A cone has height half of 16.8 cm while diameter of its base is 4.2 cm . It is melted and recast into a sphere. Find the surface area of the sphere.

Correct option is A

Given:

Height of the cone = half of 16.8 cm = 8.4 cm

Diameter of the cone's base = 4.2 cm, so the radius of the base = 2.1 cm

The cone is melted and recast into a sphere.

We need to find the surface area of the sphere.

Step 2: Volume of the Cone

The formula for the volume of a cone is:

​$$V_{\text{cone}} = \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}}$$​

Substitute the values into the formula:

​$$V_{\text{cone}} = \frac{1}{3} \pi (2.1)^2 (8.4)$$​

​$$V_{\text{cone}} = \frac{1}{3} \pi \times 4.41 \times 8.4$$​

​$$V_{\text{cone}} = \frac{1}{3} \pi \times 37.044$$​

​$$V_{\text{cone}} = 12.348 \pi \, \text{cm}^3$$​

Step 3: Volume of the Sphere

The volume of the sphere is equal to the volume of the cone.

The formula for the volume of a sphere is:

​$$V_{\text{sphere}} = \frac{4}{3} \pi r_{\text{sphere}}^3$$​

Equating the volumes of the cone and sphere:

​$$12.348 \pi = \frac{4}{3} \pi r_{\text{sphere}}^3$$​

Canceling out $$\pi$$​  from both sides:

​$$12.348 = \frac{4}{3} r_{\text{sphere}}^3$$​

Solve for  $$r_{ext{sphere}}^3$$​ :

​$$r_{\text{sphere}}^3 = \frac{12.348 \times 3}{4}$$​

​$$r_{\text{sphere}}^3 = 9.261$$​

Taking the cube root of both sides:

​$$r_{\text{sphere}} = \sqrt[3]{9.261} \approx 2.1 \, \text{cm}$$​

Step 4: Surface Area of the Sphere

The formula for the surface area of a sphere is:

​$$A_{\text{sphere}} = 4 \pi r_{\text{sphere}}^2$$​

Substitute the value of $$r_{ext{sphere}}$$​:

​$$A_{\text{sphere}} = 4 \pi (2.1)^2$$​

​$$A_{\text{sphere}} = 4 \pi \times 4.41$$​

​$$A_{\text{sphere}} = 17.64 \pi$$​

​$$A_{\text{sphere}} \approx 17.64 \times 3.1416 = 55.44 \, \text{cm}^2$$​

Final Answer:

​$$\boxed{55.44} \, \text{sq.cm}$$​