A bucket open at the top and bottom radii of circular ends as 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 21 cm. Also find the area of the tin sheet required for making the bucket.

Volume of bucket = 61544 cm³

Area of tin required = 5463.6 cm²

Step-by-step explanation:

A cone will be formed as shown in the attached photo.

The formula for determining the volume of a cone is expressed as

Volume = 1/3πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Solving the right angle triangle,

40/20 = H/(H - 21)

2 = H/(H - 21)

H = 2(H - 21)

H = 2H - 42

2H - H = 42

H = 42

For the smaller cone,

Height = 42 - 21 = 21

Volume = 1/3 × π × 20² × 21 = 2800π

For the larger cone,

Volume = 1/3 × π × 40² × 42 = 22400π

Volume of bucket = Volume of larger cone - volume of smaller cone

Volume of bucket = 22400π - 2800π = 19600π = 19600 × 3.14

Volume of bucket = 61544 cm³

Since the bucket is open, we would apply the formula for determining the curved surface area of a cone which is expressed as

Curved surface area = πrl

Where l represents the slant height

From the triangle

L² = 40² + 42² = 1600 + 1764 = 3364

L = √3364 = 58

It means that the slant height of the smaller cone is 58/2 = 29 cm

Curved surface area of larger cone is

π × 40 × 58 = 2320π cm²

Curved surface area of smaller cone is

π × 20 × 29 = 580π cm²

the area of the tin sheet required for making the bucket is

2320π - 580π = 1740π

= 1740 × 3.14 = 5463.6 cm²

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation: