A pyramid on a rectangular base measuring 8 cm by 6 cm and slant height 10 cm .

To analyze the pyramid with a rectangular base measuring 8 cm by 6 cm and a slant height of 10 cm, we can calculate a few important properties, such as the height of the pyramid, the surface area, and the volume. ### 1. Height of the Pyramid The slant height (l) is the distance from the apex of the pyramid to the midpoint of one of the sides of the base. To find the height (h) of the pyramid, we can use the Pythagorean theorem. First, we need to find the distance from the center of the base to the midpoint of one of the sides. The base is a rectangle, so the center is at: - \( \text{Center} = \left( \frac{8}{2}, \frac{6}{2} \right) = (4, 3) \) The distance from the center to the midpoint of one of the longer sides (8 cm) is: - \( \text{Distance} = \frac{8}{2} = 4 \, \text{cm} \) Now, we can apply the Pythagorean theorem: \[ l^2 = h^2 + \text{Distance}^2 \] Substituting the known values: \[ 10^2 = h^2 + 4^2 \] \[ 100 = h^2 + 16 \] \[ h^2 = 100 - 16 = 84 \] \[ h = \sqrt{84} \approx 9.17 \, \text{cm} \] ### 2. Volume of the Pyramid The volume (V) of a pyramid is given by the formula: \[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \] The area of the rectangular base is: \[ \text{Base Area} = 8 \, \text{cm} \times 6 \, \text{cm} = 48 \, \text{cm}^2 \] Now substituting the values into the volume formula: \[ V = \frac{1}{3} \times 48 \, \text{cm}^2 \times \sqrt{84} \, \text{cm} \] \[ V \approx \frac{1}{3} \times 48 \times 9.17 \approx 147.36 \, \text{cm}^3 \] ### 3. Surface Area of the Pyramid The surface area (SA) of the pyramid consists of the area of the base plus the area of the four triangular faces. - **Base Area**: \( 48 \, \text{cm}^2 \) - **Area of the Triangular Faces**: There are two triangular faces with a base of 8 cm and two with a base of 6 cm. 1. **For the triangular faces with base 8 cm**: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 8 \times 10 = 40 \, \text{cm}^2 \] Since there are two of these faces: \[ \text{Total Area} = 2 \times 40 = 80 \, \text{cm}^2 \] 2. **For the triangular faces with base 6 cm**: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 6 \times 10 = 30 \, \text{cm}^2 \] Since there are two of these faces: \[ \text{Total Area} = 2 \times 30 = 60 \, \text{cm}^2 \] Now, adding all the areas together: \[ \text{Total Surface Area} = \text{Base Area} + \text{Area of Triangular Faces} \] \[ \text{Total Surface Area} = 48 + 80 + 60 = 188 \, \text{cm}^2 \] ### Summary - Height of the pyramid: \( \approx 9.17 \, \text{cm} \) - Volume of the pyramid: \( \approx 147.36 \, \text{cm}^3 \) - Surface area of the pyramid: \( 188 \, \text{cm}^2 \