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I've seen a TV comercial that says

Our 60 inch TV have 20% more "visible area" than a 55 inch TV.

I'm curios to know if it's real or not? I need to know formula for relation between rectangle diameter and area then calculate difference between 55 and 60 inch TV.

A wide screen TV have an aspect ratio of 16:9(width:height).

Answer can be more general to say how much visible are we get for any given one inch on a wide screen TV.

This is a super simple question. I hope it is not out of this site context.

Thanks.

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1 Answer 1

up vote 4 down vote accepted

Let us suppose that the two TV screens are similar (have the same aspect ratio). If you scale the linear dimensions of an object by the factor $t$, then area is scaled by the factor $t^2$. For example, doubling linear dimensions multiplies area by $4$. In our example, $$t=\frac{60}{55},\qquad\text{and therefore}\qquad t^2=\left(\dfrac{60}{55}\right)^2.$$

It turns out (calculator) that $t^2\approx 1.19008$. So the area has increased by almost exactly $19$ percent. The $20$ percent claim is therefore not much of a stretch.

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Thanks. Is it same for 16:9 aspect ratio? –  Mohsen Nov 22 '11 at 0:33
    
Yes, this works for all postive ratios. –  The Chaz 2.0 Nov 22 '11 at 0:36
    
Can you say we get 19/5 percent for any one more inch on TV? I mean is this relation work for like 32 inch to 37 inch too? I updated my question –  Mohsen Nov 22 '11 at 0:39
2  
For $32$ to $37$, the ratio of areas is $(37/32)^2$, which is about $1.337$. So you gain almost $34$ percent in area. If you go from $100$ inch to $105$ inch, the gain in area is only about $10$ percent. Adding $5$ inches to a smallish TV has a more dramatic relative effect than adding $5$ inches to a big TV. –  André Nicolas Nov 22 '11 at 0:46

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