Pythagorean Paths (Beast Academy 5D) — Step-by-step dot-grid diagrams for a 25-year-old (√10, √10, 5, √10)

Worked examples and dot-grid diagrams showing how to use the Pythagorean Theorem to find distances between points in a coordinate grid (Beast Academy 5D practice). Includes four examples with distances √10, √10, 5, √10 and clear step-by-step methods.

PDF

Goal: Use the Pythagorean Theorem to find the distance between two points on a coordinate (dot) grid. Formula: distance d = sqrt((x2 - x1)^2 + (y2 - y1)^2).

Tip: To visualize, drop a vertical or horizontal from one point to make a right triangle. The horizontal leg length = |x2 - x1| and the vertical leg length = |y2 - y1|. Then use a^2 + b^2 = c^2.

Example 1: distance = √10

Choose A = (0, 0) and B = (3, 1).

Horizontal leg = |3 - 0| = 3. Vertical leg = |1 - 0| = 1.

Distance = sqrt(3^2 + 1^2) = sqrt(9 + 1) = sqrt(10).

   y=2  .  .  .  .  .
   y=1  .  .  .  B  .   (B = (3,1))
   y=0  A  -  -  +  .   (A = (0,0), + = projection (3,0))
        x=0 x=1 x=2 x=3 x=4

Here the horizontal run from A to the projection + has length 3 (three dashes), and the vertical rise from + to B has length 1.

Example 2: distance = √10 (different orientation)

Choose A = (2, 0) and B = (1, 3).

Horizontal leg = |1 - 2| = 1. Vertical leg = |3 - 0| = 3.

Distance = sqrt(1^2 + 3^2) = sqrt(1 + 9) = sqrt(10).

   y=4  .  .  .  .  .
   y=3  .  B  .  .  .   (B = (1,3))
   y=2  .  |  .  .  .
   y=1  .  |  .  .  .
   y=0  .  A  -  -  .   (A = (2,0), projection (1,0) shown by horizontal dashes)
        x=0 x=1 x=2 x=3 x=4

Horizontal run = 1, vertical rise = 3 → sqrt(1^2 + 3^2) = √10.

Example 3: distance = 5 (3-4-5 right triangle)

Choose A = (0, 0) and B = (3, 4).

Horizontal leg = |3 - 0| = 3. Vertical leg = |4 - 0| = 4.

Distance = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

   y=5  .  .  .  .  .
   y=4  .  .  .  B  .   (B = (3,4))
   y=3  .  .  .  |  .
   y=2  .  .  .  |  .
   y=1  .  .  .  |  .
   y=0  A  -  -  +  .   (A = (0,0), + = projection (3,0))
        x=0 x=1 x=2 x=3 x=4

Horizontal run = 3, vertical rise = 4 → distance = 5.

Example 4: distance = √10 (vertical-down orientation)

Choose A = (0, 3) and B = (1, 0).

Horizontal leg = |1 - 0| = 1. Vertical leg = |0 - 3| = 3.

Distance = sqrt(1^2 + 3^2) = sqrt(10).

   y=4  .  .  .  .  .
   y=3  A  .  .  .  .   (A = (0,3))
   y=2  |  .  .  .  .
   y=1  |  .  .  .  .
   y=0  +  B  .  .  .   (B = (1,0), + = projection (1,3))
        x=0 x=1 x=2 x=3 x=4

Horizontal run = 1, vertical drop = 3 → sqrt(1^2 + 3^2) = √10.

How to make more diagrams:

  • Pick lattice coordinates for A and B so the horizontal and vertical differences are integers (easy on dot grids).
  • Compute dx = |x2 - x1| and dy = |y2 - y1|. If dx^2 + dy^2 is a perfect square you get an integer distance (like 5).
  • Common Pythagorean pairs: (3,4,5), (1,3,√10), (6,8,10), etc.

If you want, I can generate printable dot-grid diagrams (SVG or PNG) for these four examples or create more examples from the Beast Academy practice set numbers you listed.

Ask a followup question