Math, asked by joelbiju1954, 11 months ago

A solid cylinder of silver 9 cm high and 4 cm in diameter is melted and recasted into a right circular cone of diameter 6 cm. Find height and total surface area of the come. π=3.14

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Answered by shreyabiswas133
7

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Answered by VineetaGara
3

Given,

Height of a solid cylinder = 9 cm

Diameter of the solid cylinder = 4 cm

Diameter of the recasted right circular cone = 6 cm

To find,

(a) The height of the cone

(b) Total surface area of the cone

Solution,

We can simply solve this mathematical problem using the following process:

Let us assume that the height of the cone formed is x cm.

As per mensuration;

Volume of a cylinder = π.(radius)^2.(height)

Volume of a right circular cone= 1/3×π.(radius)^2.(height)

Slant height of a cone = l = √((height)^2 + (radius)^2

Total surface area of a cone = π(radius)(slant height+ radius)

{Statement-1}

According to the question;

Diameter of the solid cylinder = 4 cm

=> Radius of the solid cylinder = (diameter)/2 = (4 cm)/2 = 2 cm

And, the diameter of the recasted right circular cone = 6 cm

=> Radius of the recasted right circular cone = diameter/2 = (6 cm)/2 = 3 cm

Now, according to the question;

Volume of the initial cylinder = Volume of the right circular cone formed

=> π.(radius of cylinder)^2.(height of the cylinder)

= 1/3×π.(radius of the cone)^2.(height of the cone)

{according to statement-1}

=> (2 cm)^2.(9 cm) = 1/3×(3 cm)^2.(x cm)

=> 4 cm^2 × 9 cm = 1/3 × 9 cm^2 × x cm

=> 36 cm^3 = 3 cm^3 × x cm

=> x = (36/3) cm = 12 cm

=> x = 12 cm

=> height of the cone = 12 cm

Now, the slant height of the cone formed

= √{(height)^2 + (radius)^2}

{according to statement-1}

= √((12 cm)^2 + (3 cm)^2

= √(144 cm^2 + 9 cm^2)

= √(153 cm^2) = 12.37 cm

Now, according to statement-1;

Total surface area of the cone

= π(radius)(slant height+ radius)

= 22/7 × 3 cm × (12.37 cm + 3 cm)

= 22/7 × 3 cm × (15.37 cm)

= 144.9 cm^2

Hence, the height of the cone is 12 cm and the total surface area of the cone is equal to 144.9 cm^2, respectively.

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