4-20-20 Grade 7 Volume of a Triangular Prism Worksheet online ... - Free Printable

Of course! Let's solve each problem step by step. The worksheet asks for the surface area of various triangular prisms.

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🔷 General Formula for Surface Area of a Triangular Prism:

A triangular prism has 2 triangular bases and 3 rectangular lateral faces.

> Surface Area = 2 × (Area of Triangle) + (Perimeter of Triangle × Height of Prism)

Or, broken down:

> SA = 2 × (½ × base × height of triangle) + (side1 + side2 + side3) × height of prism

We'll apply this to each problem.

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## 🟦 Problem 1
- Triangle base: 2 ft, height: 2 ft → So it’s an isosceles right triangle? Wait — actually, the diagram shows a triangle with base 2 ft and height 2 ft (perpendicular height), so area = ½ × 2 × 2 = 2 ft²
- Prism height (length): 4 ft
- The 3 rectangular sides: two are 2 ft × 4 ft, one is the hypotenuse × 4 ft.

Wait — let’s check: if base = 2 ft, height = 2 ft, then the two equal sides (legs) are both 2 ft? But that would make it a right triangle with legs 2 and 2, so hypotenuse = √(2² + 2²) = √8 ≈ 2.828 ft.

But looking at the diagram — it looks like an equilateral triangle? No, the height is shown as 2 ft, base 2 ft — that’s not possible for an equilateral triangle (height would be √3 ≈ 1.732). So likely, it’s a right triangle with legs 2 ft and 2 ft.

✔ Let’s proceed assuming it’s a right triangle with legs 2 ft and 2 ft.

- Area of one triangle = ½ × 2 × 2 = 2 ft²
- Two triangles = 2 × 2 = 4 ft²
- Perimeter of triangle = 2 + 2 + √(2²+2²) = 4 + √8 ≈ 4 + 2.828 = 6.828 ft
- Lateral surface area = perimeter × height = 6.828 × 4 ≈ 27.312 ft²
- Total SA = 4 + 27.312 = 31.31 ft² (rounded to nearest hundredth)

BUT — wait! Looking again at the diagram, the triangle is drawn with a base of 2 ft and a height of 2 ft, but the other two sides are not labeled. However, in many worksheets, when they show a triangle with base and height, and no side lengths, they expect you to use the given base and height for area, and assume the sides are just the base and the two slanted sides — but here, since it's a prism, we need all three side lengths for lateral area.

Actually — this might be a mistake in interpretation. Let me re-express: perhaps the triangle is equilateral? But height 2 ft for base 2 ft doesn’t fit.

Alternatively — maybe it’s a right triangle with base 2 ft, height 2 ft, and the third side is the hypotenuse. That makes sense.

✔ So final answer for #1: 31.31 ft²

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## 🟦 Problem 2
- Triangle: base 3 ft, height 3 ft → again, likely a right triangle with legs 3 ft and 3 ft.
- Area of one triangle = ½ × 3 × 3 = 4.5 ft²
- Two triangles = 9 ft²
- Hypotenuse = √(3² + 3²) = √18 ≈ 4.2426 ft
- Perimeter = 3 + 3 + 4.2426 ≈ 10.2426 ft
- Prism height = 6 ft
- Lateral SA = 10.2426 × 6 ≈ 61.4556 ft²
- Total SA = 9 + 61.4556 ≈ 70.46 ft²

✔ Answer: 70.46 ft²

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## 🟦 Problem 3
This one gives more info:
- Triangle: base = 8 ft, height = 6 ft, slant side = 7.21 ft (probably the hypotenuse)
- Also, the prism length = 14 ft
- First, confirm: if base 8, height 6, then hypotenuse should be √(8² + 6²) = √(64+36) = √100 = 10 ft — but it says 7.21 ft? That doesn't match.

Wait — 7.21 is close to √52 ≈ 7.21 — which would be if the triangle had sides 6 and something else.

Actually — look: the triangle has base 8 ft, height 6 ft, and one side is 7.21 ft — this suggests it’s not a right triangle? Or maybe the 7.21 is the slant height for lateral face? No — the diagram labels 7.21 ft as the side of the triangle.

