# Need help with a math problem?!?!

**URL:** <https://boards.straightdope.com/t/need-help-with-a-math-problem/120983>\
**Category:** Factual Questions\
**Created:** [July 26, 2002, 7:34pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983 "2002-07-26T19:34:25Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![VanLandry](https://avatars.discourse-cdn.com/v4/letter/v/eb9ed0/32.png) [@VanLandry](https://boards.straightdope.com/u/VanLandry)\
**Post date:** [July 26, 2002, 7:34pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983/1 "2002-07-26T19:34:25Z")

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I’m thinking about buying a 40 inch widescreen projection tv with a 16:9 aspect ratio.

The question is: When I’m watching standard 4:3 programming, I will have the black bars on the right and left side of the screen(as opposed to the top and bottom when veiwing letterbox format dvd’s on a standard screen), what will be the size of the picture compared to a standard television set. 27"? 32"? Is there a simple formula I could use?

Please help as I figure I’ll be watching standard 4:3 programming for sometime still before the new standard is, well, standardized.

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**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\
**Post date:** [July 26, 2002, 8:00pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983/2 "2002-07-26T20:00:07Z")

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Try this site: [http://www.cavecreations.com/tv2.cgi](http://www.cavecreations.com/tv2.cgi)

Type in 40 for the widescreen dimension and anything you want for the 4:3 dimension. On the results page, you will see that a 40" widescreen TV is the equivalent of a 32.7" 4:3 set.

Hope this helps.

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**Author:** ![KneadToKnow](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kneadtoknow/32/3999_2.png) [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)\
**Post date:** [July 26, 2002, 8:02pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983/3 "2002-07-26T20:02:02Z")

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P. S.: I’m sure you could figure it out with paper and pencil using the Pythagorean Theorem, but who has that kind of time? 😉

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**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\
**Post date:** [July 26, 2002, 8:08pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983/4 "2002-07-26T20:08:40Z")

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Well, your TV is 40", 16:9 ratio.

16^2 + 9^2 = 337

sqrt(337) = 18.357…

Scale that up to 40, and we get that your TV is 34.86" x 19.61"

So the vertical part of the screen (and image) is 19.61", the horizontal part of the image will be 4/3 of that, or 26.15", so the image is 26.15" x 19.61".

26.15^2 + 19.61^2 = 1068.37, whose square root is **32.69"** , which is the (diagonal) size of the image.

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**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\
**Post date:** [July 26, 2002, 8:14pm UTC](https://boards.straightdope.com/t/need-help-with-a-math-problem/120983/5 "2002-07-26T20:14:08Z")

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> [@](#):
>
> P. S.: I’m sure you could figure it out with paper and pencil using the Pythagorean Theorem, but who has that kind of time?

I reckon I answered your question, too. 😃