measurents of TV

**ttG** · May 7th 2008, 11:12 AM

A STANDARD picture format TV has a screen where the width is 4/3 of the height.
A WIDESCREEN TV has a screen where the width is 16/9 of the height.
Consider 2 TVs that are the same size, which means that the diagonals(28 inches= 2.54cm) of the screens are the same length, but where one of them is of the standard pic format and the other one- widescreen.

Determine screen wil have the largest area.

**Danshader** · May 7th 2008, 11:52 AM

considering th standard one first:

given that the width is 4/3 of the height

height, h = h
width, w = $\frac{4}{3}h$

also given that that size of the TV is 28 inches(71.12 cm), this is actually the hypotenuse of the TV

$ 28^2 = h^2 + w^2 $

$ 28^2 = h^2 + {(\frac{4}{3}h)}^2 $

$ 784 = h^2 + \frac{16}{9}h^2 $

$ 784 = \frac{25}{9}h^2 $

$ h^2 = 282.24 inches $

$ h = 16.8 inches $

Area of the screen
= area of a rectangle with a diagonal of 28 inches
$ =width \times height $
$ =\frac{4}{3}h\cdot h $
$ =\frac{4}{3} h^2 $
$ = \frac{4}{3} (282.24) $
$ = 376.32 inches^2 $

do the same thing for the widescreen TV and i get 335.003 inches^2. Hence concluding that the standard picture TV has a larger area.

**ttG** · May 7th 2008, 08:41 PM

Thank u!

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