Wait — perhaps the triangle is scalene. Given base 8 ft, height 6 ft, and one side 7.21 ft — we can find the other side using Pythagorean theorem on the two halves.

If height = 6 ft, base = 8 ft, then splitting the base into two parts: say x and (8-x), then:

x² + 6² = (one side)²
(8-x)² + 6² = (other side)²

But we’re given one side is 7.21 ft — let’s check:

Suppose 7.21² = x² + 36 → x² = 51.9841 - 36 = 15.9841 → x ≈ 4.0 ft

Then other side: (8-4)² + 36 = 16 + 36 = 52 → √52 ≈ 7.21 ft — oh! So both sides are 7.21 ft? That would make it isosceles!

Wait — 7.21² ≈ 52, and 52 = 4² + 6² → so if the foot of the height splits the base into 4 and 4, then yes — it’s an isosceles triangle with base 8 ft, height 6 ft, and two equal sides of √(4² + 6²) = √52 ≈ 7.21 ft.

Perfect!

So:
- Area of triangle = ½ × 8 × 6 = 24 ft²
- Two triangles = 48 ft²
- Perimeter of triangle = 8 + 7.21 + 7.21 = 22.42 ft
- Prism height = 14 ft
- Lateral SA = 22.42 × 14 = 313.88 ft²
- Total SA = 48 + 313.88 = 361.88 ft²

✔ Answer: 361.88 ft²

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## 🟦 Problem 4
Same as Problem 3? Let’s check:

- Base = 8 ft, height = 6 ft, side = 7.21 ft, prism length = 14 ft → identical to Problem 3!

So same calculation:

✔ Answer: 361.88 ft²

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## 🟦 Problem 5
- Triangle: base = 4 ft, height = 3 ft, side = 3.61 ft, prism length = 11 ft
- Check: if height 3 ft, base 4 ft, split into 2 and 2 → then side = √(2² + 3²) = √(4+9) = √13 ≈ 3.6055 ≈ 3.61 ft — perfect! So it’s isosceles.

Area of triangle = ½ × 4 × 3 = 6 ft²
Two triangles = 12 ft²

Perimeter = 4 + 3.61 + 3.61 = 11.22 ft

Lateral SA = 11.22 × 11 = 123.42 ft²

Total SA = 12 + 123.42 = 135.42 ft²

✔ Answer: 135.42 ft²

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## 🟦 Problem 6
Same as Problem 1? Base 2 ft, height 2 ft, prism height 4 ft — yes, identical.

We calculated earlier: 31.31 ft²

✔ Answer: 31.31 ft²

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## 🟦 Problem 7
- Triangle: base = 6 ft, height = 4 ft, side = 5 ft, prism length = 12 ft
- Check: if base 6, height 4, split into 3 and 3 → then side = √(3² + 4²) = √(9+16) = √25 = 5 ft — perfect! Isosceles triangle.

Area = ½ × 6 × 4 = 12 ft²
Two triangles = 24 ft²

Perimeter = 6 + 5 + 5 = 16 ft

Lateral SA = 16 × 12 = 192 ft²

Total SA = 24 + 192 = 216.00 ft²

✔ Answer: 216.00 ft²

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## 🟦 Problem 8
- Triangle: base = 2, height = 2, prism height = 4 → same as Problem 1 and 6.

✔ Answer: 31.31 ft²

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## 🟦 Problem 9
Same as above — base 2, height 2, prism height 4.

✔ Answer: 31.31 ft²

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## ✔ Final Answers:

1) 31.31 ft²
2) 70.46 ft²
3) 361.88 ft²
4) 361.88 ft²
5) 135.42 ft²
6) 31.31 ft²
7) 216.00 ft²
8) 31.31 ft²
9) 31.31 ft²

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## 💡 Note:
In problems 1, 6, 8, 9 — the triangle is assumed to be a right triangle with legs 2 ft and 2 ft, giving hypotenuse ≈ 2.828 ft. If your teacher intended it to be equilateral or something else, the answer may differ — but based on standard interpretation and given dimensions, this is correct.

Let me know if you want diagrams or step-by-step breakdowns for any specific one!