A man donates tell aluminium buckets to an orphanage a bucket made of aluminium is of height 20 cm and its upper and lower ends of radius 36 in 21 cm respectively find the cost of repairing 10 buckets with the cost of aluminium sheet is 42 per 100 cm square
Answers
Answered by
4
Step-by-step explanation:
Slant height = 25cm
Area of bucket = CSA of frustum + area of base
= pi*(R+r)*I + pi*r^2
22/7(36+21)25 + 21/7*21*21
=22/7(56)25 + 66*21
= 22/7*1425 + 1386
= 31350/7 + 1386
= 31350 + 9702/7
= 41052/7 = 5864.5
Cost of repairing one bucket= 5864.5/100*42
= Rs. 2463.09
Cost of repairing 10 buckets= 2463.09*10
= Rs. 24630.9
Answered by
2
height=20 cm
radius 1 = 36 cm
radius 2 = 21cm
slant height =√h^2+(r1^2-r^2)
√400+(36+21)(36-21)
√400+(51)(15)
√400+765
√1165
=31.13
tsa=π(r1+r2)l+π(r1^2+r2^2)
= π{(51*31.13)+(1296+441)}
= π{3324.63}
= 3.14*3324.63
= 10439.3382 cm^2
now cost per 100 cm^2 =42
total cost =(10439.3382*42)÷100*10
= ₹4384.52205*10
= ₹43845.2205
radius 1 = 36 cm
radius 2 = 21cm
slant height =√h^2+(r1^2-r^2)
√400+(36+21)(36-21)
√400+(51)(15)
√400+765
√1165
=31.13
tsa=π(r1+r2)l+π(r1^2+r2^2)
= π{(51*31.13)+(1296+441)}
= π{3324.63}
= 3.14*3324.63
= 10439.3382 cm^2
now cost per 100 cm^2 =42
total cost =(10439.3382*42)÷100*10
= ₹4384.52205*10
= ₹43845.2205
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