anonymous
  • anonymous
Use the Pythagorean Theorem to find the length of the missing side of a right triangle whose legs equal √2 and 2√15. a right triangle with legs as 2 square root of 15 and square root of 2. A. 64 B. 4 + 2√15 C. 8 D. None of the above
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathstudent55
  • mathstudent55
The Pythagorean Theorem is \(a^2 + b^2 = c^2\) where a and b are the lengths of the legs, and c is the length of the hypotenuse.
anonymous
  • anonymous
ok....
mathstudent55
  • mathstudent55
Replace a and b with the given lengths of te legs and find c, the length of the hypotenuse.

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mathstudent55
  • mathstudent55
|dw:1436938653653:dw|
mathstudent55
  • mathstudent55
Do you see how the given leg lengths are substituted into a and b?
anonymous
  • anonymous
yes.
mathstudent55
  • mathstudent55
Good. Now you need to square the two terms on the left side. When you square a product, you square every factor of the product. |dw:1436939091163:dw|
mathstudent55
  • mathstudent55
|dw:1436939186380:dw|
nincompoop
  • nincompoop
it is good to be able to memorize the pythagorean theorem, but it will be better if you learned the concept and its geometrical underpinnings so you may be able to derive any solution to any problem that relates to it.
nincompoop
  • nincompoop
|dw:1436940135801:dw|
nincompoop
  • nincompoop
|dw:1436940404114:dw|
nincompoop
  • nincompoop
** just a slight change in variables from what mathstudent55 provided you ** The Pythagorean theorem states that the longest leg or side namely hypotenuse, r, is the square root of the squared sum of the adjacent side, x, and opposite side, y. \(\huge r = \sqrt{x^2+y^2} \) The reason I use r, x and y, is because these notations carry over into analytical geometry and physics. And if you haven't already noticed, if I squared both sides of the equation it look the same as what you were provided. \(\large r^{\color{red}{2}} = (\sqrt{x^2+y^2} )\color{red}{^2} \rightarrow r^2 =x^2 + y^2\)
nincompoop
  • nincompoop
|dw:1436941361148:dw|
nincompoop
  • nincompoop
and to just give you an idea as to the application that I am talking about http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html x's and y's never go away for some reason so we might as well accept and become familiar with them
nincompoop
  • nincompoop
http://imgur.com/a/VTMUq#10
nincompoop
  • nincompoop
http://imgur.com/a/VTMUq#10
nincompoop
  • nincompoop
why circle, pi and quadrants are important http://www.analyzemath.com/unitcircle/unit_circle_applet.html
anonymous
  • anonymous
@nincompoop you've lost me
nincompoop
  • nincompoop
do you read?
nincompoop
  • nincompoop
Math is not a passive activity like what you did with mathstudent55. It requires you to actively seek out to understand patterns from simple to complex and what i just did is drew you what pythagorean theorem in its very backbone then added a few fundamental trigonometry information (and vocabulary) as you will encounter them very soon. The x, y and r are just variables and they are arbitrary chosen so they can be replaced with a, b and c respectively. As I explained, I picked those variables because they keep reappearing in other areas of mathematical and physical studies, which you will be encountering a lot. As an example, the variable r, x and y are used in a circle's standard form. \(\Longrightarrow (x-h)^2 + (y-k)^2 = r^2 \) http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php Should you ask why would you care about a circle when you're dealing with a triangle. The idea is how one concept can carry over into another as you may have already noticed that the r in a circle radius may be used as the hypotenuse of a right triangle.
anonymous
  • anonymous
i gave up. i don't get this at all
nincompoop
  • nincompoop
what part don't you get? we can begin to what you know about math

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