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Old Apr 22nd 2016, 05:18 AM   #1
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Working out dimensions of a eye-sight chart

A sight-test room is set up so that the testing chart is placed 3.5m away from a wall-mounted mirror, and a patient is sat 2.5m away from the same plane mirror.
If the mirror is 30cm tall and 20cm wide, find the maximum dimensions of the largest sight test chart needed to ensure that the patient can view the entire chart in the mirror.

To be honest, I'm having trouble visualizing this and would appreciate any help.
Even a diagram or reference to a diagram would also be great.
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Old Apr 22nd 2016, 10:05 AM   #2
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maybe the attached sketch will help. You can use similar triangles to solve for X.
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Working out dimensions of a eye-sight chart-eye-chart.jpg  

Last edited by ChipB; Apr 22nd 2016 at 10:07 AM.
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Old Apr 22nd 2016, 10:42 AM   #3
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Thanks a lot for this sketch Chip.

I will try to solve for X using a scaling method. Is that possible?

I got an answer of 28 cm (wide) for X but not sure if that's right?
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Old Apr 22nd 2016, 11:30 AM   #4
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No, that's not right. It looks like you may have tried setting similar triangles of base length 2.5 and 3.5 meters, but consider that the total distance for line of sight from patient to chart is 2.5 + 3.5 = 6 meters.
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Old Apr 22nd 2016, 01:13 PM   #5
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I think I might have it now.

I considered both triangles: base of 20 and perpendicular height 2.5, and then compared to the full triangle of base X and perpendicular height 6 and found X to be 40 cm.

Is this correct?
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Old Apr 22nd 2016, 01:21 PM   #6
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The two similar triangles height/base are:

20 cm/2.5 m= x cm/ 6 m

Solve for x.

Last edited by ChipB; Apr 22nd 2016 at 02:11 PM.
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Old Apr 22nd 2016, 01:34 PM   #7
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X = 48 cm

Thanks so much Chip
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