Question

frustum is mounted on a hemisphere total height is 7 cm and diameters of the frustum are 5 & 2 cm calculate the external surface area

Answer 1

Total ht of the object= 7cm
ht of frustum =7-5=2cm

slant ht of frustum l =
$sqrt( h ^{2} + (r1 - r2) ^{2} )$
l=
$\sqrt{13}$

external surface area is CSA
CSA = CSA of frustum + CSA of hemisphere
=
$\pi \times l(r1 + r2) + 2\pi \times r {}^{2}$
=22/7×√13×(5+2) + 2×22/7×5×5
=22/7×√13×7 + 44×25÷7
=22√13 + 1100/7
=(154√13+1100)÷7
=(555.25+1100)÷7
=1655.25÷7
=236.464

so, the external surface area is 236.464cmsq