All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. The total area of all its surfaces (in `cm^(2)`) is

Options

A. 8
B. 10
C. 12
D. 16

Answer

To find the total surface area of a regular pyramid with a square base where all five faces have the same area, we can follow these steps:

### Step 1: Understand the Pyramid Structure
A regular pyramid with a square base consists of:
- 1 square base
- 4 triangular faces

### Step 2: Define Variables
Let the side length of the square base be \( s \). The area of the square base is given by:
\[ \text{Area of base} = s^2 \]

### Step 3: Calculate the Area of Triangular Faces
Each triangular face has a base equal to the side of the square, which is \( s \), and a height that we need to determine. The height of the pyramid is given as \( h = 3 \, \text{cm} \).

### Step 4: Relate the Areas of Faces
Since all five faces have the same area, we can denote the area of the square base as \( A \) and the area of each triangular face as \( A_t \). Therefore, we have:
\[ s^2 = A_t \]

The area of one triangular face can be expressed as:
\[ A_t = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times s \times l \]
where \( l \) is the slant height of the triangular face.

### Step 5: Find the Slant Height
To find the slant height \( l \), we can use the Pythagorean theorem in the right triangle formed by the height of the pyramid, half the base of the square, and the slant height:
\[ l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2} \]
Substituting \( h = 3 \):
\[ l = \sqrt{3^2 + \left(\frac{s}{2}\right)^2} = \sqrt{9 + \frac{s^2}{4}} \]

### Step 6: Set Areas Equal
Since \( s^2 = A_t \), we can set the area of the triangular face equal to the area of the base:
\[ s^2 = \frac{1}{2} \times s \times l \]
Substituting for \( l \):
\[ s^2 = \frac{1}{2} \times s \times \sqrt{9 + \frac{s^2}{4}} \]
Dividing both sides by \( s \) (assuming \( s \neq 0 \)):
\[ s = \frac{1}{2} \sqrt{9 + \frac{s^2}{4}} \]

### Step 7: Solve for \( s \)
Squaring both sides:
\[ s^2 = \frac{1}{4} \left(9 + \frac{s^2}{4}\right) \]
Multiplying through by 4 to eliminate the fraction:
\[ 4s^2 = 9 + \frac{s^2}{4} \]
Rearranging gives:
\[ 16s^2 - s^2 = 36 \]
\[ 15s^2 = 36 \]
\[ s^2 = \frac{36}{15} = \frac{12}{5} \]

### Step 8: Calculate Total Surface Area
Now we can find the total surface area \( A_{total} \):
\[ A_{total} = \text{Area of base} + 4 \times \text{Area of triangular faces} \]
\[ A_{total} = s^2 + 4 \times s^2 = 5s^2 \]
Substituting \( s^2 = \frac{12}{5} \):
\[ A_{total} = 5 \times \frac{12}{5} = 12 \, \text{cm}^2 \]

### Final Answer
The total area of all its surfaces is \( 12 \, \text{cm}^2 \).

---