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

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Math Expert
Joined: 02 Sep 2009
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The aspect ratio of an image is the ratio of its width to its height.  [#permalink]

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26 Nov 2019, 00:31
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35% (medium)

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73% (02:30) correct 27% (02:30) wrong based on 21 sessions

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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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26 Nov 2019, 04:30
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
The aspect ratio of an image is the ratio of its width to its height.   [#permalink] 26 Nov 2019, 04:30
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