A right pyramid stands on a square base of a diagonal `10sqrt(2) cm`. If the height of the pyramid is 12 cm, the area (in `cm^(2)`) of its slant surface is

Options

A. 520
B. 420
C. 360
D. 260

Answer

To find the area of the slant surface of a right pyramid with a square base, we can follow these steps:

### Step 1: Find the side length of the square base
Given the diagonal of the square base is \(10\sqrt{2}\) cm. The relationship between the side length \(A\) of a square and its diagonal \(d\) is given by:
\[ d = A\sqrt{2} \]
Thus, we can rearrange this to find the side length:
\[ A = \frac{d}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \text{ cm} \]

### Step 2: Calculate the perimeter of the base
The perimeter \(P\) of a square is given by:
\[ P = 4 \times A \]
Substituting the value of \(A\):
\[ P = 4 \times 10 = 40 \text{ cm} \]

### Step 3: Find the slant height of the pyramid
To find the slant height \(L\), we can use the formula:
\[ L = \sqrt{\left(\frac{A}{2}\right)^2 + H^2} \]
where \(H\) is the height of the pyramid. We know \(H = 12\) cm and \(A = 10\) cm, so:
\[ L = \sqrt{\left(\frac{10}{2}\right)^2 + 12^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} \]

### Step 4: Calculate the area of the slant surface
The area \(A_s\) of the slant surface of the pyramid is given by:
\[ A_s = \frac{1}{2} \times P \times L \]
Substituting the values we found:
\[ A_s = \frac{1}{2} \times 40 \times 13 = 20 \times 13 = 260 \text{ cm}^2 \]

### Final Answer
The area of the slant surface of the pyramid is \(260 \text{ cm}^2\).

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