Question

the top of a solid cone of height 12 cm and base radius 6 cm, a cone of

height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid

Answer 1

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

Answer 2

Step-by-step explanation:

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