Q. 75.0( 2 Votes )

# A cylindrical buc

Answer :

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 = πr^{2}h

Volume of cone = πr^{2}h

Let the Radius of conical heap be x

Volume of bucket = πr^{2}h

= π × (21 cm)^{2} × 44 cm

= 19404π cm^{3}

Volume of conical heaps = πr^{2}h

= π × x^{2} × 33 cm

= 11π × x^{2} 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 × x^{2} cm = 19404π cm^{3}

x^{2} =

x^{2} = 1764 cm^{2}

x = √ (1764 cm^{2})

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 cm^{2} + 1764 cm^{2}

(Slant height)^{2} = 2853 cm^{2}

Slant height = √(2853 cm^{2})

= 53.41 cm

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