Question

Derive all Surface Area and Volume Formulae of Frustum of a Cone from that of a Cone mathematically.


Answer with detailed explanations and steps. NO SPAM!

NCERT [OPTIONAL] | CLASS 10 | MATH

Answer 1

Heya *GovindKrishnan* !! ^_^

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Refer to the attachment for the figure of frustum of the cone.
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Let h be the height, l the slant height and r1 and r2 the radii of the circular bases of the frustum ABDC shown in the figure such that r1 > r2. Clearly, this frustum can be reviewed as the difference of the two right circular cones PAB and PCD.

Let the height of the cone PAB be h1 and its slant height be l1 i.e., PO = h1 and PA = PB = l1

Therefore, PC = PA–AC = l1–l
and PQ = PO–OQ = h1–h

Clearly, right ∆s POA and PQC are similar.
Therefore, PO/PQ = OA/QC = PA/PC

=> h1/(h1-h) = r1/r2 = l1/(l1-l)
=> (h1-h)/h1 = r2/r1 = (l1-l)/l1
=> 1 - (h/h1) = r2/r1 =1 - (l/l1)
=> h/h1 = 1 - (r2/r1) and l/l1 = 1 - (r2/r1)
=> h/h1 = (r1-r2)/r1 and l/l1 = (r1-r2)/r1
=> h1 = hr1/(r1-r2) and l1 = lr1/(r1-r2) ...(i)

Now, height of the cone PAB = h1-h = hr1/(r1-r2) – h = hr2/(r1-r2) ...(ii)

Slant height of the cone PAB = l1-l = lr1/(l1-l2) – l = r2/(l1-l2) ...(iii)

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⭐Let V be the volume of the frustum of the cone. Then,
V = Volume of cone PAB - Volume of cone PCD
=> V = 1/3×πr1²h1 – 1/3×πr2²(h1-h)
=> V = π/3 [h1r1² – (h1-h)r2²]
=> V = π/3 { [ hr1³/(r1-r2) ] – [ hr2³/(r1-r2) ] }
=> V = π/3 { h/(r1-r2) × (r1³–r2³) }
=> V = π/3 { h/(r1-r2) × (r1–r2) × (r1²+r1r2+r2²) }
=> V = π/3 × h ×( r1²+r1r2+r2²)

Thus, the volume V of the frustum of the cone is given by V =1/3 × π ×( r1²+r1r2+r2²)h.

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⭐Curved surface area if the frustum of the cone = C.S.A. of cone PAB – C.S.A. of cone PCD
=> πr1l1 – πr2(l1–l)
=> πr1 × lr1/(r1–r2) – πr1 × (lr2/(r1–r2)
[Using (i) and (iii) ]
=> π [ (r1²–r2²)/(r1–r2) ] l
=> π (r1 + r2) l

Thus, Curved Surface Area of the frustum = π(r1+r2) l

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⭐Total surface area of the frustum = C.S.A. + Surface area of circular bases
=> π (r1 + r2) l + πr1² + πr2²
=> π { (r1 + r2) l + r1² + r2² }

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Hope my ans.'s satisfactory. ☺

Answer 2

13 πH (r2 + r′r + r′2)

This gives the required