Question

13. The ratio of the heights of two right cones having the same radius of the base is 6:5. Find the ratio
of their (a) volumes and (b) curved surface areas, when each radius of the base is half the height of
the shorter cone.
​

Answer 1

Given :-

◉ Ratio of heights of two right cones having the same radius of the base is 6:5

To Find :-

◉ Ratio of thier:

1. Volume
2. Curved Surface Area

Solution :-

Let the common factor of the ratio be k.

Then, Height of both the cones would be 6k & 5k respectively.

Also, It is given that the radius of base of both the cones is same and it is also equal to half of the height of the shorter cone ( i.e., 5k )

Now, We know

➞ Volume of Cone = 1/3 πr²h

Where, r = radius & h = height

⇒ Volume of Cone₁ = 1/3 π(5k/2)² 6k

⇒ V₁ = π/3 ( 25k²/4 × 6k)

⇒ V₁ = π (25k³ / 2)

⇒ V₁ = 1/2 (25π k³) ...(1)

Similarly,

⇒ Volume of Cone₂ = 1/3 π(5k/2)² 5k

⇒ V₂ = π/3 (25k² / 4 × 5k )

⇒ V₂ = 1/12 (125π k³) ...(2)

So,

⇒ Ratio of Volume = (1) / (2)

⇒ V₁ / V₂ = { 1/2 (25π k³) } / { 1/12 (125π k³) }

⇒ V₁ / V₂ = { 6(25π k³) } / (125π k³)

⇒ V₁ / V₂ = 6 / 5

Again,

➞ Curved Surface Area = πrl

Where, r = radius, l = √(h² + r²) , h = height

⇒ CSA of cone₁ = π(5k/2) √{ (6k)² + (5k/2)²}

⇒ CSA₁ = 5πk / 2 { √169k² / 2 }

⇒ CSA₁ = 5πk/2 × 13k/2

⇒ CSA₁ = 65π k² / 4 ...(3)

Similarly,

⇒ CSA of cone₂ = π(5k/2) √{ (5k)² + (5k/2)² }

⇒ CSA₂ = 5πk/2 { √125k² / 2 }

⇒ CSA₂ = 5πk/2 × 5k√5 / 2

⇒ CSA₂ = 25πk²√5 / 4 ...(4)

Hence,

⇒ Ratio of CSA = (3) / (4)

⇒ CSA₁ / CSA₂ = 1/4(65πk²) / 1/4(25πk²√5)

⇒ CSA₁ / CSA₂ = 65 / 25√5

⇒ CSA₁ / CSA₂ = 13 / 5√5

Therefore,

• Ratio of Volume = 6 : 5
• Ratio of CSA = 13 : 5√5