Please create an illustration of a geometric concept. Show a right circular cone with a noticeable marker indicating a section one third from the very top (the vertex). The cone is divided by this marker into two distinct parts. The cone's total volume is symbolized as 5 litres. There's no calculation or specific text within the image. The upper and lower sections of the cone signify two different volumes. The image focuses on the visual representation of the mathematical principles rather than the calculation details.

the volumeof a right circular cone is 5 litres. calculate the volumes of the two parts into which the cone is divided by a plane parallel to the base one third of the way down from the vertex to the base to the nearest ml

volume of the right circular cone=1/3pi*r2*h=5

pi*r2*h=15
from the third part of the statement,the radius(r)=1/3r.the height(h)=1/3h
the volume of the cone=1/3*pi*(1/3r)2*1/3h
=1/3pi*1/9r2*1/3h
=1/3*1/9*1/3*pi*r2*h
From equation one above pi*r2*h=15
=1/81*15=~0.185
~185mL
volume of the frustum=5 - 0.185=~4.815=~4815mL

Volume of the right circular cone= 1/3pi *r^2×h = 5

pi x r^2 ×h = 15
: r=1/3, h= 1/3
The volume of the cone = 1/3 × pi ×(1/3r) 2×1/3h
= 1/3 pi x1/9r^2x1/3h
From eqn 1
Pi x r^2 x h = 15
1/81× 15 = 0.185 or 185ml.
Volume of frustum = 5 - 0.185= 4.815ml

Thanks guys

Good

Thanks

Can you please explain clearly how you got the volume of the top cone?

U are good in mathematics wow

I wish I could do this

please how did you get 15

How do you derive the formula

The solution is:

Volume of right circular cone = 1/3 * pi * r^2 * h = 5 L

Therefore, pi * r^2 * h = 15 (multiplying both sides by 3)

Now, the plane bisects the cone into 2 similar cones. Let the height of the smaller cone be h1 and its radius be r1.

So, h1/h = r1/r = 1/3 (given)

Now, the volume of the top cone can be found as:

V1 = 1/3 * pi * r1^2 * h1 = 1/3 * pi * (r/3)^2 * (h/3) = (1/81) * pi * r^2 * h

V1 = (1/81) * 15 * pi = 0.185 L or 185 mL (approx)

The volume of the frustum can then be found as:

Vfrustum = Vcone - V1

Vfrustum = 5 - 0.185 = 4.815 L or 4815 mL (approx)

Therefore, the volumes of the top cone and the frustum are 185 mL and 4815 mL (approx) respectively.