Minniiiii bottttt

Answer 1

Hello! It looks like you’re excited! How can I assist you today?

Answer 2

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

Answer 3

To find the surface area of a rectangular pyramid, we need to calculate the area of the base and the area of the triangular sides (lateral surface area).

1. Area of the base: The base is rectangular, so the area \( A_{\text{base}} \) is given by: \[ A_{\text{base}} = l \times w = 15 , \text{m} \times 12 , \text{m} = 180 , \text{m}^2 \]

2. Area of the triangular sides: Each triangular face has a base equal to the length or the width of the pyramid and a height that can be found using the slant height (\( l_s \)).

   To find the slant height, we need to use the Pythagorean theorem on the triangle formed by half the base and the height of the pyramid. The slant height dimensions differ for the two triangular faces:

   • For triangles with base \( l \):

     • Half of \( l \) = \( \frac{l}{2} = \frac{15}{2} = 7.5 , \text{m} \)
     • Height = \( h = 10 , \text{m} \)
     • Slant height \( l_s \): \[ l_s = \sqrt{ \left( \frac{l}{2} \right)^2 + h^2 } = \sqrt{(7.5)^2 + (10)^2} = \sqrt{56.25 + 100} = \sqrt{156.25} = 12.5 , \text{m} \]
     • Area of two triangular faces with base \( l \) (2 triangles): \[ A_{\text{triangular, l}} = 2 \times \left( \frac{1}{2} \times l \times l_s \right) = 2 \times \left( \frac{1}{2} \times 15 \times 12.5 \right) = 15 \times 12.5 = 187.5 , \text{m}^2 \]

   • For triangles with base \( w \):

     • Half of \( w \) = \( \frac{w}{2} = \frac{12}{2} = 6 , \text{m} \)
     • Height = \( h = 10 , \text{m} \)
     • Slant height: \[ l_s = \sqrt{ \left( \frac{w}{2} \right)^2 + h^2 } = \sqrt{(6)^2 + (10)^2} = \sqrt{36 + 100} = \sqrt{136} \approx 11.62 , \text{m} \]
     • Area of two triangular faces with base \( w \) (2 triangles): \[ A_{\text{triangular, w}} = 2 \times \left( \frac{1}{2} \times w \times l_s \right) = 2 \times \left( \frac{1}{2} \times 12 \times 11.62 \right) = 12 \times 11.62 = 139.44 , \text{m}^2 \]

3. Total surface area: \[ A_{\text{total}} = A_{\text{base}} + A_{\text{triangular, l}} + A_{\text{triangular, w}} \] \[ A_{\text{total}} = 180 , \text{m}^2 + 187.5 , \text{m}^2 + 139.44 , \text{m}^2 = 506.94 , \text{m}^2 \]

Thus, the surface area of the rectangular pyramid, rounded to the nearest hundredth, is \( \boxed{506.94} , \text{m}^2 \).

Answer 4

R u sure

Answer 5

Let's double-check the calculations step by step for clarity.

Given:

• Length \( l = 15 , m \)
• Width \( w = 12 , m \)
• Height \( h = 10 , m \)

Step 1: Area of the Base

The area of the base (which is a rectangle): \[ A_{\text{base}} = l \times w = 15 , m \times 12 , m = 180 , m^2 \]

Step 2: Area of the Triangular Faces

Each triangular face's area depends on its slant height.

For triangular faces with base \( l = 15 , m \):

1. Half of the base: \[ \frac{l}{2} = \frac{15}{2} = 7.5 , m \]

2. Calculate the slant height \( l_s \): \[ l_s = \sqrt{\left(\frac{l}{2}\right)^2 + h^2} = \sqrt{(7.5)^2 + (10)^2} = \sqrt{56.25 + 100} = \sqrt{156.25} = 12.5 , m \]

3. Area of one triangular face: \[ A_{\text{triangular, length}} = \frac{1}{2} \times l \times l_s = \frac{1}{2} \times 15 , m \times 12.5 , m = 93.75 , m^2 \] Since there are 2 such triangles: \[ A_{\text{triangular, length}} = 2 \times 93.75 , m^2 = 187.5 , m^2 \]

For triangular faces with base \( w = 12 , m \):

1. Half of the base: \[ \frac{w}{2} = \frac{12}{2} = 6 , m \]

2. Calculate the slant height \( l_s \): \[ l_s = \sqrt{\left(\frac{w}{2}\right)^2 + h^2} = \sqrt{(6)^2 + (10)^2} = \sqrt{36 + 100} = \sqrt{136} \approx 11.62 , m \]

3. Area of one triangular face: \[ A_{\text{triangular, width}} = \frac{1}{2} \times w \times l_s = \frac{1}{2} \times 12 , m \times 11.62 , m \approx 69.72 , m^2 \] Since there are 2 such triangles: \[ A_{\text{triangular, width}} = 2 \times 69.72 , m^2 \approx 139.44 , m^2 \]

Step 3: Total Surface Area

Sum of the areas: \[ A_{\text{total}} = A_{\text{base}} + A_{\text{triangular, length}} + A_{\text{triangular, width}} \] \[ A_{\text{total}} = 180 , m^2 + 187.5 , m^2 + 139.44 , m^2 \approx 506.94 , m^2 \]

Conclusion

The final surface area calculation seems consistent and accurate. Thus, the surface area of the rectangular pyramid is indeed \( \boxed{506.94} , \text{m}^2 \). If you have a different method or result, please share it so we can compare!

Answer 6

What is the surface area of a square pyramid where the base edge is 1.5 cm and the slant height is 4 cm? Round your answer to the nearest hundredth.

Answer 7

To find the surface area of a square pyramid, we need to calculate the area of the base and the area of the triangular sides.

Given:

• Base edge \( b = 1.5 , \text{cm} \)
• Slant height \( l_s = 4 , \text{cm} \)

Step 1: Area of the Base

The base of the pyramid is square, so the area \( A_{\text{base}} \) is given by: \[ A_{\text{base}} = b^2 = (1.5 , \text{cm})^2 = 2.25 , \text{cm}^2 \]

Step 2: Area of the Triangular Faces

Each triangular face has a base equal to the edge length of the square base and a height equal to the slant height. Since a square pyramid has 4 triangular faces, we can find the area of one triangular face first and then multiply by 4.

1. Area of one triangular face: \[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times l_s = \frac{1}{2} \times 1.5 , \text{cm} \times 4 , \text{cm } \] \[ A_{\text{triangle}} = \frac{1}{2} \times 1.5 \times 4 = 3 , \text{cm}^2 \]

2. Area of all four triangular faces: \[ A_{\text{triangles}} = 4 \times A_{\text{triangle}} = 4 \times 3 , \text{cm}^2 = 12 , \text{cm}^2 \]

Step 3: Total Surface Area

Now we can find the total surface area \( A_{\text{total}} \) by adding the area of the base and the area of the triangular faces: \[ A_{\text{total}} = A_{\text{base}} + A_{\text{triangles}} = 2.25 , \text{cm}^2 + 12 , \text{cm}^2 = 14.25 , \text{cm}^2 \]

Final Answer

Thus, the surface area of the square pyramid, rounded to the nearest hundredth, is \( \boxed{14.25} , \text{cm}^2 \).