Answer to Question #149237 in Geometry for marshy taray

Question

The areas of the bases of a frustum of a regular pyramid is 250 m2 and 640 m2 respectively. Find the altitude of the pyramid if the altitude of its frustum is 130 m?

Answer 1

Let $A_1,A_2$ and $h_1$ be the areas of upper base,lower base and height of the frustum.

Also let $H$ be the height of the pyramid.

Then,Volume of pyramid is $=(1/3)×A_2×H$

and Volume of frustum is $=(1/3)×h_1×[A_1+A_2+\sqrt{(A_1.A_2)}]$

We know that,

Volume of pyramid $=$ Volume of frustum $+$Volume of pyramid containing upper base of frustum as its base. ..........(1)

Let $h_2$ be the height of the pyramid containing upper base of frustum as its base.

Then we have , $H=h_1+h_2$

$\implies h_2=H-h_1$ .......(2)

Here we have given, $A_1=250$ $m^2$ , $A_2=640$ $m^2$ , $h_1=130$ $m$

From (2) we have, $h_2=H-130$

From (1) we have,

$(1/3)×A_2×H=[(1/3)×h_1×[A_1+A_2+\sqrt {(A_1.A_2)}]]+[(1/3)×A_1×h_2]$

Therefore,

$(1/3)×640×H=[(1/3)×130×[640+250+\sqrt {(640×250)}]]+[(1/3)×(H-130)×250]$

$\implies 640H-250H=130(1290-250)$

$\implies 390H=130×1040$

$\implies H=(1040/3)$

Therefor the required height of the pyramid is

$=(1040/3)$ $m$