1. Surface Area of a Rectangular Pyramid (l = 8 cm, w = 4 cm, h = 2 cm)
The formula for the surface area \( SA \) of a rectangular pyramid is:
\[ SA = lw + \frac{1}{2} l s_1 + \frac{1}{2} w s_2 \]
Where:
- \( l \) = length
- \( w \) = width
- \( s_1 \) = slant height of the triangular face with base \( l \)
- \( s_2 \) = slant height of the triangular face with base \( w \)
First, we need to find the slant heights \( s_1 \) and \( s_2 \):
\[ s_1 = \sqrt{\left(\frac{l}{2}\right)^2 + h^2} = \sqrt{\left(\frac{8}{2}\right)^2 + 2^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.47 , \text{cm} \]
\[ s_2 = \sqrt{\left(\frac{w}{2}\right)^2 + h^2} = \sqrt{\left(\frac{4}{2}\right)^2 + 2^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83 , \text{cm} \]
Now plug in the values into the surface area formula:
\[ SA = (8)(4) + \frac{1}{2}(8)(4.47) + \frac{1}{2}(4)(2.83) \]
Calculating each part:
\[ SA = 32 + 17.88 + 5.66 = 55.54 , \text{cm}^2 \]
Answer: 55.54 cmΒ²
2. Surface Area of a Rectangular Pyramid (l = 13 cm, w = 11 cm, h = 8 cm)
Using the same formula, we first find the slant heights \( s_1 \) and \( s_2 \):
\[ s_1 = \sqrt{\left(\frac{l}{2}\right)^2 + h^2} = \sqrt{\left(\frac{13}{2}\right)^2 + 8^2} = \sqrt{6.5^2 + 8^2} = \sqrt{42.25 + 64} = \sqrt{106.25} \approx 10.31 , \text{cm} \]
\[ s_2 = \sqrt{\left(\frac{w}{2}\right)^2 + h^2} = \sqrt{\left(\frac{11}{2}\right)^2 + 8^2} = \sqrt{5.5^2 + 8^2} = \sqrt{30.25 + 64} = \sqrt{94.25} \approx 9.71 , \text{cm} \]
Now, substitute into the surface area formula:
\[ SA = (13)(11) + \frac{1}{2}(13)(10.31) + \frac{1}{2}(11)(9.71) \]
Calculating each part:
\[ SA = 143 + 66.015 + 53.505 = 262.52 , \text{cm}^2 \]
Answer: 262.52 cmΒ²
3. Surface Area of a Square Pyramid with Base Length 10 cm and Slant Height 20 cm
The formula for the surface area \( SA \) of a square pyramid is:
\[ SA = l^2 + 2 \left( \frac{1}{2} l s \right) \]
Where \( l \) is the length of the base, and \( s \) is the slant height.
Calculating:
\[ SA = 10^2 + 2 \left( \frac{1}{2} \times 10 \times 20 \right) = 100 + 10 \times 20 = 100 + 200 = 300 , \text{cm}^2 \]
Answer: 300 cmΒ²
4. Finding the Slant Height of a Square Pyramid Given Surface Area
Given the base length \( l = 40 \) in and \( SA = 2800 , in^2 \):
Using the formula for the surface area:
\[ SA = l^2 + 2 \left( \frac{1}{2} l s \right) \]
Rewriting it gives:
\[ 2800 = 40^2 + 2 \left( \frac{1}{2} \times 40 \times s \right) \] \[ 2800 = 1600 + 40s \] \[ 2800 - 1600 = 40s \] \[ 1200 = 40s \] \[ s = \frac{1200}{40} = 30 , \text{in.} \]
Answer: 30 in.
5. Formula for Finding the Surface Area of a Regular Triangular Pyramid
The correct formula for the surface area of a regular triangular pyramid is:
\[ SA = \frac{1}{2} ab + 3 \left( \frac{1}{2} l b \right) \]
Where:
- \( a \) = base length
- \( b \) = height of the base triangle
- \( l \) = slant height
Answer: \( SA = \frac{1}{2} ab + 3 \left( \frac{1}{2} lb \right) \)
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