Answered

Find the best answers to your questions with the help of IDNLearn.com's expert contributors. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.

Find the volume of a right circular cone that has a height of 19.9 cm and a base with a radius of 9.6

Sagot :

We are given –

  • Height (h) of cone is = 19.9cm
  • Radius (r) of cone is = 9.6 cm

According to the question, we are asked to find out volume of the cone. As we know –

  • Volume of cone = ⅓ πr²h

Substituting values –

[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf Volume_{(Cone)} = \dfrac{1}{3} π r^2h}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf Volume_{(Cone)} = \dfrac{1}{3} π\times (9.6)² \times 19.9[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf Volume_{(Cone)} = \dfrac{1}{3} π \times 92.16 \times 19. 9[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\sf Volume_{(Cone)} = 96.4608 \times 19.9[/tex]

[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf Volume_{(Cone)} = 1919.57\:cm³}\\[/tex]

[tex]\therefore{\underline{\textsf{ Volume of cone is \textbf{1919.57cm³}}}} [/tex]

Know More –

  • C.S.A of cone = πrl
  • T.S.A of cone = πrl + πr²
  • Volume of cuboid = l × b × h
  • C.S.A of cuboid = 2(l + b)h
  • T.S.A of cuboid = 2(lb + bh + lh)
  • C.S.A of cube = 4a²
  • T.S.A of cube = 6a²
  • Volume of cube = a³
  • Volume of sphere = 4/3πr³
  • Surface area of sphere = 4πr²
  • Volume of hemisphere = ⅔ πr³
  • C.S.A of hemisphere = 2πr²
  • T.S.A of hemisphere = 3πr²

_________________________________________

SupremeStrangeidn

Step-by-step explanation:

Volume of cone = 1/3πr²h

here,

  • Height = 19.9 cm
  • base radius = 9.6cm

Putting the known values ,

Volume = [tex] \frac{1}{3} \times 3.14 \times 9.6 {}^{2} \times 19.9cm {}^{3} [/tex]

Volume = 1919.56 cm³

Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.