anonymous
  • anonymous
can some one help me?
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
https://media.glynlyon.com/g_geo_2012/9/q415a.gif
1 Attachment
anonymous
  • anonymous
Prove: In an equilateral triangle the three medians are equal.
jim_thompson5910
  • jim_thompson5910
what do you have so far?

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anonymous
  • anonymous
absolutely nothing ive been staring at the page for an hour
jim_thompson5910
  • jim_thompson5910
|dw:1439083430829:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1439083442888:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1439083487262:dw|
jim_thompson5910
  • jim_thompson5910
do you see how P is the midpoint of AB?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
they use the midpoint formula to find point P
jim_thompson5910
  • jim_thompson5910
that means you add up the x coordinates and divide by 2 |dw:1439083618310:dw|
jim_thompson5910
  • jim_thompson5910
make sense?
anonymous
  • anonymous
somewhat
jim_thompson5910
  • jim_thompson5910
I added the x coordinates 0 and 2a to get 2a then I divided 2a by 2 to get 'a'
jim_thompson5910
  • jim_thompson5910
|dw:1439083746502:dw|
anonymous
  • anonymous
ok got it
jim_thompson5910
  • jim_thompson5910
the same happens with the y coordinates |dw:1439083780104:dw|
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
|dw:1439083807218:dw|
jim_thompson5910
  • jim_thompson5910
so hopefully you see how (a,0) was found
anonymous
  • anonymous
ya think so
jim_thompson5910
  • jim_thompson5910
the same idea is applied to find Q and R
UsukiDoll
  • UsukiDoll
feels like a proof... like verify (a,0) using A and B
anonymous
  • anonymous
okay
anonymous
  • anonymous
i think i understand.. thanks
UsukiDoll
  • UsukiDoll
|dw:1439083816461:dw|
jim_thompson5910
  • jim_thompson5910
for the other stuff, you use the distance formula \[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2}\]
anonymous
  • anonymous
kk
anonymous
  • anonymous
|dw:1439083949599:dw|
UsukiDoll
  • UsukiDoll
???
UsukiDoll
  • UsukiDoll
this assignment looks painful D: but I sort of see where they are going.
anonymous
  • anonymous
it is very painful i have 7 more
UsukiDoll
  • UsukiDoll
clicked close by accident.
UsukiDoll
  • UsukiDoll
so we need Q and R hmmm Q is (3a/2,b/2) R is (a/2, b/2)
anonymous
  • anonymous
lol i wish. It only lets me do that for quixs and test
UsukiDoll
  • UsukiDoll
no I said I accidentally closed the tab on my end XD
anonymous
  • anonymous
ohhhhh lol
UsukiDoll
  • UsukiDoll
so I couldn't reach you for a bit XD
UsukiDoll
  • UsukiDoll
if Q ( 3a/2, b/2) then there are points that can lead us to this. if R ( a/2,b/2) then there are points that lead to this result points A and C may lead us to R
UsukiDoll
  • UsukiDoll
A (0,0) C (a,b) |dw:1439084550670:dw|
anonymous
  • anonymous
okay
UsukiDoll
  • UsukiDoll
|dw:1439084709293:dw|
UsukiDoll
  • UsukiDoll
a HA! B AND C can make Q!
UsukiDoll
  • UsukiDoll
B(2a,0) C( a,b) (2a+a/2, 0+b/2) Q(3a/2,b/2)
UsukiDoll
  • UsukiDoll
|dw:1439084868914:dw|
UsukiDoll
  • UsukiDoll
|dw:1439084931274:dw|
anonymous
  • anonymous
ohh!!! i understand now thank you sooo much!!!!
UsukiDoll
  • UsukiDoll
I think midpoint formula and distance formula is bring used here.
UsukiDoll
  • UsukiDoll
QA hmm Q(3a/2, b/2) A (0,0)
UsukiDoll
  • UsukiDoll
|dw:1439085179111:dw|
UsukiDoll
  • UsukiDoll
\[\Large d = \sqrt{\left(x_{2}-x_{1}\right)^2+\left(y_{2}-y_{1}\right)^2} \] \[\Large d = \sqrt{(0-\frac{3a}{2})^2+(0-\frac{b}{2})^2} \] \[\Large d = \sqrt{(-\frac{3a}{2})^2+(-\frac{b}{2})^2} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{b^2}{4}} \] if we let \[\large b = a \sqrt{3} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(a \sqrt{3})^2}{4}} \] \[\Large d = \sqrt{(\frac{9a^2}{4})+\frac{(3a^2)}{4}} \] \[\Large d = \sqrt{\frac{12a^2}{4}}\] \[\Large d = \sqrt{3a^2} \] \[\Large d = a\sqrt{3} \] Now do the same process for RB for R ( a/2,b/2) and B (2a,0)

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