Question

A flower vase is in the form of a frustum of a cone.The perimeters of its bases are 44cm and 8.4πCM. if the depth is 14cm, then find how much soil it can hold​

Answer 1

If the depth of the flower vase is 14cm and the perimeters of the bases are 44 cm and 8.4π cm, then the amount of soil it can hold is 1408.58 cm³.

Step-by-step explanation:

Step 1:

Let the perimeter and radius of the bases be “P₁” & “R₁” and “P₂” & “R₂” respectively.

It is given that,  

Perimeter, P₁ = 2πR₁ = 44 cm

∴ R₁ = 44/2π = 22 * 7/22 = 7 cm

And,

Perimeter, P₂ = 2πR₂ = 8.4π cm

∴ R₂ = 8.4π / 2π = 4.2 cm

Step 2:

The depth of the frustum cone, h = 14 cm

We know,  

The volume of the flower vase which is in the form of frustum cone is,

= $\frac{1}{3}$ * π * [R₁² + R₂² + (R₁ * R₂)] * h

= $\frac{1}{3} * \frac{22}{7}$ * [(7)² + (4.2)² + (7 * 4.2)] * 14

= $\frac{1}{3} * \frac{22}{7}$ * [49 + 17.64 + 29.4] * 14  

= 2 * $\frac{22}{3}$ * 96.04

= 1408.58 cm³

Thus, the flower vase can hold 1408.58 cm³ of soil.

-----------------------------------------------------------------------------------------

Also View:

The slant height of the frustum of a cone is 5 cm. If the difference the radii of its two circular ends is 4 cm, write the height of the frustum.

https://brainly.in/question/7474636

Derive the formula of volume of frustum of cone

https://brainly.in/question/6779988

Answer 2

Answer:

Step-by-step explanation: