Question

the diameter of the lower and upper hands of a bucket is in the form of frustum of a cone are 10 cm and 30 cm respectively if its height is 24 cm find the area of the metal sheet used to make the bucket

Answer 1

GIVEN :
Diameter of upper end of bucket =30cm
Radius of the upper end of the frustum of cone( r1) = 15cm
Diameter of lower end of bucket = 10 cm
radius of the lower end of the frustum of cone( r2) = 5 cm
H of the frustum of Cone = 24 cm
Slant height of bucket ( L)= √(h² + (r1- r2)²
L =√24² + (15 - 5)² = √576 + 10²
L =√(576+(100)= √676 = 26cm
L = 26 cm
Area of metal sheet require to make bucket = Curved surface Area of frustum + area of base
= π(r1 + r2)L + πr2²
= 3.14(15 + 5) × 26 + π(5)²
= 3.14 × 20 × 26 + 25 × 3.14
= 3.14 (520+ 25)
= 545 × 3.14
= 1711.3 cm²
Hence, the Area of metal sheet used to make the bucket is 1711.3 cm².

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Answer 2

Given:

Diameter of the upper end = 30 cm

Diameter of the lower end = 10 cm

Height = 24 cm

To find:

Area

Solution:

Radius of the upper end of the cone = 15 cm

Radius of the lower end of the cone = 5 cm

In order to find the area,

The slant height should be calculated.

Slant height = √ ( h^2 + ( Radius of the upper end of the cone - Radius of the lower end of the cone )^2

√24^2 + ( 15 - 5 )^2  

√576 + 10^2

Slant height = 26 cm

Area of the metal  = Curved surface Area + Area of base of the cone.

Curved surface area = π ( Radius of the upper end of the cone - Radius of the lower end of the cone ) * Slant height

Area of the base = πr2^2

Hence,

π ( Radius of the upper end of the cone - Radius of the lower end of the cone ) * Slant height +  πr2^2

3.14 ( 15 + 5 ) × 26 + π( 5 )^2  

3.14 × 20 × 26 + 25 × 3.14

545 × 3.14

Area = 1711.3 sq.cm

Hence, the area of metal sheet used to make the bucket is 1711.3 sq.cm

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