Surface Area & Volume Word Problems worksheet with geometric figures and real-world scenarios.

Problem Analysis and Solution

The image contains several word problems related to surface area, volume, and practical applications. Below, I will solve each problem step by step.

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#### Problem 1: Classroom Dimensions
A classroom has the dimensions shown in the diagram:
- Length = 8 m
- Width = 6 m
- Height = 3 m

Find:
(a) The area of the floor
(b) The volume of the room
(c) The total area of the four walls

##### Solution:

(a) Area of the floor:
The floor is a rectangle with length = 8 m and width = 6 m.
\[
\text{Area of the floor} = \text{Length} \times \text{Width} = 8 \, \text{m} \times 6 \, \text{m} = 48 \, \text{m}^2
\]

(b) Volume of the room:
The volume of a rectangular room is given by:
\[
\text{Volume} = \text{Length} \times \text{Width} \times \text{Height} = 8 \, \text{m} \times 6 \, \text{m} \times 3 \, \text{m} = 144 \, \text{m}^3
\]

(c) Total area of the four walls:
The four walls consist of two pairs of opposite walls:
- Two walls with dimensions \( \text{Height} \times \text{Length} \)
- Two walls with dimensions \( \text{Height} \times \text{Width} \)

The area of the two longer walls:
\[
2 \times (\text{Height} \times \text{Length}) = 2 \times (3 \, \text{m} \times 8 \, \text{m}) = 2 \times 24 \, \text{m}^2 = 48 \, \text{m}^2
\]

The area of the two shorter walls:
\[
2 \times (\text{Height} \times \text{Width}) = 2 \times (3 \, \text{m} \times 6 \, \text{m}) = 2 \times 18 \, \text{m}^2 = 36 \, \text{m}^2
\]

Total area of the four walls:
\[
\text{Total area} = 48 \, \text{m}^2 + 36 \, \text{m}^2 = 84 \, \text{m}^2
\]

Final Answers for Problem 1:
\[
\boxed{(a) \, 48 \, \text{m}^2, \, (b) \, 144 \, \text{m}^3, \, (c) \, 84 \, \text{m}^2}
\]

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#### Problem 2: Iron Bar with Rectangular Cross-Section
An iron bar has a mass of 7.52 g. The density of iron is \( 7.8 \, \text{g/cm}^3 \). The dimensions of the bar are:
- Length = 10 cm
- Width = 2 cm
- Height = 1 cm

Find:
The volume of the iron bar.

##### Solution:
The volume \( V \) of a rectangular prism is given by:
\[
V = \text{Length} \times \text{Width} \times \text{Height}
\]
Substitute the given dimensions:
\[
V = 10 \, \text{cm} \times 2 \, \text{cm} \times 1 \, \text{cm} = 20 \, \text{cm}^3
\]

Final Answer for Problem 2:
\[
\boxed{20 \, \text{cm}^3}
\]

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#### Problem 3: Water Supply for Puppies
One puppy requires 5 L of water to live happily. A water tank has the following dimensions:
- Length = 20 cm
- Width = 10 cm
- Height = 10 cm

Find:
The maximum number of puppies that can be kept in this tank.

##### Solution:
First, calculate the volume of the water tank:
\[
\text{Volume of the tank} = \text{Length} \times \text{Width} \times \text{Height} = 20 \, \text{cm} \times 10 \, \text{cm} \times 10 \, \text{cm} = 2000 \, \text{cm}^3
\]

Convert the volume from cubic centimeters to liters (1 liter = 1000 cm³):
\[
\text{Volume in liters} = \frac{2000 \, \text{cm}^3}{1000} = 2 \, \text{L}
\]

Each puppy requires 5 L of water. The maximum number of puppies that can be kept is:
\[
\text{Number of puppies} = \frac{\text{Volume of the tank}}{\text{Water required per puppy}} = \frac{2 \, \text{L}}{5 \, \text{L/puppy}} = 0.4
\]

Since the number of puppies must be a whole number, the maximum number of puppies that can be kept is:
\[
\boxed{0}
\]

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#### Problem 4: Excavation for a House Foundation
The foundation of a house is being excavated. The dimensions of the excavation are:
- Length = 10 m
- Width = 8 m
- Depth = 2 m

Trucks are used to carry away the soil, and each truck can carry 5 m³ of soil.

Find:
The number of trips the trucks need to make to remove all the soil.

##### Solution:
First, calculate the volume of the excavation:
\[
\text{Volume of excavation} = \text{Length} \times \text{Width} \times \text{Depth} = 10 \, \text{m} \times 8 \, \text{m} \times 2 \, \text{m} = 160 \, \text{m}^3
\]

Each truck can carry 5 m³ of soil. The number of trips required is:
\[
\text{Number of trips} = \frac{\text{Volume of excavation}}{\text{Capacity of one truck}} = \frac{160 \, \text{m}^3}{5 \, \text{m}^3/\text{trip}} = 32 \, \text{trips}
\]

Final Answer for Problem 4:
\[
\boxed{32}
\]

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Final Summary of Answers:

1. \(\boxed{(a) \, 48 \, \text{m}^2, \, (b) \, 144 \, \text{m}^3, \, (c) \, 84 \, \text{m}^2}\)
2. \(\boxed{20 \, \text{cm}^3}\)
3. \(\boxed{0}\)
4. \(\boxed{32}\)