perl
  • perl
Maggie has a kite with the dimensions shown below. What is the width of the kite?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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perl
  • perl
http://assets.openstudy.com/updates/attachments/4f33f61de4b0fc0c1a0bbc09-ihavemathquestions-1328805410611-screenshot20120209at11.32.56am.png
perl
  • perl
im not sure what theorem is used here
perl
  • perl
whats up rooma

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perl
  • perl
rumoor
Directrix
  • Directrix
|dw:1328862799911:dw|To find x, this theorem is used. If an altitude is drawn to the hypotenuse of a right triangle, the length of the altitude is the geometric mean between the lengths of the segments of the hypotenuse.
Directrix
  • Directrix
11 is to x as x is to 25. 11/x = x/25 x^2 = 25*11 x = 5 √ 11
Directrix
  • Directrix
The width of the kite is 2 times ( 5 √ 11 = 10 √ 11
perl
  • perl
woops, didnt see that right angle there, yes makes sense now
anonymous
  • anonymous
You don't need that theorem just use pyhtagoras theorem on the three triangles. If the remainng side lenghts are y and z. x^2+25^2=y^2 and x^2+11^2=z^2. and y^2+z^2=36^2. 2x^2+25^2+11^2=36^2.
perl
  • perl
where are you getting y^2 + z^2 = 36^2
perl
  • perl
oh i see, yes you can prove it that way
perl
  • perl
so you actually proved the general result |dw:1328881279082:dw|
perl
  • perl
and after doing some algebra we have h = sqrt (ab)
perl
  • perl
so this geometric mean theorem falls out from pythagorean theorem
radar
  • radar
Or properties of similar triangles (proportions) a:h :: h:b 11:h :: h:25 \[h ^{2}=11 X 25\]\[h = \sqrt{11X25}=5\sqrt{11}\]

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