Worksheet on calculating volume and surface area of prisms with diagrams and formulas.

Problem Overview:

The task involves calculating the volume and total surface area (T.S.A.) of various prisms. The formulas provided are:

1. Volume of a Prism:
\[
\text{Volume} = \text{Base Area} \times \text{Height}
\]

2. Total Surface Area of a Prism:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height of Prism}) + 2 \times \text{Base Area}
\]

We will solve each part step by step.

---

Q1: Calculate the volume and total surface area of each prism.

#### (a) Rectangular Prism
- Dimensions: Length = 6 cm, Width = 4 cm, Height = 3 cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 6 \times 4 = 24 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 24 \times 3 = 72 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (6 + 4) = 20 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (20 \times 3) + 2 \times 24 = 60 + 48 = 108 \, \text{cm}^2
\]

Answer for (a):
\[
\boxed{72 \, \text{cm}^3, 108 \, \text{cm}^2}
\]

#### (b) Triangular Prism
- Base: Right triangle with legs 5 cm and 6 cm
- Height of Prism: 20 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 5 \times 6 = 15 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 15 \times 20 = 300 \, \text{cm}^3
\]
- Perimeter of Base:
The hypotenuse of the right triangle is calculated using the Pythagorean theorem:
\[
\text{Hypotenuse} = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \, \text{cm}
\]
\[
\text{Perimeter} = 5 + 6 + \sqrt{61} = 11 + \sqrt{61} \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (11 + \sqrt{61}) \times 20 + 2 \times 15
\]
\[
\text{T.S.A.} = 220 + 20\sqrt{61} + 30 = 250 + 20\sqrt{61} \, \text{cm}^2
\]

Answer for (b):
\[
\boxed{300 \, \text{cm}^3, 250 + 20\sqrt{61} \, \text{cm}^2}
\]

#### (c) Triangular Prism
- Base: Isosceles triangle with base 7 m, height 5 m, and equal sides 6.1 m
- Height of Prism: Not explicitly given, assume it is the same as the height of the triangular base (5 m)
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 7 \times 5 = 17.5 \, \text{m}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 17.5 \times 5 = 87.5 \, \text{m}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 7 + 6.1 + 6.1 = 19.2 \, \text{m}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (19.2 \times 5) + 2 \times 17.5
\]
\[
\text{T.S.A.} = 96 + 35 = 131 \, \text{m}^2
\]

Answer for (c):
\[
\boxed{87.5 \, \text{m}^3, 131 \, \text{m}^2}
\]

#### (d) Rectangular Prism
- Dimensions: Length = 40 cm, Width = 1.1 m = 110 cm, Height = 2 m = 200 cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 40 \times 110 = 4400 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 4400 \times 200 = 880000 \, \text{cm}^3 = 0.88 \, \text{m}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (40 + 110) = 300 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (300 \times 200) + 2 \times 4400
\]
\[
\text{T.S.A.} = 60000 + 8800 = 68800 \, \text{cm}^2 = 6.88 \, \text{m}^2
\]

Answer for (d):
\[
\boxed{0.88 \, \text{m}^3, 6.88 \, \text{m}^2}
\]

#### (e) Trapezoidal Prism
- Base: Trapezoid with parallel sides 7 cm and 5.5 cm, height 8 cm
- Height of Prism: 8.38 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times (\text{Sum of Parallel Sides}) \times \text{Height} = \frac{1}{2} \times (7 + 5.5) \times 8
\]
\[
\text{Base Area} = \frac{1}{2} \times 12.5 \times 8 = 50 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 50 \times 8.38 = 419 \, \text{cm}^3
\]
- Perimeter of Base:
The non-parallel sides are not given, so we cannot calculate the perimeter directly. Assume the problem provides enough information to compute it if needed.
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area}
\]
Without the perimeter, we cannot complete this part.

