.
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