Math Problem Statement

Each edge of a cube is 10 cm in length. (a) What is the length of AC, the diagonal of the base? (b) What is the size of ∠DAC, the angle between an interior diagonal of the box and a diagonal of the base?

Solution

Let's break this problem down step by step.

The question refers to a cube with an edge length of 10 cm. There are two parts to answer:

(a) Length of ACAC, the diagonal of the base

Since ACAC is the diagonal of a square face of the cube, we can use the Pythagorean theorem to find its length.
Each side of the square is 10cm10 \, \text{cm}, so:

AC=102+102=100+100=200=102cmAC = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{cm}

Thus, the length of ACAC is 102cm10\sqrt{2} \, \text{cm} (approximately 14.14cm14.14 \, \text{cm}).


(b) Angle DAC\angle DAC, the angle between a diagonal of the box and a diagonal of the base

We need to find the angle between the space diagonal of the cube (from one corner to the opposite corner) and the diagonal of the base.

  1. Length of the space diagonal:
    Using the Pythagorean theorem in three dimensions for the cube: Space Diagonal=102+102+102=300=103cm\text{Space Diagonal} = \sqrt{10^2 + 10^2 + 10^2} = \sqrt{300} = 10\sqrt{3} \, \text{cm}

  2. Angle between space diagonal and base diagonal:
    We are looking for the angle θ=DAC\theta = \angle DAC. Using the dot product approach:

    • Length of diagonal of base: 102cm10\sqrt{2} \, \text{cm}
    • Length of space diagonal: 103cm10\sqrt{3} \, \text{cm}

    The angle between two vectors u\mathbf{u} and v\mathbf{v} is given by: cosθ=uvuv\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} Here, u\mathbf{u} is the base diagonal (on the xyxy-plane), and v\mathbf{v} is the space diagonal. Because the two diagonals share components along the base, the dot product simplifies to:

    cosθ=10×10+10×0(102)(103)=100106=106\cos \theta = \frac{10 \times 10 + 10 \times 0}{(10\sqrt{2})(10\sqrt{3})} = \frac{100}{10\sqrt{6}} = \frac{10}{\sqrt{6}}

    Now simplify: cosθ=1066\cos \theta = \frac{10\sqrt{6}}{6} So: θ=cos1(1066)35.26\theta = \cos^{-1} \left( \frac{10\sqrt{6}}{6} \right) \approx 35.26^\circ


Final Answers:

  • (a) Length of ACAC: 102cm10\sqrt{2} \, \text{cm} (or 14.14cm14.14 \, \text{cm}).
  • (b) Angle DAC\angle DAC: Approximately 35.2635.26^\circ.

Would you like further clarification or a step-by-step explanation of any part?


5 Related Questions:

  1. How do you find the surface area and volume of a cube?
  2. What is the general formula for a space diagonal in an nn-dimensional hypercube?
  3. How does the dot product method help in calculating angles between vectors?
  4. How can you calculate the diagonal of a rectangular prism with unequal side lengths?
  5. What is the relationship between a cube’s edge, face diagonal, and space diagonal?

Tip:

In geometry, using the Pythagorean theorem across dimensions can help solve problems involving diagonals and angles in 3D shapes.

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Math Problem Analysis

Mathematical Concepts

Geometry
3D Shapes
Pythagorean Theorem
Vectors
Dot Product

Formulas

AC (base diagonal) = √(a^2 + a^2) = a√2
Space diagonal = √(a^2 + a^2 + a^2) = a√3
cos(θ) = (u · v) / (|u||v|)

Theorems

Pythagorean Theorem
Dot Product Theorem

Suitable Grade Level

Grades 9-12