A cylindrical buc

Given.

Height of bucket is 44 cm

Radius of bucket is 21 cm

Height of conical heap is 33 cm

Formula used/Theory.

Volume of cylinder = πr²h

Volume of cone = πr²h

Let the Radius of conical heap be x

Volume of bucket = πr²h

= π × (21 cm)² × 44 cm

= 19404π cm³

Volume of conical heaps = πr²h

= π × x² × 33 cm

= 11π × x² cm

**Note we will not put value of π as it will be divided in next step

As we put bucket of sand on ground it will form a conical heap volume of conical heaps will be equal to volume of bucket

∴ equating both we will get the Radius of conical heap

Volume of bucket = Volume of conical heaps

11π × x² cm = 19404π cm³

x² =

x² = 1764 cm²

x = √ (1764 cm²)

x = 42 cm

∴ Radius of conical heap is 42 cm

In cone;

As the radius , height and slant height makes Right angled triangle where hypotenuse is slant height

Then by Pythagoras theorem

(Slant height)² = (height)² + (radius)²

(Slant height)² = (33 cm)² + (42 cm)²

(Slant height)² = 1089 cm² + 1764 cm²

(Slant height)² = 2853 cm²

Slant height = √(2853 cm²)

= 53.41 cm