Question

A semicircular sheet of metal of diameter 56 cm is bent to form an open conical Cup. Find the capacity of the cup. ​

Answer 1

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Answer 2

Answer:

The volume of the conical cup is 4979.34 cm³

Step-by-step explanation:

Given the diameter of the semicircular sheet = 56 cm

We know that diameter = 2 * radius.

Therefore, the radius of the semicircular sheet = 56/2 = 28 cm

When the semicircular sheet is bent to form a cup,

the radius becomes the slant height (L) = 28 cm

Let the radius of the cone be R

We know that the circumference of the semicircle excluding the straight line = Circumference of the cone.

=> πr = 2πR

=> r = 2R

=> 28 = 2 * R

=> R = 14

Let the height of the cone be H

We know that L² = H² + R²

=> 28² = H² + 14²

=> 784 = H² + 196

=> H² = 588

=> H = 24.248

We have to find the capacity of the cup

= Volume of the conical cup

= $\frac{1}{3}$ * π R² H

= $\frac{1}{3}$ * $\frac{22}{7}$ * 14² * 24.248

= 4979.34 cm³

Therefore, the volume of the conical cup is 4979.34 cm³