Answer for (e):
\[
\boxed{419 \, \text{cm}^3, \text{Perimeter needed for T.S.A.}}
\]

#### (f) Rectangular Prism
- Dimensions: Length = 12 cm, Width = 9 cm, Height = \(2\sqrt{10}\) cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 12 \times 9 = 108 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 108 \times 2\sqrt{10} = 216\sqrt{10} \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (12 + 9) = 42 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (42 \times 2\sqrt{10}) + 2 \times 108
\]
\[
\text{T.S.A.} = 84\sqrt{10} + 216 \, \text{cm}^2
\]

Answer for (f):
\[
\boxed{216\sqrt{10} \, \text{cm}^3, 84\sqrt{10} + 216 \, \text{cm}^2}
\]

#### (g) Triangular Prism
- Base: Right triangle with legs 12 cm and 5 cm
- Height of Prism: 22 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 12 \times 5 = 30 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 30 \times 22 = 660 \, \text{cm}^3
\]
- Perimeter of Base:
The hypotenuse of the right triangle is:
\[
\text{Hypotenuse} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{cm}
\]
\[
\text{Perimeter} = 12 + 5 + 13 = 30 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (30 \times 22) + 2 \times 30
\]
\[
\text{T.S.A.} = 660 + 60 = 720 \, \text{cm}^2
\]

Answer for (g):
\[
\boxed{660 \, \text{cm}^3, 720 \, \text{cm}^2}
\]

#### (h) Rectangular Prism
- Dimensions: Length = 40 cm, Width = 8 cm, Height = 10 cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 40 \times 8 = 320 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 320 \times 10 = 3200 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (40 + 8) = 96 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (96 \times 10) + 2 \times 320
\]
\[
\text{T.S.A.} = 960 + 640 = 1600 \, \text{cm}^2
\]

Answer for (h):
\[
\boxed{3200 \, \text{cm}^3, 1600 \, \text{cm}^2}
\]

#### (i) Triangular Prism
- Base: Triangle with sides 14 cm, 15 cm, and 16 cm
- Height of Prism: 10.82 cm
- Base Area: Use Heron's formula.
\[
s = \frac{14 + 15 + 16}{2} = 22.5 \, \text{cm}
\]
\[
\text{Base Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{22.5(22.5-14)(22.5-15)(22.5-16)}
\]
\[
\text{Base Area} = \sqrt{22.5 \times 8.5 \times 7.5 \times 6.5} \approx 84 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 84 \times 10.82 \approx 911.28 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 14 + 15 + 16 = 45 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (45 \times 10.82) + 2 \times 84
\]
\[
\text{T.S.A.} = 486.9 + 168 = 654.9 \, \text{cm}^2
\]

Answer for (i):
\[
\boxed{911.28 \, \text{cm}^3, 654.9 \, \text{cm}^2}
\]

---

Q2: Calculate the volume and total surface area of each prism.

#### (a) Triangular Prism
- Base: Isosceles triangle with base 10 cm, height 5 cm, and equal sides 6 cm
- Height of Prism: 20 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 10 \times 5 = 25 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 25 \times 20 = 500 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 10 + 6 + 6 = 22 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (22 \times 20) + 2 \times 25
\]
\[
\text{T.S.A.} = 440 + 50 = 490 \, \text{cm}^2
\]

Answer for (a):
\[
\boxed{500 \, \text{cm}^3, 490 \, \text{cm}^2}
\]

#### (b) Rectangular Prism
- Dimensions: Length = 20 cm, Width = 15 cm, Height = 12 cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 20 \times 15 = 300 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 300 \times 12 = 3600 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (20 + 15) = 70 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (70 \times 12) + 2 \times 300
\]
\[
\text{T.S.A.} = 840 + 600 = 1440 \, \text{cm}^2
\]

Answer for (b):
\[
\boxed{3600 \, \text{cm}^3, 1440 \, \text{cm}^2}
\]

#### (c) Trapezoidal Prism
- Base: Trapezoid with parallel sides 15 cm and 9 cm, height 8 cm
- Height of Prism: 6.02 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times (\text{Sum of Parallel Sides}) \times \text{Height} = \frac{1}{2} \times (15 + 9) \times 8
\]
\[
\text{Base Area} = \frac{1}{2} \times 24 \times 8 = 96 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 96 \times 6.02 = 577.92 \, \text{cm}^3
\]
- Perimeter of Base:
The non-parallel sides are not given, so we cannot calculate the perimeter directly.
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area}
\]
Without the perimeter, we cannot complete this part.

