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Whisprin_Eye t1_ivx4i2l wrote

I understand arithmetic. The article never mention 14.4 inches. Just 14 inches. Just read the article.

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Orion113 t1_ivx9wt2 wrote

I did read the article. It was like one paragraph. 14.4 was a Freudian slip, as I saw it in somebody else's comment as I went to make mine.

But none of that changes the fact that a screen or other rectangle with a 14 (or 14.4) inch diagonal can be 20% larger than a rectangle with a 12 inch diagonal. Just as an example, plugging these formulas into Wolfram alpha:

12=(x^2 + y^2 )^(1/2) (formula for the sides of a rectangle with a 12 inch diagonal)

14=(w^2 + z^2 )^(1/2) (formula for the sides of a rectangle with a 14 inch diagonal)

wz=1.2xy (formula stating that the area of rectangle wz is 20% larger than the area of rectangle xy)

And picking a random reasonable value for w, let's say 7, gives us this as a result (all numbers rounded to two decimal places):

Rectangle xy has sides 7.64 and 9.24, a diagonal length of 12 inches, and an area of 70.72.

Rectangle wz has sides of 7 and 12.12, a diagonal length of 14 inches, and an area of 84.87.

Rectangle wz is exactly 20% larger than rectangle xy.

This would work for a range of values of w.

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Whisprin_Eye t1_ivxjhi1 wrote

You're spending a lot of time arguing about information that wasn't in the article.

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Orion113 t1_ivz5c9a wrote

Nowhere in the article does it say that 2 is 20% of 12, either, so what are you arguing about?

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