# Archimedes parabola/ Triangles

• Jun 15th 2009, 06:35 AM
abccbaabc014
Archimedes parabola/ Triangles
The Quadrature of the Parabola - Wikipedia, the free encyclopedia

I can follow the geometric proof on this page, fine, but I can't figure out why the second triangles height is 1/4th of the first. Can anyone give an explanation for this?
• Jun 15th 2009, 09:56 PM
VonNemo19
Note that each successive triangle has 1/8 the area of the one before, and the each width is half of the one before

$\displaystyle \frac{1}{8}(\frac{1}{2}b_Lh_L)=\frac{1}{2}\frac{b_ L}{2}h_s$

since we know that the area of each succesive triangle is 1/8 of the previous one, and that $\displaystyle b_s=b_L/2$, then we can just set up the above equation and ask what would make the statement true?

well?

taking $\displaystyle h_s=h_L/4$

Understand that htis only works because everything in the equation is understood to be constant except for b_s

s=smaller triangle L=larger
• Jun 15th 2009, 10:13 PM
Properties of a parabola
Hello abccbaabc014

Welcome to Math Help Forum!
Quote:

Originally Posted by abccbaabc014
The Quadrature of the Parabola - Wikipedia, the free encyclopedia

I can follow the geometric proof on this page, fine, but I can't figure out why the second triangles height is 1/4th of the first. Can anyone give an explanation for this?

As it says in footnote 1, you can use analytical geometry to prove this. If you're going to try, you could begin by assuming that the parabola is $\displaystyle y^2=4ax$, and that the ends of the first chord (the one that forms the longest side of the original triangle) are at $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$. The mid-point of this chord is then $\displaystyle \Big(\tfrac12(x_1+x_2), \tfrac12(y_1+y_2)\Big)$. From this you can work out the coordinates of the third vertex of the triangle (using $\displaystyle y^2=4ax$), and hence its height.

In a similar way, you can then calculate the coordinates of the third vertex of one of the smaller triangles, and hence its height.

How good is your analytical geometry? Can you do this?

And Grandad, I did follow what you said but I used a different equation $\displaystyle (y=(32-X^2)/4)$. Every time I worked out the actual height of the triangles, the ratio was 1:4, but I was wondering if there is any proof of this?