To find the
surface area of a square pyramid, we add the area of the
square base and the areas of the
four identical isosceles triangular faces.
The formula is:
>
Surface Area = Area of Square Base + 4 × Area of One Triangular Face
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🔹 Problem 1)
-
Square base side = 6 cm
-
Triangle height (slant height) = 5 cm *(This is the height from base to apex of each triangle, perpendicular to the base edge)*
✔ Step 1: Area of square base
> $ A_{\text{base}} = \text{side}^2 = 6 \times 6 = 36 \text{ cm}^2 $
✔ Step 2: Area of one triangle
> $ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 5 = 15 \text{ cm}^2 $
✔ Step 3: Total surface area
> $ SA = 36 + 4 \times 15 = 36 + 60 = \boxed{96} \text{ cm}^2 $
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🔹 Problem 2)
-
Square base side = 12 in
-
Triangle height (slant height) = 8 in
✔ Step 1: Area of square base
> $ A_{\text{base}} = 12 \times 12 = 144 \text{ in}^2 $
✔ Step 2: Area of one triangle
> $ A_{\text{triangle}} = \frac{1}{2} \times 12 \times 8 = 48 \text{ in}^2 $
✔ Step 3: Total surface area
> $ SA = 144 + 4 \times 48 = 144 + 192 = \boxed{336} \text{ in}^2 $
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🔹 Problem 3)
-
Square base side = 9 in
-
Triangle height (slant height) = 10 in
✔ Step 1: Area of square base
> $ A_{\text{base}} = 9 \times 9 = 81 \text{ in}^2 $
✔ Step 2: Area of one triangle
> $ A_{\text{triangle}} = \frac{1}{2} \times 9 \times 10 = 45 \text{ in}^2 $
✔ Step 3: Total surface area
> $ SA = 81 + 4 \times 45 = 81 + 180 = \boxed{261} \text{ in}^2 $
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🔹 Problem 4)
-
Square base side = 4 ft
-
Triangle height (slant height) = 3½ ft = 3.5 ft
✔ Step 1: Area of square base
> $ A_{\text{base}} = 4 \times 4 = 16 \text{ ft}^2 $
✔ Step 2: Area of one triangle
> $ A_{\text{triangle}} = \frac{1}{2} \times 4 \times 3.5 = 2 \times 3.5 = 7 \text{ ft}^2 $
✔ Step 3: Total surface area
> $ SA = 16 + 4 \times 7 = 16 + 28 = \boxed{44} \text{ ft}^2 $
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##
✔ Final Answers:
1)
96 cm²
2)
336 in²
3)
261 in²
4)
44 ft²
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📌
Note: In all problems, the given "height" for the triangles is the
slant height (perpendicular height of the triangular face), not the vertical height of the pyramid — which is correct for calculating lateral surface area.
Let me know if you’d like diagrams or step-by-step visuals!
Parent Tip: Review the logic above to help your child master the concept of 6th grade surface area worksheet.
