The radii of the ends of a frustum of a cone 45 cm high are 28cm and 7cm. Find its volume, the curved surface area. [Take π=22/7]

Frustum = difference of two right circular cones OAB and OCD

Let the height of the cone OAB be h₁ and its slant height l₁

i.e. OA = OB = l₁ and OP = h₁.

Let h₂ be the height of cone OCD and l₂ its slant height

i.e. OC = OD = l₂ and OQ = h₂

We have , r₁ = 28cm and r₂ = 7cm

and height of frustum (h) = 45cm

Also,

h₁ = 45 + h₂ …(i)

Now, we first need to determine the h₁ and h₂

∵ ΔOPB and OQD are similar, we have

⇒ h₁ = 4h₂ …(ii)

From (i) and (ii), we get

4h₂ = 45 + h₂

⇒ 3h₂ = 45

⇒ h₂ = 15cm

Putting the value of h₂ in eq. (ii), we get

h₁ = 4×15 = 60cm

So, h₁ = 60cm and h₂ = 15cm

Now, the volume of frustum = Vol. of cone OAB – Vol. of cone OCD

= 5 × 22 × 441

= 48510 cm³

Now, we first have to find the slant height l₁ and l₂

l₁ = √{(28)²+(60)²}

⇒ l₁ = √{(4×7)² + (4×15)²

⇒ l₁ = 4√{(7)² + (15)²}

⇒ l₁ = 4√49 + 225

⇒ l₁ = 4√(274)

⇒ l₁ = 4 × 16.55

⇒ l₁ = 66.20cm

and l₂ = √{(7)² + (15)²}

⇒ l₂ = √49 + 225

⇒ l₂ = √(274)

⇒ l₂ = 16.55 cm

So, CSA of frustum = CSA of cone OAB – CSA of cone OCD

= πr₁l₁ – πr₂l₂

= 22 [4×66.20 – 16.55]

= 22 × 248.25

= 5461.5cm²