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Math Help - Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

  1. #1
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    Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

    I calculated that given a rectangle with 16:9 dimensions and one with 4:3 that the 4:3 one will have 12 percent more area given the same diagonal length. If anyone's interested, please check my result.

    Here's my work:
    (16x)^2+(9x)^2=d^2
    (4y)^2+(3y)^2=d^2

    A(d)=16x*9x=144/337*d^2 and the other: A(d)=4y*3y=12/25*d^2

    thus the ratio = 12/25*337/144 ~ 1.12
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  2. #2
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    Re: Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

    Hey lamp23.

    If you are looking at setting the smaller sides equal for both sets then the closest monitor to a square will have the highest area and its got to do with the square being the optimal choice for area given a set of two sides (i.e max area with minimal perimeter has a square as a solution).

    This is in line with your working out.
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    Re: Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

    Quote Originally Posted by lamp23 View Post
    I calculated that given a rectangle with 16:9 dimensions and one with 4:3 that the 4:3 one will have 12 percent more area given the same diagonal length. If anyone's interested, please check my result.

    Here's my work:
    (16x)^2+(9x)^2=d^2
    (4y)^2+(3y)^2=d^2

    A(d)=16x*9x=144/337*d^2 and the other: A(d)=4y*3y=12/25*d^2

    thus the ratio = 12/25*337/144 ~ 1.12
    That is nothing to do with CRT technology, you can/could get a 16:9 CRT TV as well as 4:3. But that for a given advertised screen size it would be smaller that the same advertised screen size LCD, plasma, ... TV because of about a 25mm strip around the edge of the screen was covered by the bezel.

    .
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    Re: Do CRT's have 12 percent more area than widescreens (given the same diagonal)?

    Quote Originally Posted by chiro View Post
    Hey lamp23.

    If you are looking at setting the smaller sides equal for both sets then the closest monitor to a square will have the highest area and its got to do with the square being the optimal choice for area given a set of two sides (i.e max area with minimal perimeter has a square as a solution).

    This is in line with your working out.
    Thanks! I forgot about that fact from calculus.
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