anonymous
  • anonymous
The ratio of the lengths of two equilateral triangles is 4:9. what is the ratio of their areas?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
Lets see if the area of triangles are ratioed as well. 3-4. 3*4 = 12/2 = 6 are of 3-4 is 6 triple the side 9-12-15; 9*12 = 108/2 = 54 but 6*3 is not 54, so we cant assume that the areas are of equal ration to the sides...
amistre64
  • amistre64
4^2 = h^2 + 2^2 16 - 4 = h^2 12 = h^2 h = 2sqrt(3); b=4/2 Area of 4sides equilateral Triangle is 4sqrt(3) 9^2 = h^2 + 4.5^2 81 - 20.25 = h^2 60.75 = h^2 h = sqrt(60.75) = 2sqrt(15) the ration of the areas is 4sqrt(3) : 2sqrt(15) if I havent tripped over myself :)
amistre64
  • amistre64
yep, I tripped alright...let me try finishing that up alittle :)

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amistre64
  • amistre64
h/2 times 9 The Area of the 9side Triangle is: 9sqrt(15) We got a ratio of 4sqrt(3) : 9sqrt(15).....maybe :)
radar
  • radar
This is obviously confusing. The area of a triangle is given as A=1/2bh where base would be one side but the heights h would also be related to the lenght of the side. In a equilateral triangle, what would that relationship be? Without a diagram you have to imagine the height as h=sqrt(b^2-(b/2)^2)
radar
  • radar
\[h=\sqrt{b ^{2}-(b/2)^{2}}\]
radar
  • radar
So Area expressed in terms of its base (which is a side) would be: \[A=1/2b(\sqrt{b ^{2}-(b/2)^{2}}\] I need to think this further. I'll be back
radar
  • radar
After doing some more ciphering! The above works out to be:\[A=(b ^{2}\sqrt{3})\div4\]
radar
  • radar
substituting 4 and then substituting 9 shows a ratio between areas 1:5 I am not comfortable with this answer, hopefully you will get the school solution and post it here
anonymous
  • anonymous
the answers i have are 4:9, 9:4, 2:3, 16:81, and 81:16
radar
  • radar
Hey DavisAshkey it looks like you came up with the proportion squared, or the square root of the proportion of the sides. You very well may be correct. I will look at it again. The squared version looks reasonable!

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