Question

An expandable cone-shaped funnel consists of two sections as shown. ( Question A) What is the volume of the bottom section? ( Question B) What is the volume of the top section? " Will Mark Brainliest. Please don't copy and rewrite someone else's answer ​

Answer 1

Image of the cone is attached.

Answer:

a) 24π cm³

b) 168π cm³

Step-by-step explanation:

We know the formula for volume of a cone is expressed as:

$V = \frac{1}{3} \pi r^2 h$

a) To find the volume of the bottom section, given the values:

diameter, d = 6 cm

radius, r = $\frac{d}{2} = \frac{6}{2}= 3 cm$

height, h = 8cm

Let's use the formula,

$V = \frac{1}{3} \pi r^2 h$

Substituting values, we have:

$V = \frac{1}{3} \pi 3^2 * 8$

Vtop = 24π cm³

Therefore, volume of the bottom section of the cone is 24π cm³

b) volume of the top section.

Volume of the top section would be derived by subtracting the volume of the bottom section from the volume of the cone.

Vtop = V - Vbottom

Now, let's find the volume of the cone.

Given:

diameter, d = 12 cm

radius, r = $\frac{d}{2} = \frac{12}{2}= 6 cm$

height, h = 16cm

Let's use the formula,

$V = \frac{1}{3} \pi r^2 h$

Substituting values, we have:

$V = \frac{1}{3} \pi 6^2 * 16$

V = 196π cm³

Therefore, volume of the bottom section =

Vtop = V - Vbottom

= 192π - 24π

= 168π cm³

Therefore volume of the top section of the cone is 168π cm³

Answer 2

Answer:

Step-by-step explanation:

The bottom section of the funnel is in the shape of cone with radius = 6 cm and height = 8 cm

The volume of a cone is given by

$\frac{1}{3} \pi r^{2} h\\\\V= \frac{1}{3} \pi ( 3^{2} )(8)=24\pi$

The volume of the top part of the cone, called frustum, is the volume of big cone subtract volume of smaller cone

$\text {Volume of big cone} = \frac{1}{3} \pi ( 6^{2} )(16)=192 \pi Volume of frustum = 192 \pi -24 \pi =168 \pi$

We can leave the answer as an exact answer $(168 \pi )$

or as 527.79