High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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]