## measure distance of diag. of monitor to nearest 1/10 cm

Geometric formulae, word problems, theorems and proofs, etc.
pjubera
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### measure distance of diag. of monitor to nearest 1/10 cm

Measure the distance of the diagonal (from one corner to the opposite corner) of the screen on your computer monitor to the nearest tenth of a centimeter or sixteenth of an inch. Measure the height of the screen along the verical as well. Use the Pythagorean theorem to find the width along the horizontal. I used sixteenth of an inch. The diagonal is 304/16 and the height is 192 sixteenths. My issue is that I can't at this time find what the pythagorean theorm is to get started with it. I know the formula should be a^2 + b^2 = c^2, which would be 304sixteenths^2 plus 192 sixteenths ^2 = c^2.

Thanks

stapel_eliz
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I can't at this time find what the pythagorean theorm is...
Google is your friend...

sdbielz
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Joined: Sun Jul 19, 2009 4:53 pm
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### Re: measure distance of diag. of monitor to nearest 1/10 cm

You had the pythagorean theorem right, but you plugged your values in wrong. C is always the hypotenuse (the longest side of the triangle). the diagonal mesurement should be C in the pythagorean, not A as you had it.

jordierocks94
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Joined: Thu Aug 13, 2009 10:26 pm
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### Re: measure distance of diag. of monitor to nearest 1/10 cm

If I write this problem plugging in the values (304/16) ^2 = c^2; and a^2 = ( 192/16^2) + b^2 = n^2, then do the rules
for equations apply -- meaning what value is applies to one side of an equation if that same value is applied to the other side, then the answer remains the same? If that rule is applicable here, then I can reduce both 304/16 before squaring it as per the rule Parenthesis, Exponents, Mulitiply, Divide, Add & Subtract, as well as reducing 192/16 before squaring it by the same factor of 16? And, then solve the problem? The problem would start like this:

(304/16)^2 = (192/16)^2 + x^2
Reduce values within parenthesis -- if allowed
Square
Isolate x^2
Add or Subtract Values representing a^2 and b^2
Reduce with SQRT

Thanks for help setting this up.

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