Answer to Question #225071 in Geometry for LEI

Question

A frustum of a square pyramid has an upper base edge of 6 cm and a lower base edge of 4 cm if the distance between these 2 bases is 10 cm. Find the total surface area of the frustum.

Answer 1

let small height, h = xcm

let big height, H=(10+x)

$\frac{10+x}{x}=\frac{6}{4}$

$40+4x=6x$

$2x=40$

$x=h=20$

H$=10+20=30cm$

for the small pyramid, height of one face $=\sqrt{20^2+2^2}=20.1cm$

$=\sqrt{20^2+2^2}=20.1cm$

$area\space of \space face=1/2\times 4\times 20.1=40.2cm^2$

area of the four faces $=40.2\times 4=160.8cm^2$

for the big pyramid, height of one face=$\sqrt{30^2+3^2}=30.15cm$

$\sqrt{30^2+3^2}=30.15cm$

area of the face $=1/2\times 6\times 30.15=90.45cm^2$

area of four faces$=90.45\times 4=361.8cm^2$

area of sides of frustum=361.8-160.8=201

assuming the frustum is closed both sides area of upper base $=6\times 6=36cm^2$

area of lower base$=4\times 4=16cm^2$

total area of frustum=$201+36+16=253cm^2$