Find the Longest Internal Diagonal of a Cube

This article will demonstrate that the lowest to highest and opposing corners diagonal of a cube is equal to the side times the square root of 3.

Steps

  1. Sketch and label a diagram of a cube. Specify the long (internal) diagonal of a cube as line AD.
  2. Open a new Excel workbook and worksheet and draw a unit-cube using the Media Browser "Shapes" tool option. That means the the length of the sides must be equal to 1 unit; that is side s = 1 unit.
    • The six square shaped exterior surfaces (faces) are equal in dimensions, size, area and have the same shape. Therefore all faces are congruent.
  3. Label 3 consecutive corners (vertices) of the bottom face (the base) as A, B and C, thus forming triangle ABC.

    • See the figure: label as point D the corner (vertex) above C, at the top of the cube. The segment CD is at a right angle (90 degrees) to the base.
  4. Use the Pythagorean theorem: a2 + b2 = c2, for the right triangle ABC where: `
    • Let [AB]2 + [BC]2 = [AC]2
    • Then let = [1]2 + [1]2 = 1 + 1 = 2, for the "left hand side" (LHS) = 2 thus:
      • Examine the length of the RHS = AC squared: [AC]2 = 2.
      • Let [AC]2 = [sqrt(2)]2. Simplify that; you will find the length of the diagonal of the base, AC. We have AC = sqrt(2).
  5. Find the length of the long internal diagonal by using the Pythagorean theorem for right triangle ACD: [AC]2 + [CD]2 = [AD]2, where AD is the long internal diagonal we seek.
    • Use AC = sqrt(2) and knowing that CD = 1, we substitute these known values into the Pythagorean formula and have the following equation:

      [sqrt(2)]2 + 12 = [AD]2
    • Then let [sqrt(2)]2 + 12 = 2 + 1 = 3, then [AD]2 = [sqrt(3)]2.
    • Then realize that, [AD] the length of the internal diagonal from bottom to top and between opposing corners equals sqrt(3), because [sqrt(3)]2 = 3 (square root of the squared number) is just that number; let's call the number a, such as [sqrt(a)]2 = a ) and lengths are always positive numbers.
  6. Find the internal diagonal of a cube with a different side length: modify the formula to side s equaling a different number, as not for the unit cube but any length of side s; so that each side of the triangle is a multiple of the parts of the unit cube:
    • Let [s*AC]2 + [s*CD]2 = [s*AD]2 , by multiplication for sides of rt triangle ACD,

      and [s*sqrt(2)]2 + [s*1]2 = [s*sqrt(3)]2, by substitution.
    • You also can modify the earlier formula to [s*AB]2 + [s*BC]2 = [s*AC]2.

      [s*1]2 + [s*1]2 = [s*sqrt(2)]2, to convert from the unit cube with sides equaling 1, into a multiple of the sides of right triangle ABC with two legs = s*1, and its hypotenuse = s*sqrt(2).
    • In both cases, the absolute value of s (your cube's side length) is used as the multiplier.

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