# A truncated pyramid.

• May 3rd 2011, 05:52 AM
IoNForce
A truncated pyramid.
I have been trying to solve this one but i simply don't know how to. I asked my teacher about it and he said I should try to use the coefficient of similarity, but still no results. The task:

The bases of a truncated pyramid have the surfaces (areas) of 98cm^2 and 2cm^2. Whats the surface (area) of a section that cuts trough the truncated pyramid and goes trough the center of the truncated pyramid's height and is parallel with the bases.

Every help is welcome!
• May 3rd 2011, 06:00 AM
TheEmptySet
I have just one question. What is the shape of the base? Is is a square pyramid?
• May 3rd 2011, 07:16 AM
IoNForce
It isn't specified so it can be either one of them but i guess its a triangle in the base.
• May 4th 2011, 11:07 PM
earboth
Quote:

Originally Posted by IoNForce
I have been trying to solve this one but i simply don't know how to. I asked my teacher about it and he said I should try to use the coefficient of similarity, but still no results. The task:

The bases of a truncated pyramid have the surfaces (areas) of 98cm^2 and 2cm^2. Whats the surface (area) of a section that cuts trough the truncated pyramid and goes trough the center of the truncated pyramid's height and is parallel with the bases.

Every help is welcome!

1. Draw a sketch (see attachment)

2. Complete the frustrum to a pyramide as indicated in my sketch. The proportion of similar areas equals the proportion of the square of corresponding lengthes.

3. Let t denote the top area (2 cm²), b the base area (98 cm²) and s the cross-section area.
Then you know:

$\dfrac{y^2}{x^2} = \dfrac{98}2=49~\implies~y=7x$

4. The cross-section has the distance
$d = \dfrac{x+y}2$
from the top of the completed pyramide.
Then you know:

$\dfrac{\left(\frac{x+y}2\right)^2}{x^2} = \dfrac{s}2$

Replace y by 7x:

$\dfrac{\left(\frac{x+7x}2\right)^2}{x^2} = \dfrac{s}2~\implies~\dfrac{(4x)^2}{x^2}=\dfrac s2~\implies~s=32$
• May 8th 2011, 12:00 AM
IoNForce
Quote:

Originally Posted by earboth
1. Draw a sketch (see attachment)

2. Complete the frustrum to a pyramide as indicated in my sketch. The proportion of similar areas equals the proportion of the square of corresponding lengthes.

3. Let t denote the top area (2 cm²), b the base area (98 cm²) and s the cross-section area.
Then you know:

$\dfrac{y^2}{x^2} = \dfrac{98}2=49~\implies~y=7x$

4. The cross-section has the distance
$d = \dfrac{x+y}2$
from the top of the completed pyramide.
Then you know:

$\dfrac{\left(\frac{x+y}2\right)^2}{x^2} = \dfrac{s}2$

Replace y by 7x:

$\dfrac{\left(\frac{x+7x}2\right)^2}{x^2} = \dfrac{s}2~\implies~\dfrac{(4x)^2}{x^2}=\dfrac s2~\implies~s=32$

Thanks, I managed to do it with the new height (3x) and I forgot to reply here, sorry ><
Anyway thanks a lot for the help!