The volume of the frustum of a cone is `(1)/(3) pih [r_(1)^(2) +r_(2)^(2)- r_(1)r_(2)]`, where h is vertical height of the frustum and `r_(1)r_(2)` are the radii of the ends.

What is volume of the frustum of a cone with height h and radii r_(1),r_(2) ?

r+r_(3)+r_(1)-r_(2)=

The curved surface area of frustum of a cone is pi//(r_(1) +r_(2)) , where l = sqrt(h^(2) +(r_(1) + r_(2))^(2)),r_(1) and r_(2) are the radii of two ends of the frustum and h s the vertical height.

The total surface area of a frustum-shaped glass nuber is (r_(1)gtr_(2))

Value of 1/(r_(1)^2)+ 1/(r_(2)^2)+ 1/(r_(3)^2)+ 1/(r_()^2) is :

If r_1 and r_2 denote the radii of the circular bases of the frustum of a cone such that r_1> r_2 , then write the ratio of the height of the cone of which the frustum is a part to the height of the frustum.

If h is the height, I the slant height and r_(1), r_(2) radii of the circular bases of the frustum of a cone, then slant height of the frustum = sqrt(( r_(1) - r_(2))^(2) + h^(2)) . Find the height of the cone of which the frustum is a part = (h r_(1) /( r_(1) - r_(2) ).We have a bucket in the form of frustum of a cone in which h = 8 cm, r_(1) = 9 cm and r_(2) = 3 cm.

show that r_(2)r_(3) +r_(3) r_(1)+ r _(1) r_(2)=s^(2)

If h is the height, I the slant height and r_(1), r_(2) radii of the circular bases of the frustum of a cone, then slant height of the frustum = sqrt(( r_(1) - r_(2))^(2) + h^(2)) . Height of the cone of which the frustum is a part = (h r_(1) /( r_(1) - r_(2) ).We have a bucket in the form of frustum of a cone in which h = 8 cm, r_(1) = 9 cm and r_(2) = 3 cm. Find its volume.