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by **little_dragon** on Tue Jun 30, 2009 7:48 pm

The area of a triangular face of a square-based pyramid is equal to the square of the vertical height of a pyramid. Show how the ratio of the height of each of the faces to one half of the length of the base of the face is the Golden Mean.

ive looked up the golden mean but how do i get started??? :wave:

**Sponsor**

by **stapel_eliz** on Sat Jul 04, 2009 8:34 pm

Sorry to be so long in replying, but it took me a while to "see" this the right way....

I'm sure you've drawn the square-based right pyramid. Let's label the height as "h", the slant height (that is, the height of a triangular face) as "s", and the length of a side of the base as "b".

Looking at a cross-sectional view of the pyramid, so you're looking at an isosceles triangle with base b and height h, the height-line splits this triangle into two right triangles, each having a base with length b/2 and a height of h. The Pythagorean Theorem gives you the length of the hypotenuse:

. . . . . $\sqrt{h^2\, +\, \left(\frac{b}{2}\right)^2}\, =\, \frac{\sqrt{4h^2\, +\, b^2}}{2}$

If you look at the pyramid from another angle, you'll note that the above also stands for the slant-height s, being the distance from the peak of the pyramid, right down the middle of a side, to the midpoint of that side of the base.

Then the area of a triangular face is given by:

. . . . . $\left(\frac{1}{2}\right)(b)\left(\frac{\sqrt{4h^2\, +\, b^2}}{2}\right)\, =\, \left(\frac{b}{4}\right)\sqrt{4h^2\, +\, b^2}$

You are given that this area (above) is equal to the square of the height h, so:

. . . . . $\left(\frac{b}{4}\right)\sqrt{4h^2\, +\, b^2}\, =\, h^2$

. . . . . $\sqrt{4h^2\, +\, b^2}\, =\, \frac{4h^2}{b}$

. . . . . $4h^2\, +\, b^2\, =\, \frac{16h^4}{b^2}$

. . . . . $4h^2b^2\, +\, b^4\, =\, 16h^4$

. . . . . $b^4\, +\, 4h^2b^2\, -\, 16h^4$

. . . . . $b^2\, =\, \frac{-4h^2\, \pm\, \sqrt{16h^4\, -\, 4(-16h^4)}}{2}\, =\, \frac{4h^2(-1\, \pm\,\sqrt{5})}{2}$

But of course the square must be positive, so you can ignore the "minus" on the square root, giving you:

. . . . . $b^2\, =\, \frac{4h^2(-1\, +\,\sqrt{5})}{2}$

Leaving that aside for now, let's turn to the ratio, using the radical expression found previously for the slant-height:

. . . . . $\frac{\left(\frac{\sqrt{4h^2\, +\, b^2}}{2}\right)}{\left(\frac{b}{2}\right)$

. . . . . $\frac{\sqrt{4h^2\, +\, b^2}}{b}$

The square root in the numerator can be replaced, because we saw above that:

. . . . . $\sqrt{4h^2\, +\, b^2}\, =\, \frac{4h^2}{b}$

So we get:

. . . . . $\frac{\sqrt{4h^2\, +\, b^2}}{b}\, =\, \frac{\left(\frac{4h^2}{b}\right)}{b}\, =\, \frac{4h^2}{b^2}$

Now replace b² with the solution value (from the quadratic) to get:

. . . . . $\frac{4h^2}{\left(\frac{4h^2(-1\, +\,\sqrt{5})}{2}\right)$

. . . . . $=\, \frac{2}{-1\, +\,\sqrt{5}}$

The ratio is now strictly numerical, which is promising, but you need to "rationalize" to get rid of the radicals in the denominator, so:

. . . . . $\left(\frac{2}{-1\, +\,\sqrt{5}}\right)\left(\frac{-1\, -\, sqrt{5}}{-1\, -\, \sqrt{5}}\right)$

Simplify, and see what you get. :wink:

by **little_dragon** on Fri Jul 24, 2009 4:53 pm

now i dont feel so bad abt not getting it
thnx!! :wave:

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area of face = square of ht; show ratio is golden mean

area of face = square of ht; show ratio is golden mean

Re: area of face = square of ht; show ratio is golden mean