Question

from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diameter is hollowed out find the total surface area of the remaining solid to the nearest CM square​

Answer 1

Given :- from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diameter is hollowed out find the total surface area of the remaining solid to the nearest CM square ?

Solution :-

from given data we have :-

• Height of cylinder = Height of conical cavity = h = 3. 6 cm.
• Radius of cylinder = Radius of conical cavity = r = Diameter / 2 = 2.1/2 = r = 1.05 cm.

we know that,

• Slant height of cone(l) = √[(radius)² + (height)²]
• CSA of cylinder = 2 * π * r * h .
• CSA of cone = π * r * l
• Base Area = π * r² .

Putting values we get,

→ Slant height of conical cavity = √[(radius)² + (height)²]

→ l = √[(1.05)² + (3.6)²]

→ l = √[1.1025 + 12.96]

→ l = √(14.0625)

→ l = 3.75 cm.

Now, we know that,

• Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base = 2πrh + πrl + πr²

Putting all values we get :-

→ 2πrh + πrl + πr²

→ πr[2h + l + r]

→ (22/7) * 1.05 * [2 * 3.6 + 3.75 + 1.05]

→ 3.3 * [7.2 + 4.8]

→ 3.3 * 12

→ 39.6 cm² ≈ 40 cm².

Hence, The total surface area of the remaining solid to the nearest cm² is 40 cm².

Learn more :-

A hemisphere of radius 21 cm is completely filled with milk. There is a hole in

the bottom whose radius is 0.1 cm. If ra...

https://brainly.in/question/25349591?utm_source=android&utm_medium=share&utm_campaign=question