Question

A solid cone of height 12cm and base radius 6cm has the top 4cm removed. find the whole surface of the remaining frustum of the cone?

Answer 1




Here there are two similar triangle AFD and AGC
So AF = 4cm And AG = 12
Hence AF/AG = FD/GC = 1/3
As GC = 6, So FD = 2cm
Hence the top and base radius of frustum are known and height is also known.
Hence Total surface area of the frustum :
TSA = πl(r1 +r2) +πr12 +πr22πl(r1 +r2) +πr12 +πr22
Here l ==√h2+(r1−r2)2=h2+(r1-r2)2
So l = √82+(6−2)2=4√5So TSA =π×4√5(6 +2) +π62+π22=π(32√5+36+4)=350.59cm2

Answer 2

In Δ ABE and Δ ACD,
BE || CD
∠AEB= ∠ADC [each 90°]
∠ABE= ∠ACD [ corresponding angles]
Δ ABE ∼ Δ ACD
AB/AC = BE/CD
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
= 22/7(111.552)= 2,454.144 /7 = 350.59 cm²
Hence, the Total surface area of the remaining cone = 350.59 cm²