Question #225071

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.


1
Expert's answer
2021-08-19T08:00:51-0400


let small height, h = xcm

let big height, H=(10+x)

10+xx=64\frac{10+x}{x}=\frac{6}{4}

40+4x=6x40+4x=6x

2x=402x=40

x=h=20x=h=20

H=10+20=30cm=10+20=30cm

for the small pyramid, height of one face =202+22=20.1cm=\sqrt{20^2+2^2}=20.1cm

=202+22=20.1cm=\sqrt{20^2+2^2}=20.1cm

area of face=1/2×4×20.1=40.2cm2area\space of \space face=1/2\times 4\times 20.1=40.2cm^2

area of the four faces =40.2×4=160.8cm2=40.2\times 4=160.8cm^2

for the big pyramid, height of one face=302+32=30.15cm\sqrt{30^2+3^2}=30.15cm

302+32=30.15cm\sqrt{30^2+3^2}=30.15cm

area of the face =1/2×6×30.15=90.45cm2=1/2\times 6\times 30.15=90.45cm^2

area of four faces=90.45×4=361.8cm2=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 =6×6=36cm2=6\times 6=36cm^2

area of lower base=4×4=16cm2=4\times 4=16cm^2

total area of frustum=201+36+16=253cm2201+36+16=253cm^2

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