The internal and external diameter of a hollow hemispherical vessel are 24cm and 25cm respectively
Answers
Answer:
given,
internal diameter=24cm
internal radius=24/2=12cm
external diameter=25 cm
external radius =25/2=12.5cm
surface area of internal bowl=2πr²
=2(22/7)(12²)
=2×22×12×12/7
=905.14cm²
surface area of external bowl=2πR²
=2(22/7)(12.5²)
=2×22×12.5×12.5/7
=982.14cm²
surface area of ring=π(R²-r²)
=22/7(12.5²-12²)
=22/7(12.25)
=22×12.25/7
=38.5cm²
cost of internal bowl=905.14×0.05=Rs.45.25
cost of external bowl=982.14×0.05=Rs.49.1
cost of ring=38.5×0.05=Rs.1.925
cost of bowl=45.25+49.1+1.9=96.25 rupees approx
I THINK IT IS CORRECT ANSWER
given,
given,internal diameter=24cm
given,internal diameter=24cminternal radius=24/2=12cm
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cm
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cm
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25cost of external bowl=982.14×0.05=Rs.49.1
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25cost of external bowl=982.14×0.05=Rs.49.1cost of ring=38.5×0.05=Rs.1.925
given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25cost of external bowl=982.14×0.05=Rs.49.1cost of ring=38.5×0.05=Rs.1.925cost of bowl=45.25+49.1+1.9=96.25 rupees approx...........!!!!!!!!!