gillgill said:
1) A=x^2 and 2y-3x=12;x=2/3y-4
A=(2/3y-4)^2 - (2/3y-4)
=4/9y^2 + 16 - 2/3y + 4
=4/9y^2-2/3y +20
Where is my mistake?
2) 3 equal sides...all 60 degrees angle...B x H/2
1) okay A=x^2 and 2y-3x=12 is given
you solved the right expression for x,
x= \frac{2}{3}y-4 which looks fine
Next I see you plugging in your value for x into an equation:
A = x^2-x Why?
Where did you get that equation from?
If you follow my meaning, next be careful when you expand (\frac{2}{3}y-4)^2
it is the same as this operation (\frac{2}{3}y-4)(\frac{2}{3}y-4)
2) equilateral \Delta, 3 equal sides, all angles 60 deg, Area = 1/2 base*ht .. good
Let each side of your \Delta be length
s
If you draw your \Delta with a vertex at the top, drop a vertical
from that vertex to the bottom of \Delta (so it bisects it).
You now have two equal right \Delta's, inside your equilateral \Delta. And the vertical line is now your height.
Can you see where I'm leading?
What are the angles inside your two right \Deltas?
Is there something special about them such that you can determine the
sides of each right triangle in terms of
s ?
With that information you should be able to find the area of the equilateral
\Delta as a function of its side
s.
