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 :

To solve the problem, we need to calculate the lateral area and the surface area of a right cylinder given specific information:

1. The cylinder has a circumference of [tex]\( 16 \pi \)[/tex] cm.
2. The height of the cylinder is half the radius.

Let's follow these steps:

### Step 1: Find the Radius
The formula for the circumference [tex]\( C \)[/tex] of a cylinder is:
[tex]\[ C = 2 \pi r \][/tex]

Given:
[tex]\[ C = 16 \pi \][/tex]

We can solve for the radius [tex]\( r \)[/tex]:
[tex]\[ 16 \pi = 2 \pi r \][/tex]
[tex]\[ r = \frac{16 \pi}{2 \pi} \][/tex]
[tex]\[ r = 8 \text{ cm} \][/tex]

### Step 2: Find the Height
We know the height is half the radius.

[tex]\[ h = \frac{r}{2} \][/tex]
[tex]\[ h = \frac{8 \text{ cm}}{2} \][/tex]
[tex]\[ h = 4 \text{ cm} \][/tex]

### Step 3: Calculate the Lateral Area
The formula for the lateral area [tex]\( L \)[/tex] of a cylinder is:
[tex]\[ L = 2 \pi r h \][/tex]

Plugging in the values:
[tex]\[ L = 2 \pi (8 \text{ cm})(4 \text{ cm}) \][/tex]
[tex]\[ L = 64 \pi \text{ cm}^2 \][/tex]

Approximating the value (since [tex]\(\pi \approx 3.14\)[/tex]):
[tex]\[ L \approx 64 \times 3.14 \][/tex]
[tex]\[ L \approx 201.1 \text{ cm}^2 \][/tex]

### Step 4: Calculate the Surface Area
The formula for the total surface area [tex]\( S \)[/tex] of a cylinder is:
[tex]\[ S = L + 2 \pi r^2 \][/tex]

We already have the lateral area [tex]\( L = 64 \pi \text{ cm}^2 \)[/tex] and need to add the area of the two circular bases ([tex]\( 2 \pi r^2 \)[/tex]).

Calculating the area of the two circular bases:
[tex]\[ 2 \pi r^2 = 2 \pi (8 \text{ cm})^2 \][/tex]
[tex]\[ 2 \pi (64 \text{ cm}^2) \][/tex]
[tex]\[ 2 \pi \times 64 \text{ cm}^2 \][/tex]
[tex]\[ 128 \pi \text{ cm}^2 \][/tex]

Now, add this to the lateral area:
[tex]\[ S = 64 \pi + 128 \pi \][/tex]
[tex]\[ S = 192 \pi \text{ cm}^2 \][/tex]

Approximating the value:
[tex]\[ S \approx 192 \times 3.14 \][/tex]
[tex]\[ S \approx 603.2 \text{ cm}^2 \][/tex]

So, the lateral area and the surface area of the cylinder are:
[tex]\[ \text{L} \approx 201.1 \text{ cm}^2 \][/tex]
[tex]\[ \text{S} \approx 603.2 \text{ cm}^2 \][/tex]

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