from a solid cylinder whose height is 3.6 cm and diameter 2.1 cm, a conical same height and same diameter is hollowed out. Find the total surface area of the
solid to the nearest cm^2
Answers
Answer: 40 cm²
Step-by-step explanation:
Height of the solid cylinder, h = 3.6 cm
The diameter of the solid cylinder, d = 2.1 cm
∴ The radius of the solid cylinder, r = 2.1/2 = 1.05 cm
Since the height and diameter of the conical part that is hollowed out from the solid cylinder is given to be same.
∴ The slant height of the cone,
l = √[r² + h²] = √[(1.05)² + (3.6)²] = √[14.0625] = 3.75 cm
Since the conical part is cut out from the solid cylinder, therefore,
The Total Surface Area of the remaining solid
= (Curved surface area of the cylinder) + (Curved surface area of the cone) + (Area of the cylindrical base)
= (2πrh) + (πrl) + (πr²)
= [2*(22/7)*1.05*3.6] + [(22/7)*1.05*3.75] + [(22/7)* (1.05)²]
= 23.76 + 12.375 + 3.465
= 39.6
≈ 40 cm²
Thus, the total surface area of the solid is 40 cm².
I hope that It would be helpful for you.
