Math Problem Statement

Find the surface area and volume of each solid in problems 63 and 64.

Solution

Let's solve each of the problems shown in the image.

Problem 63:

You are asked to find the surface area and volume of a triangular prism.

  • Dimensions:
    • Base of the triangle: 4 m
    • Height of the triangle: 3 m
    • Length of the prism: 10 m
    • Hypotenuse of the triangle: 5 m

1. Volume of the triangular prism:

The volume of a prism is calculated as: V=Base Area×Height of PrismV = \text{Base Area} \times \text{Height of Prism}

For this triangular prism, the base is a right triangle with:

  • Base = 4 m
  • Height = 3 m

The area of the triangular base is: Base Area=12×base×height=12×4×3=6m2\text{Base Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6 \, \text{m}^2

Now, the height of the prism (or length) is 10 m, so the volume is: V=6×10=60m3V = 6 \times 10 = 60 \, \text{m}^3

2. Surface Area of the triangular prism:

The surface area consists of:

  • 2 triangular faces
  • 3 rectangular faces (formed by the sides of the triangle and the length of the prism)
a) Triangular faces:

Each triangular face has an area of 6 m² (as calculated before), so for both triangular faces: Area of both triangles=2×6=12m2\text{Area of both triangles} = 2 \times 6 = 12 \, \text{m}^2

b) Rectangular faces:
  • One rectangle formed by the base of the triangle and the length of the prism: Area=Base×Length=4×10=40m2\text{Area} = \text{Base} \times \text{Length} = 4 \times 10 = 40 \, \text{m}^2

  • One rectangle formed by the height of the triangle and the length of the prism: Area=Height×Length=3×10=30m2\text{Area} = \text{Height} \times \text{Length} = 3 \times 10 = 30 \, \text{m}^2

  • One rectangle formed by the hypotenuse of the triangle and the length of the prism: Area=Hypotenuse×Length=5×10=50m2\text{Area} = \text{Hypotenuse} \times \text{Length} = 5 \times 10 = 50 \, \text{m}^2

Total surface area:

Total Surface Area=12+40+30+50=132m2\text{Total Surface Area} = 12 + 40 + 30 + 50 = 132 \, \text{m}^2

Problem 64:

This is a rectangular prism (box).

  • Dimensions:
    • Length = 15 in
    • Width = 6 in
    • Height = 8 in

1. Volume of the rectangular prism:

The volume is calculated as: V=Length×Width×Height=15×6×8=720in3V = \text{Length} \times \text{Width} \times \text{Height} = 15 \times 6 \times 8 = 720 \, \text{in}^3

2. Surface Area of the rectangular prism:

The surface area consists of 6 rectangular faces. The formula for the surface area of a rectangular prism is: A=2(Length×Width+Length×Height+Width×Height)A = 2(\text{Length} \times \text{Width} + \text{Length} \times \text{Height} + \text{Width} \times \text{Height})

Substituting the values: A=2(15×6+15×8+6×8)A = 2(15 \times 6 + 15 \times 8 + 6 \times 8) A=2(90+120+48)=2(258)=516in2A = 2(90 + 120 + 48) = 2(258) = 516 \, \text{in}^2

Final Answers:

  1. Problem 63:

    • Volume: 60m360 \, \text{m}^3
    • Surface Area: 132m2132 \, \text{m}^2
  2. Problem 64:

    • Volume: 720in3720 \, \text{in}^3
    • Surface Area: 516in2516 \, \text{in}^2

Let me know if you need more details or have any questions! Here are some related questions:

  1. How is the surface area of prisms calculated in general?
  2. What other 3D shapes share a similar formula for volume calculation?
  3. Can the Pythagorean theorem be used to calculate the hypotenuse of any right triangle in prisms?
  4. How would you calculate the surface area if the prism had slanted sides?
  5. What is the significance of finding surface area versus volume in real-world applications?

Tip: Understanding how to break down complex shapes into simpler parts (like triangles and rectangles) helps make surface area calculations easier!

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Math Problem Analysis

Mathematical Concepts

Geometry
Surface Area
Volume
Prisms

Formulas

Volume of a triangular prism: V = (Base Area) × (Height of Prism)
Surface Area of a prism: Add areas of all faces
Volume of a rectangular prism: V = Length × Width × Height
Surface Area of a rectangular prism: A = 2(L×W + L×H + W×H)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 6-8