Find the surface area of a rectangular pyramid with these measurements: l=15 m, w=12 m, and h=10 m. Express your answer as a decimal rounded to the nearest hundredth.

1 answer

To find the surface area of a rectangular pyramid, we need to consider both the area of the base and the areas of the triangular faces. The rectangular pyramid has a rectangular base, and in this case:

  • \( l = 15 \) meters (length of the base)
  • \( w = 12 \) meters (width of the base)
  • \( h = 10 \) meters (height of the pyramid)

Step 1: Calculate the area of the base.

The area of the rectangular base is given by:

\[ \text{Area of the base} = l \times w = 15 , \text{m} \times 12 , \text{m} = 180 , \text{m}^2 \]

Step 2: Calculate the area of the triangular faces.

The pyramid has four triangular faces. We have two pairs of triangles: one pair with a base equal to the length \( l \) and the other pair with a base equal to the width \( w \).

Face triangles with base \( l = 15 \) m:

To find the height of the triangular face, we use the Pythagorean theorem:

  • The height \( h_t \) of the triangle from the apex of the pyramid to the base can be found by first calculating the distance from the center of the base to the midpoint of the base of the triangle:

\[ \text{Half of } l = \frac{l}{2} = \frac{15}{2} = 7.5 , \text{m} \]

Using Pythagorean theorem: \[ h_t = \sqrt{h^2 + \left(\frac{l}{2}\right)^2} = \sqrt{10^2 + 7.5^2} = \sqrt{100 + 56.25} = \sqrt{156.25} = 12.5 , \text{m} \]

Now calculate the area of one triangular face of base \( l \):

\[ \text{Area of one triangular face} = \frac{1}{2} \times b \times h_t = \frac{1}{2} \times 15 \times 12.5 = \frac{1}{2} \times 187.5 = 93.75 , \text{m}^2 \]

Thus, the area for the pair of triangular faces with base \( l \):

\[ \text{Total area for these two triangular faces} = 2 \times 93.75 = 187.5 , \text{m}^2 \]

Face triangles with base \( w = 12 \) m:

Now, we do the same for the triangular faces with base \( w \):

Find the height to calculate the area:

\[ \text{Half of } w = \frac{w}{2} = \frac{12}{2} = 6 , \text{m} \]

Using Pythagorean theorem: \[ h_t = \sqrt{h^2 + \left(\frac{w}{2}\right)^2} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.62 , \text{m} \]

Now calculate the area of one triangular face of base \( w \):

\[ \text{Area of one triangular face} = \frac{1}{2} \times b \times h_t = \frac{1}{2} \times 12 \times 11.62 \approx \frac{1}{2} \times 139.44 = 69.72 , \text{m}^2 \]

Thus, the area for the pair of triangular faces with base \( w \):

\[ \text{Total area for these two triangular faces} = 2 \times 69.72 \approx 139.44 , \text{m}^2 \]

Step 3: Calculate total surface area.

Now we can sum the areas:

\[ \text{Total Surface Area} = \text{Base area} + \text{Area of triangular faces} = 180 + (187.5 + 139.44) \] \[ = 180 + 326.94 = 506.94 , \text{m}^2 \]

Thus, the surface area of the rectangular pyramid is:

\[ \boxed{506.94} , \text{m}^2 \]