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A right pyramid has a square base of sides 8 cm and slant edge 12 cm long. a) Draw the net of the pyramid. b) Use the net to determine the surface area of the pyramid.

Mathematics

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Answer

b

64(1+22) cm2 or approximately 245.0 cm264(1 + 2\sqrt{2}) \text{ cm}^2 \text{ or approximately } 245.0 \text{ cm}^2

Step-by-step solution

5 steps

Part a

Draw the net of the pyramid.

The net of this right square pyramid consists of a square base and four identical isosceles triangular faces.

  • The base is a square with side length 8 cm8 \text{ cm}.
  • Each of the four triangular faces has a base of 8 cm8 \text{ cm} (which is a side of the square base) and two equal sides of 12 cm12 \text{ cm} (which are the slant edges of the pyramid).
  • To draw the net, place the 8 cm×8 cm8 \text{ cm} \times 8 \text{ cm} square in the center. Then, attach one isosceles triangle to each of the four sides of the square. The height of each triangle (the slant height of the pyramid) can be calculated as 82 cm8\sqrt{2} \text{ cm}.

Part b

64(1+22) cm2 or approximately 245.0 cm264(1 + 2\sqrt{2}) \text{ cm}^2 \text{ or approximately } 245.0 \text{ cm}^2

Use the net to determine the surface area of the pyramid.

  1. Calculate the area of the square base.

    The side length of the square base is s=8 cms = 8 \text{ cm}.

    Abase=s2=(8 cm)2=64 cm2A_{base} = s^2 = (8 \text{ cm})^2 = 64 \text{ cm}^2

  2. Calculate the slant height (ll) of the triangular faces.

    Each triangular face has a base of 8 cm8 \text{ cm} and slant edges of 12 cm12 \text{ cm}. We can form a right-angled triangle using half of the base of the triangular face (s/2s/2), the slant height (ll), and the slant edge (ee).

    (s/2)2+l2=e2(s/2)^2 + l^2 = e^2 (8/2)2+l2=122(8/2)^2 + l^2 = 12^2 42+l2=1224^2 + l^2 = 12^2 16+l2=14416 + l^2 = 144 l2=144−16l^2 = 144 - 16 l2=128l^2 = 128 l=128=64×2=82 cml = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2} \text{ cm}

  3. Calculate the area of one triangular face.

    Atriangle=12×base×slant heightA_{triangle} = \frac{1}{2} \times \text{base} \times \text{slant height} Atriangle=12×8 cm×82 cmA_{triangle} = \frac{1}{2} \times 8 \text{ cm} \times 8\sqrt{2} \text{ cm} Atriangle=4×82 cm2A_{triangle} = 4 \times 8\sqrt{2} \text{ cm}^2 Atriangle=322 cm2A_{triangle} = 32\sqrt{2} \text{ cm}^2

  4. Calculate the total lateral surface area (area of the four triangular faces).

    Alateral=4×Atriangle=4×322 cm2A_{lateral} = 4 \times A_{triangle} = 4 \times 32\sqrt{2} \text{ cm}^2 Alateral=1282 cm2A_{lateral} = 128\sqrt{2} \text{ cm}^2

  5. Calculate the total surface area of the pyramid.

    The total surface area is the sum of the base area and the lateral surface area.

    Atotal=Abase+AlateralA_{total} = A_{base} + A_{lateral} Atotal=64 cm2+1282 cm2A_{total} = 64 \text{ cm}^2 + 128\sqrt{2} \text{ cm}^2 Atotal=64(1+22) cm2A_{total} = 64(1 + 2\sqrt{2}) \text{ cm}^2

    Using 2≈1.414\sqrt{2} \approx 1.414:

    Atotal≈64(1+2×1.414) cm2A_{total} \approx 64(1 + 2 \times 1.414) \text{ cm}^2 Atotal≈64(1+2.828) cm2A_{total} \approx 64(1 + 2.828) \text{ cm}^2 Atotal≈64(3.828) cm2A_{total} \approx 64(3.828) \text{ cm}^2 Atotal≈244.992 cm2A_{total} \approx 244.992 \text{ cm}^2

    The surface area of the pyramid is 64(1+22) cm2 or approximately 245.0 cm2\boxed{64(1 + 2\sqrt{2}) \text{ cm}^2 \text{ or approximately } 245.0 \text{ cm}^2}.

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