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Tom is buying a 42 inch plasma high definition 1080p television, which is on sale for...

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loishy | Student, Grade 10 | (Level 1) Salutatorian

Posted May 16, 2013 at 4:42 PM via web

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Tom is buying a 42 inch plasma high definition 1080p television, which is on sale for 40% off. If the aspect ratio for this widescreen is 16:9 and the frame around the screen is 1 inch wide, will it fit in the entertainment center he bought last week for $753 if its opening is 36 inches by 24 inches? Show work that supports your conclusion and remember that televisions are measured diagonally.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted May 25, 2013 at 1:48 AM (Answer #1)

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Television aspect ratio refers to the screen's width compared to its height.  Traditional television aspect ratio is 4x3. For example, a 32-inch traditional television screens would be 25 1/2 inches wide and 19 inches tall. Widescreen aspect ratio is 16x9. A 32-inch widescreen television screen would be 28 inches wide and 19 inches tall.

The screen specifications are given by their diagonal length. The following formulae can be used to find the height (h) and width (w), where r stands for aspect ratio and d for diagonal length.

`h = d/sqrt(r^2+1)` and `w = d/sqrt(1/r^2+1)`

For the widescreen set, Tom is thinking of buying at 40%off,  the screen dimensions are:

`w = 42/sqrt(1/r^2+1)`

`= 42/sqrt((9/16)^2+1)`

`=42/1.147347`

`= 36.6`

Considering the 1 inch wide frame, the screen with frame should be (36.6+1+1) = 38.6 inches wide.

 Again, height

`h = 42/sqrt(r^2+1)`

`= 42/sqrt((16/9)^2+1)`

`=42/2.039729`

= 20.591 inches

= 20.6 inches (approximately)

Considering the 1 inch wide frame, the screen with frame should be (20.6+1+1) = 22.6 inches tall.

But the space in the entertainment center Tom had bought last week for $753 has dimensions of 36 by 24 inches (width × height).

Comparison of the dimensions reveals that the width of the existing entertainment center will fall short of the widescreen TV and hence it would not fit in there.

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