# Pyramid

• Mar 5th 2006, 07:33 AM
totalnewbie
Pyramid
The basis of a regular pyramid is "a". Perpendicular plane which crosses the centres of two basis, forms smaller pyramid which volume is V.
Declare the followings through "a" and "V":
1)the height of the bigger pyramid
2)the angle between the side edge and base

1)h=[64*V*sqrt(3)]/[3*a^2]
2)arctan[64*V]/[a^3]
• Apr 4th 2006, 06:05 AM
totalnewbie
It appears to me that I am alone with the problem.
• Apr 4th 2006, 06:59 PM
ThePerfectHacker
There is something wrong with your problem. There is no way to determine the the height of the larger pyramid.
• Apr 6th 2006, 05:34 AM
totalnewbie
There is nothing wrong with my problem.
The larger and the bigger triangle are both isosceles and it is possible to solve it by using the area formula for isosceles tirangle. The lenght of the pyramid fells into the center of the median which relates to the median as 1:2.
There is a formula which says that the areas of similar triangle relate as square of something but I don't know it exactly.
• Apr 6th 2006, 09:15 PM
earboth
Quote:

Originally Posted by totalnewbie
The basis of a regular pyramid is "a". Perpendicular plane which crosses the centres of two basis, forms smaller pyramid which volume is V.
Declare the followings through "a" and "V":
1)the height of the bigger pyramid
2)the angle between the side edge and base

1)h=[64*V*sqrt(3)]/[3*a^2]
2)arctan[64*V]/[a^3]

Hello,

to 1.:
let $A_b$ the base area of the bigger pyramid.
let $A_s$ the base area of the smaller pyramid.

$A_b= \frac{a^2}{4} \cdot \sqrt{3}$. Then it is
$A_s= \frac{1}{4} \cdot \frac{a^2}{4} \cdot \sqrt{3}$

let $h_s$ the height of the smaller pyramid. Then the Volume of the smaller pyramid is:
$V=\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{a^2}{4} \cdot \sqrt{3} \cdot h_s$

Because $h\ \parallel \ h_s$ you get the proportion:
$\frac{h_s}{h}=\frac{\frac{1}{2} \frac{1}{2} a \sqrt{3}}{\frac{2}{3} \frac{1}{2} a \sqrt{3}}$

That means: $h_s=\frac{3}{4} \cdot h$.

The volume of the smaller pyramid is: $V=\frac{1}{3} \cdot A_s \cdot h_s$
$V=\frac{1}{3} \cdot \frac{1}{16} \cdot a^2 \sqrt{3} \cdot \frac{3}{4} \cdot h$. Solve this equation for h and you'll get exactly your answer.

to 2.:

the length of the median is $\frac{1}{2} \cdot a \sqrt{3}$. So the tangens of the angle is:
$\tan (\alpha)=\frac{h}{\frac{2}{3} \cdot \frac{1}{2} \cdot a\sqrt{3}}=\frac{3h}{a\sqrt{3}}$

$\tan (\alpha)=\frac{3 \cdot \frac{64 V \sqrt{3}}{3a^2}}{a\sqrt{3}}=\frac{64V}{a^3}$

Greetings

EB
• Apr 7th 2006, 02:00 AM
totalnewbie
I got these answers:
1)h=[64*V*sqrt(3)]/[3*a^2]
2)arctan[64*V]/[a^3]
Thank you anyway.