Question

The ratio of the lengths of two equilateral triangles is 4:9. what is the ratio of their areas?

Answer 1

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...

Answer 2

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 :)

Answer 3

yep, I tripped alright...let me try finishing that up alittle :)

Answer 4

h/2 times 9

The Area of the 9side Triangle is:

9sqrt(15)

We got a ratio of 4sqrt(3) : 9sqrt(15).....maybe :)

Answer 5

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)

Answer 6

\[h=\sqrt{b ^{2}-(b/2)^{2}}\]

Answer 7

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

Answer 8

After doing some more ciphering!

The above works out to be:\[A=(b ^{2}\sqrt{3})\div4\]

Answer 9

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

Answer 10

the answers i have are 4:9, 9:4, 2:3, 16:81, and 81:16

Answer 11

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!