Question

The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If it's volume be 1/125 of the volume of the original cone determine at what height above the base was the section made ​

Answer 1

$\huge\underline\mathfrak\blue{\:♡Solution\:♡}$

Given : Height of the cone = 20 cm

Let the small cone is cut off at a height 'h' from the top.

Let the radius of the big cone be 'R' and of small cone be 'r'

Let the volume of the big cone be V1 and of small cone be V2

Volume of the big cone = 1/3R²h

= 1/3πR²*20

= 20/3πR² cu cm

Volume of small cone = 1/3πr²h

⇒ V2 = 1/125th of the volume of big cone.

⇒ V2/V1 = 1/125

⇒ (1/3πr²h)/(20/3πR²) = 1/125

⇒ r²h/20R² = 1/125

⇒ r²/R² × h/20 = 1/125 ...(1)

⇒ From the figure. Δ ACD ~ Δ AOB  (By AA similarity criterion)

⇒ r/R = h/20

Putting this value of r/R = h/20 in equation (1), we get.

(h/20)² × (h/20) = 1/125

(h/20)³ = 1/125

(h/20)³ = (1/5)³

h/20 = 1/5

h = 20/5

h = 4 cm

So, the height above the base where the section is made is 20 - 4 = 16 cm

Answer.

Answer 2

Answer:

Let the small cone is cut off at a height 'h' from the top.

Let the radius of the big cone be 'R' and of small cone be 'r'

Let the volume of the big cone be V1 and of small cone be V2

Volume of the big cone = 1/3R²h

= 1/3πR²*20

= 20/3πR² cu cm

Volume of small cone = 1/3πr²h

⇒ V2 = 1/125th of the volume of big cone.

⇒ V2/V1 = 1/125

⇒ (1/3πr²h)/(20/3πR²) = 1/125

⇒ r²h/20R² = 1/125

⇒ r²/R² × h/20 = 1/125 ...(1)

⇒ From the figure. Δ ACD ~ Δ AOB (By AA similarity criterion)

⇒ r/R = h/20

Putting this value of r/R = h/20 in equation (1), we get.

(h/20)² × (h/20) = 1/125

(h/20)³ = 1/125

(h/20)³ = (1/5)³

h/20 = 1/5

h = 20/5

h = 4 cm

So, the height above the base where the section is made is 20 - 4 = 16 cm

Answer.