Question

3. Find the volume and lateral surface area of the frustum of a regular square pyramid whose altitude is 38 cm and whose base edges are 10 cm and 20 cm respectively.

Answer 1

We know that ,if $a,b$ and $h$ are the base edge,top edge and height of a frustum of a regular square pyramid,then formula for volume and lateral surface area are -

Volume$(V) =(1/3)(a^2+b^2+ab)×h$

Lateral surface area$(F)=2(a+b)\sqrt(((a-b)/2)^2+h^2)$

Now according to the given problem,

$a=20 cm$ , $b=10cm$ ,$h=38cm$

Therefore required volume of the frustum of square pyramid is

$V=(1/3)×((20)^2+(10)^2+(20×10))×38$

$=(1/3)×(400+100+200)×38$ $cm^3$

$=(26600/3)$ $cm^3$

and Lateral surface area is

$F=2×(20+10)×\sqrt(((20-10)/2)^2+38^2)$

$=2×30×\sqrt(25+1444)$ $cm^2$

$=60×\sqrt1469$ $cm^2$