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Given a square-based pyramid with a base side length of 2 meters and a side ridge (slant height) of 32 meters, we need to find the height of the pyramid and its volume.
### Step 1: Find the Height of the Pyramid
The slant height (\( h_s \)) forms the hypotenuse of a right triangle where one leg is half the base length (\( b/2 \)) and the other leg is the height of the pyramid (\( h \)).
Using the Pythagorean theorem:
\[
h_s^2 = h^2 + \left(\frac{b}{2}\right)^2
\]
Given:
- \( h_s = 32 \) meters
- \( b = 2 \) meters
- \( \frac{b}{2} = 1 \) meter
Substituting the values:
\[
32^2 = h^2 + 1^2
\]
\[
1024 = h^2 + 1
\]
\[
h^2 = 1023
\]
\[
h = \sqrt{1023}
\]
Approximating the square root:
\[
h \approx 31.984 \text{ meters}
\]
Converting to millimeters (1 meter = 1000 millimeters):
\[
h \approx 31984 \text{ millimeters}
\]
Rounded to the nearest millimeter:
\[
h \approx 31984 \text{ mm}
\]
### Step 2: Calculate the Volume of the Pyramid
The volume (V) of a pyramid is given by:
\[
V = \frac{1}{3} b^2 h
\]
Given:
- \( b = 2 \) meters
- \( h \approx 31.984 \) meters
Substituting the values:
\[
V = \frac{1}{3} (2)^2 (31.984)
\]
\[
V = \frac{1}{3} \times 4 \times 31.984
\]
\[
V = \frac{1}{3} \times 127.936
\]
\[
V \approx 42.6453 \text{ cubic meters}
\]
Rounded to three decimal places:
\[
V \approx 42.645 \text{ m}^3
\]
### Final Results
- **Height**: \( 31984 \) mm
- **Volume**: \( 42.645 \) m\(^3\)
