New Blue Book, Test 1, Section 3, problem 19

oblivi0n
Mon, 16 Nov 2009 18:59:37 +0000

this is really hard to explain without drawing but i know how agonizing it can be to not know how to solve a question so here goes nothing…

first of all we need to find the slant height and slant height here isn’t “e”. A slant height is the height that’s slanted and rises up the the vertex of the pyramid. It’s not necessarily the edge of the pyramid. (http://www.mathematicsdictionary.com/english/vmd/images/s/antheightofaregularpyramid.gif)

So to find this we make a slanted triangle that includes the slant height, “e”, and half of a side of the base which is equal to “m/2” (because one side is “m”)

after getting this triangle we can find the slant height because it forms a right triangle with “e” being our hypotenuse. Since e=m, i’m going to use “m” for “e”. Using the Pythagorean Theorem (S is our slant height):

m^2 = S^2 + (m/2)^2
So slant height is equal to
S^2 = m^2 - (m/2)^2
S^2 = (3m^2)/4
S = (sqrt(3)*m) / 2

since we have slant height, we can form another right triangle with “h”, the slant height, and again half of the square base (m/2)…

so this time i’ll skip some steps and here the slant height is our hypotenuse.

(sqrt(3)m / 2)^2 = h^2 + (m/2)^2
h^2 = 3m^2/4 - m^2/4
h^2 = 2m^2 / 4
h^2 = m^2/2
h = m/sqrt(2)

So basically you need to find slant height using the edge then you need to find “h” in terms of m, using the slant height.