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Reasoning A cone with radius 6 and height 9 has its radius quadrupled. How many times greater is the
volume of the larger cone than the smaller cone?l

Sagot :

Let's see

For initial cone

  • r=6
  • h=9

Volume

  • 1/3πr²h
  • 1/3π(6)²(9)
  • 3(36)π
  • 108π units³

For new cone

  • r=4(6)=24
  • h=9

Volume

  • 1/3π(24)²(9)
  • 3(576)π
  • 1728π

So

  • 1728π/108π
  • 16times
semsee45

Answer:

16 times

Step-by-step explanation:

[tex]\textsf{Volume of a cone}=\sf \dfrac{1}{3} \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}[/tex]

If only the radius is changed, the change in volume will be proportionate to the multiplicative factor squared.

[tex]\sf \implies Volume =\dfrac{1}{3} \pi (ar)^2h=\dfrac{1}{3} \pi (a^2)r^2h[/tex]

Therefore, if the cone is quadrupled (multiplied by 4), the volume of the larger cone will be 4² times greater than the volume of the smaller cone, so 16 times greater than the smaller cone.

Proof

Given:

  • radius = 6
  • height = 9

Substituting the given values into the formula:

[tex]\sf \implies Volume =\dfrac{1}{3} \pi (6)^2(9)=108 \pi \: \:cubic\:units[/tex]

If the radius is quadrupled:

  • radius = 6 × 4 = 24
  • height = 9

Substituting the new given values into the formula:

[tex]\sf \implies Volume =\dfrac{1}{3} \pi (24)^2(9)=1728 \pi \: \:cubic\:units[/tex]

To find the number of times greater the volume of the large cone is than the volume of the smaller cone, divide their volumes:

[tex]\sf \implies \dfrac{V_{large}}{V_{small}}=\dfrac{1728\pi}{108\pi}=16[/tex]

So the volume of the larger cone is 16 times greater than the volume of the smaller cone.

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