Surface area of solids worksheet with 10 geometric figures and dimensions for calculation.

Let’s solve each surface area problem one by one. We’ll use standard formulas for surface area of common solids, and round to the nearest tenth as instructed.

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1) Sphere — radius = 4 m

Formula: Surface Area = \( 4\pi r^2 \)

\( SA = 4 \pi (4)^2 = 4 \pi (16) = 64\pi \approx 64 \times 3.1416 = 201.06 \)

✔ Answer: 201.1 m²

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2) Rectangular Prism — dimensions 5 cm × 6 cm × 5 cm

Formula: SA = 2(lw + lh + wh)

Here, l=6, w=5, h=5

SA = 2(6×5 + 6×5 + 5×5) = 2(30 + 30 + 25) = 2(85) = 170

✔ Answer: 170.0 cm²

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3) Cube — side = 6 m

Formula: SA = 6s²

SA = 6 × (6)² = 6 × 36 = 216

✔ Answer: 216.0 m²

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4) Cone — radius = 8 cm, slant height = 17.9 cm

Formula: SA = πr² + πrℓ (base + lateral)

SA = π(8)² + π(8)(17.9) = 64π + 143.2π = 207.2π ≈ 207.2 × 3.1416 ≈ 650.9

✔ Answer: 650.9 cm²

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5) Square Pyramid — base = 4 in, slant height = 5 in, height of triangular face = 5 in (given)

Wait — actually, the diagram shows a square pyramid with:

- Base edge = 4 in
- Slant height (height of triangular face) = 5 in (labeled on triangle)
- Also shows height of pyramid = 3.7 in — but we don’t need that for surface area.

Formula: SA = base area + lateral area = s² + 4 × (½ × base × slant height)

Base area = 4 × 4 = 16

Lateral area = 4 × (½ × 4 × 5) = 4 × 10 = 40

Total SA = 16 + 40 = 56

✔ Answer: 56.0 in²

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6) Square Pyramid — base edge = 7 ft, slant height = 8.7 ft

Same formula:

Base area = 7 × 7 = 49

Lateral area = 4 × (½ × 7 × 8.7) = 4 × (30.45) = 121.8

Total SA = 49 + 121.8 = 170.8

✔ Answer: 170.8 ft²

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7) Triangular Pyramid (Tetrahedron) — all edges 3 yd? But diagram shows base is 3 yd, height of triangle 2.6 yd, and slant height 2.2 yd? Wait — let’s interpret carefully.

Actually, it looks like a triangular pyramid with equilateral triangle base? But labeled:

- Base triangle: sides 3 yd, 3 yd, 3 yd? (equilateral)
- Height of base triangle = 2.6 yd (given)
- Slant height (height of lateral face) = 2.2 yd? That doesn't make sense because slant height should be longer than height of base.

Wait — re-examining: The diagram shows:

- Base is a triangle with base 3 yd and height 2.6 yd → so area of base = ½ × 3 × 2.6 = 3.9 yd²
- The three lateral faces are triangles with base 3 yd and height (slant height) = 2.2 yd? But 2.2 < 2.6? That’s impossible if it’s a pyramid — slant height must be ≥ height of base.

Actually, likely the 2.2 yd is the slant height, and the 2.6 yd is the height of the base triangle (which is fine).

So:

Base area = ½ × 3 × 2.6 = 3.9 yd²

Each lateral face = ½ × 3 × 2.2 = 3.3 yd²

There are 3 lateral faces → 3 × 3.3 = 9.9 yd²

Total SA = 3.9 + 9.9 = 13.8 yd²

✔ Answer: 13.8 yd²

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8) Trapezoidal Prism — bases are trapezoids

Dimensions:

- Two parallel sides of trapezoid: 7 m and 3 m
- Height of trapezoid = 4 m (vertical distance between bases)
- Length of prism (depth) = 5 m

First, find area of trapezoid base:

Area = ½ × (sum of bases) × height = ½ × (7 + 3) × 4 = ½ × 10 × 4 = 20 m²

Two bases → 2 × 20 = 40 m²

Now lateral surface area = perimeter of base × length of prism

Perimeter of trapezoid: need all 4 sides.

We have top = 3 m, bottom = 7 m, height = 4 m.

The non-parallel sides: since it's symmetric? Diagram shows two slanted sides — we can calculate using Pythagorean theorem.

Difference in bases: 7 - 3 = 4 m → split equally on both sides → 2 m overhang per side.

So each slanted side = √(2² + 4²) = √(4 + 16) = √20 ≈ 4.472 m

So perimeter = 3 + 7 + 4.472 + 4.472 ≈ 18.944 m

Lateral SA = 18.944 × 5 ≈ 94.72 m²

Total SA = 40 + 94.72 = 134.72 → rounded to nearest tenth: 134.7 m²

✔ Answer: 134.7 m²

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9) Sphere — diameter = 4 ft → radius = 2 ft

SA = 4πr² = 4π(2)² = 16π ≈ 50.265

✔ Answer: 50.3 ft²

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10) Cone — radius = 13 in, slant height = 29.1 in

SA = πr² + πrℓ = π(13)² + π(13)(29.1)

= 169π + 378.3π = 547.3π ≈ 547.3 × 3.1416 ≈ 1719.3

✔ Answer: 1719.3 in²

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## ✔ Final Answers:

1) 201.1 m²
2) 170.0 cm²
3) 216.0 m²
4) 650.9 cm²
5) 56.0 in²
6) 170.8 ft²
7) 13.8 yd²
8) 134.7 m²
9) 50.3 ft²
10) 1719.3 in²

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