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The aspect ratio of an image is the ratio of its width to its height.

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The aspect ratio of an image is the ratio of its width to its height.  [#permalink]

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New post 26 Nov 2019, 01:31
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The aspect ratio of an image is the ratio of its width to its height. If the aspect ratio of an image on a particular television is 16:9, and the diagonal of the image measures x inches, which of the following is an expression for the height of the image, in terms of x?

A. \(\frac{16x}{9}\)

B. \(\sqrt{\frac{x^2}{256}+81}\)

C. \(\frac{9x}{16}\)

D. \(\frac{9x\sqrt{337}}{337}\)

E. \(25\sqrt{x}\)


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The aspect ratio of an image is the ratio of its width to its height.  [#permalink]

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New post 26 Nov 2019, 05:30
Bunuel wrote:
The aspect ratio of an image is the ratio of its width to its height. If the aspect ratio of an image on a particular television is 16:9, and the diagonal of the image measures x inches, which of the following is an expression for the height of the image, in terms of x?

A. \(\frac{16x}{9}\)

B. \(\sqrt{\frac{x^2}{256}+81}\)

C. \(\frac{9x}{16}\)

D. \(\frac{9x\sqrt{337}}{337}\)

E. \(25\sqrt{x}\)


Let the height = \(h\) & width = \(w\)
aspect ratio = \(\frac{16}{9} = \frac{w}{h}\)
--> Values of \(w = 16k, h = 9k\) for some positive value k

Given diagonal = \(x\)
--> \(x = \sqrt{w^2 + h^2}\)
--> \(x = \sqrt{(16k)^2 + (9k)^2}\)
--> \(x = \sqrt{256*k^2 + 81*k^2} = \sqrt{337*k^2}\)
--> \(x = (\sqrt{337})k\)
--> \(k = \frac{x}{\sqrt{337}}\)

So, height, \(h = 9k = \frac{9x}{\sqrt{337}}\) = \(\frac{9x\sqrt{337}}{337}\)

IMO Option D
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The aspect ratio of an image is the ratio of its width to its height.   [#permalink] 26 Nov 2019, 05:30
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