Volume of a Triangular Pyramid Worksheets - Free Printable

Let's solve each problem step by step using the formula for the volume of a triangular pyramid (tetrahedron):

$$
\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
$$

The base area is the area of the triangular base, which can be calculated using:
$$
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}
$$

We'll go through each problem.

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Problem 1

- Triangle base: base = 12 in, height = 14 in → Area = $ \frac{1}{2} \times 12 \times 14 = 84 $ in²
- Height of pyramid = 12 in
- Volume = $ \frac{1}{3} \times 84 \times 12 = 336 $ in³

✔ Answer: 336.00 in³

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Problem 2

- Triangle base: base = 9 ft, height = 11 ft → Area = $ \frac{1}{2} \times 9 \times 11 = 49.5 $ ft²
- Height of pyramid = 17 ft
- Volume = $ \frac{1}{3} \times 49.5 \times 17 = 280.5 $ ft³

✔ Answer: 280.50 ft³

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Problem 3

- Triangle base: base = 23 yd, height = 10 yd → Area = $ \frac{1}{2} \times 23 \times 10 = 115 $ yd²
- Height of pyramid = 20 yd
- Volume = $ \frac{1}{3} \times 115 \times 20 = 766.666... $ ≈ 766.67 yd³

✔ Answer: 766.67 yd³

---

Problem 4

- Triangle base: base = 20 yd, height = 13 yd → Area = $ \frac{1}{2} \times 20 \times 13 = 130 $ yd²
- Height of pyramid = 15 yd
- Volume = $ \frac{1}{3} \times 130 \times 15 = 650 $ yd³

✔ Answer: 650.00 yd³

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Problem 5

- Triangle base: base = 5 ft, height = 8 ft → Area = $ \frac{1}{2} \times 5 \times 8 = 20 $ ft²
- Height of pyramid = 11 ft
- Volume = $ \frac{1}{3} \times 20 \times 11 = 73.333... $ ≈ 73.33 ft³

✔ Answer: 73.33 ft³

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Problem 6

- Triangle base: base = 8 in, height = 11 in → Area = $ \frac{1}{2} \times 8 \times 11 = 44 $ in²
- Height of pyramid = 10 in
- Volume = $ \frac{1}{3} \times 44 \times 10 = 146.666... $ ≈ 146.67 in³

✔ Answer: 146.67 in³

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Problem 7

- Base is an isosceles triangle with sides: 19 in, 19 in, 25 in.
- We need to find the area of this triangle first.

Use Heron’s formula:

Semi-perimeter:
$$
s = \frac{19 + 19 + 25}{2} = \frac{63}{2} = 31.5 \text{ in}
$$

Area:
$$
A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{31.5(31.5 - 19)(31.5 - 19)(31.5 - 25)}
$$
$$
= \sqrt{31.5 \times 12.5 \times 12.5 \times 6.5}
$$

Calculate step-by-step:

- $ 31.5 \times 6.5 = 204.75 $
- $ 12.5 \times 12.5 = 156.25 $
- $ 204.75 \times 156.25 = ? $

Let’s compute:

First, approximate:
$$
\sqrt{31.5 \times 12.5 \times 12.5 \times 6.5} = \sqrt{(12.5^2) \times (31.5 \times 6.5)} = 12.5 \times \sqrt{31.5 \times 6.5}
$$

$ 31.5 \times 6.5 = 204.75 $

$ \sqrt{204.75} \approx 14.31 $

So area ≈ $ 12.5 \times 14.31 = 178.875 $ in²

Alternatively, use height from base method:

In an isosceles triangle, drop perpendicular from apex to base (25 in). This splits base into two 12.5 in parts.

Now use Pythagorean theorem:
$$
h = \sqrt{19^2 - 12.5^2} = \sqrt{361 - 156.25} = \sqrt{204.75} \approx 14.31 \text{ in}
$$

Area = $ \frac{1}{2} \times 25 \times 14.31 = 178.875 $ in²

Now, height of pyramid = 3 inches

Volume = $ \frac{1}{3} \times 178.875 \times 3 = 178.875 $ in³

✔ Answer: 178.88 in³ (rounded to two decimal places)

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Problem 8

- Triangle base: sides 9 ft, 8 ft, 15 ft
- Height of pyramid = 21 ft

Again, use Heron’s formula for area.

Semi-perimeter:
$$
s = \frac{9 + 8 + 15}{2} = \frac{32}{2} = 16 \text{ ft}
$$

Area:
$$
A = \sqrt{16(16 - 9)(16 - 8)(16 - 15)} = \sqrt{16 \times 7 \times 8 \times 1} = \sqrt{896}
$$

$ \sqrt{896} = \sqrt{64 \times 14} = 8\sqrt{14} \approx 8 \times 3.7417 = 29.9336 $ ft²

So base area ≈ 29.93 ft²

Volume = $ \frac{1}{3} \times 29.93 \times 21 = \frac{1}{3} \times 628.53 = 209.51 $ ft³

✔ Answer: 209.51 ft³

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✔ Final Answers:

1) 336.00 in³
2) 280.50 ft³
3) 766.67 yd³
4) 650.00 yd³
5) 73.33 ft³
6) 146.67 in³
7) 178.88 in³
8) 209.51 ft³

Let me know if you'd like a visual explanation or how to draw the diagrams!