Question

a solid cone of height 12 cm and base radius 6 cm has the top 4 cm removed. Find the whole surface area of remaining figure.

Answer 1

A cone of radius CD and height AD as shown in figure is cut from the top of 4cm at point E.
now, AD = 12cm , CD = 6cm , AE = 4cm

here it is clear that ∆ABE ~ ∆ACD
so, AE/AD = BE/CD
4cm/12cm = BE/6cm
BE = 2 cm { it is the radius of small circular part , Let r }

now, whole surface area of remaining part of cone = lateral surface area of frustum + area of above circular part + area of below circular part
= πl(R + r) + πr² + π R²
where, l = √{h² + (R - r)²}
here, h = 12cm - 4cm = 8 cm
so, l = √{8² + (6-2)²} = √{64 + 16} = 4√5cm

now, whole surface area = π × 4√5 × (6 +2) + π × (6)² + π × (2)²
= 32√5π + 36π + 4π cm²
= (32√5 + 40)π cm²
= (32 × 2.236 + 40) × 22/7 cm²
= 350.59 cm²

Answer 2

Given:
AB= 4 cm, AC= 12 cm, CD = 6 cm
In Δ ABE and Δ ACD,
BE || CD
∠AEB= ∠ADC [each 90°]
∠ABE= ∠ACD [ corresponding angles]
Δ ABE ∼ Δ ACD [By AA Similarity]
AB/AC = BE/CD
[Corresponding sides of a similar triangles are proportional]
4/12 = BE /6
1/3 = BE/6
1 = BE/2
BE = 2
In ∆ACD
AD² = AC² + CD²
AD² = 12² + 6²
AD² = 144 + 36
AD²= 180
AD = √180 = √36×5 = 6√5 =6×2.236
Slant height of bigger cone AD = 13.416 cm
Total surface area of bigger cone with radius 6 cm = πr(l + r)
= π×6(6 + 13.416)
= π×6×19.416= π(116.496) cm²
Slant height of smaller cone (l) =√h²+r² √(AB²+BE² )
l = √(4²+ 2²)
l = √(16 + 4)
l = √20 =√4×5=2×2.236
l = 4.472 cm
Curved surface area of smaller cone of height 4 cm and radius 2 cm = πrl
= π×2×4.472 = π(8.944) cm
Total surface area of the remaining cone = Total surface area of bigger cone - curved surface area of smaller cone + area of base of smaller cone
= π(116.496) - π(8.944) + πr²
= π(116.496) - π(8.944) + π(2)²
= π(116.496 - 8.944 +4)
= π(107.552 +4) = π (111.552) cm
= 22/7(111.552)= 2,454.144 /7 = 350.59 cm²
Hence, the Total surface area of the remaining cone = 350.59 cm²