Answer for (c):
\[
\boxed{577.92 \, \text{cm}^3, \text{Perimeter needed for T.S.A.}}
\]

#### (d) Triangular Prism
- Base: Isosceles triangle with base 11 cm, height 8 cm, and equal sides 5 cm
- Height of Prism: 17 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 11 \times 8 = 44 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 44 \times 17 = 748 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 11 + 5 + 5 = 21 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (21 \times 17) + 2 \times 44
\]
\[
\text{T.S.A.} = 357 + 88 = 445 \, \text{cm}^2
\]

Answer for (d):
\[
\boxed{748 \, \text{cm}^3, 445 \, \text{cm}^2}
\]

#### (e) Rectangular Prism
- Dimensions: Length = 10 cm, Width = 6 cm, Height = 3 cm
- Base Area:
\[
\text{Base Area} = \text{Length} \times \text{Width} = 10 \times 6 = 60 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 60 \times 3 = 180 \, \text{cm}^3
\]
- Perimeter of Base:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (10 + 6) = 32 \, \text{cm}
\]
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area} = (32 \times 3) + 2 \times 60
\]
\[
\text{T.S.A.} = 96 + 120 = 216 \, \text{cm}^2
\]

Answer for (e):
\[
\boxed{180 \, \text{cm}^3, 216 \, \text{cm}^2}
\]

#### (f) Trapezoidal Prism
- Base: Trapezoid with parallel sides 14 cm and 8 cm, height 3 cm
- Height of Prism: 9 cm
- Base Area:
\[
\text{Base Area} = \frac{1}{2} \times (\text{Sum of Parallel Sides}) \times \text{Height} = \frac{1}{2} \times (14 + 8) \times 3
\]
\[
\text{Base Area} = \frac{1}{2} \times 22 \times 3 = 33 \, \text{cm}^2
\]
- Volume:
\[
\text{Volume} = \text{Base Area} \times \text{Height} = 33 \times 9 = 297 \, \text{cm}^3
\]
- Perimeter of Base:
The non-parallel sides are not given, so we cannot calculate the perimeter directly.
- Total Surface Area:
\[
\text{T.S.A.} = (\text{Perimeter of Base} \times \text{Height}) + 2 \times \text{Base Area}
\]
Without the perimeter, we cannot complete this part.

Answer for (f):
\[
\boxed{297 \, \text{cm}^3, \text{Perimeter needed for T.S.A.}}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Q1 (a)} & 72 \, \text{cm}^3, 108 \, \text{cm}^2 \\
\text{Q1 (b)} & 300 \, \text{cm}^3, 250 + 20\sqrt{61} \, \text{cm}^2 \\
\text{Q1 (c)} & 87.5 \, \text{m}^3, 131 \, \text{m}^2 \\
\text{Q1 (d)} & 0.88 \, \text{m}^3, 6.88 \, \text{m}^2 \\
\text{Q1 (e)} & 419 \, \text{cm}^3, \text{Perimeter needed for T.S.A.} \\
\text{Q1 (f)} & 216\sqrt{10} \, \text{cm}^3, 84\sqrt{10} + 216 \, \text{cm}^2 \\
\text{Q1 (g)} & 660 \, \text{cm}^3, 720 \, \text{cm}^2 \\
\text{Q1 (h)} & 3200 \, \text{cm}^3, 1600 \, \text{cm}^2 \\
\text{Q1 (i)} & 911.28 \, \text{cm}^3, 654.9 \, \text{cm}^2 \\
\text{Q2 (a)} & 500 \, \text{cm}^3, 490 \, \text{cm}^2 \\
\text{Q2 (b)} & 3600 \, \text{cm}^3, 1440 \, \text{cm}^2 \\
\text{Q2 (c)} & 577.92 \, \text{cm}^3, \text{Perimeter needed for T.S.A.} \\
\text{Q2 (d)} & 748 \, \text{cm}^3, 445 \, \text{cm}^2 \\
\text{Q2 (e)} & 180 \, \text{cm}^3, 216 \, \text{cm}^2 \\
\text{Q2 (f)} & 297 \, \text{cm}^3, \text{Perimeter needed for T.S.A.} \\
\end{array}
}
\]