High School

A right cylinder has a circumference of [tex]$16 \pi \, \text{cm}$[/tex]. Its height is half the radius. Identify the lateral area and the surface area of the cylinder.

A. [tex]L \approx 100.5 \, \text{cm}^2 ; S \approx 401.8 \, \text{cm}^2[/tex]

B. [tex]L \approx 201.1 \, \text{cm}^2 ; S \approx 401.8 \, \text{cm}^2[/tex]

C. [tex]L \approx 100.5 \, \text{cm}^2 ; S \approx 603.2 \, \text{cm}^2[/tex]

D. [tex]L \approx 201.1 \, \text{cm}^2 ; S \approx 603.2 \, \text{cm}^2[/tex]

Answer :

Let's solve the problem step-by-step.

1. Identify the Given Information:
- Circumference of the base of the right cylinder: [tex]\(16\pi \text{ cm}\)[/tex]
- Height of the cylinder is half the radius.

2. Find the Radius of the Cylinder:
- The formula for the circumference [tex]\(C\)[/tex] of a circle is [tex]\(C = 2\pi r\)[/tex], where [tex]\(r\)[/tex] is the radius.
- Given [tex]\(C = 16\pi\)[/tex]:
[tex]\[
2\pi r = 16\pi
\][/tex]
- Divide both sides by [tex]\(2\pi\)[/tex]:
[tex]\[
r = \frac{16\pi}{2\pi} = 8 \text{ cm}
\][/tex]

3. Find the Height of the Cylinder:
- It is given that the height [tex]\(h\)[/tex] is half the radius.
[tex]\[
h = \frac{r}{2} = \frac{8}{2} = 4 \text{ cm}
\][/tex]

4. Calculate the Lateral Area:
- The formula for the lateral area [tex]\(L\)[/tex] of a cylinder is [tex]\(L = 2\pi rh\)[/tex].
[tex]\[
L = 2\pi \times 8 \times 4 = 64\pi
\][/tex]
- Converting to a numerical value (using [tex]\(\pi \approx 3.14159\)[/tex]):
[tex]\[
L \approx 64 \times 3.14159 \approx 201.1 \text{ cm}^2
\][/tex]

5. Calculate the Surface Area:
- The formula for the surface area [tex]\(S\)[/tex] of a cylinder is [tex]\(S = 2\pi r (r + h)\)[/tex].
[tex]\[
S = 2\pi \times 8 \times (8 + 4) = 2\pi \times 8 \times 12 = 192\pi
\][/tex]
- Converting to a numerical value:
[tex]\[
S \approx 192 \times 3.14159 \approx 603.2 \text{ cm}^2
\][/tex]

Based on these calculations, the correct answers are:
- Lateral Area (L): [tex]\(\approx 201.1 \text{ cm}^2\)[/tex]
- Surface Area (S): [tex]\(\approx 603.2 \text{ cm}^2\)[/tex]

Thus, the correct option is:
[tex]\[ L \approx 201.1 \text{ cm}^2 ; S \approx 603.2 \text{ cm}^2 \][/tex]