Question

Please answer me fast....​

Answer 1

Answer:

I know this answer.

but I am very busy so tomorrow I will answer this question.

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Answer 2

Answer:

the amount of soil held in the frustum is 1408.59 $cm^{3}$(approx)

Step-by-step explanation:

in a frustum small radius/original radius = small height/ original height

then from given info about perimeter if the two surfaces the radius are 4.2 cm and 7 cm

so, 42./7= h-14/h

from here we get the total height as 35 cm

then we take out the total volume of the cone

1/3$\pi *R^{2} *H$= 1/3*(22/7)*7*7*35= 1/3*154*35

then volume of the frustum cut

1/3$\pi* r^{2} *h$= 1/3*(22/7)*4.2*4.2*21=1/3*66*4.2*4.2

then amount of soil held in the frustum is (1796.67-388.08)=1408.59 $cm^{3}$