Beast Academy 5D — Pythagorean Paths Enrichment Worksheet (Age 15, 8.G.B.8, Geometry)

Instructions

Use the Pythagorean theorem (a² + b² = c²) to find the lengths of the lines and paths described below. The distance between two adjacent dots on a grid (horizontally or vertically) is 1 unit. Express your answers in simplest radical form unless the result is a whole number.

Part 1: Finding Segment Lengths

For each grid, determine the exact length of the straight line segment connecting the two marked endpoints (*). The dashed lines represent the legs of a right triangle (the "rise" and "run").

1. Find the length of the segment.

   . . . *
         |
         | 1 unit
         |
   * - - -
     3 units
   

2. Find the length of the segment.

   *
   |
   | 3 units
   |
   . - *
     1 unit
   

3. Find the length of the segment.

         . . . *
             . |
             . | 3 units
             . |
   * - - - - - .
       4 units
   

4. Find the length of the segment.

   * - - - .
     3 units |
           | 1 unit
           |
           *
   

Part 2: Path Comparison

An ant is at point A(2, 1) and wants to travel to a crumb at point B(7, 13).

5. a) What is the total distance if the ant only travels along the grid lines (first horizontally, then vertically)?
   
   b) What is the distance if the ant travels in a straight, direct line from A to B?
   
   c) How much shorter is the direct path?

Part 3: Multi-Step Journeys

Solve the following problems involving complex paths.

6. A drone flies in a straight line from its starting point at S(-4, -2) to its destination at D(4, 4). What is the total distance of its flight?


7. A delivery robot travels from the warehouse at W(-3, 5) to its first stop at P(3, -3), and then continues to its second stop at S(8, 9). What is the total distance the robot traveled? (Assume it travels in straight lines between points).


8. Challenge: A hiker starts at a base camp located at (0, 0) and needs to get to a lookout point at L(10, 5). A straight river flows along the horizontal line y=3. The hiker must go to the river to fill her water bottle at some point R on the riverbank before continuing to the lookout point. What is the minimum possible distance for her entire trip (from base camp to the river, then to the lookout)? Hint: Consider reflecting the lookout point across the river.

Answer Key

1. The legs of the right triangle are 3 units and 1 unit.
   3² + 1² = c²
   9 + 1 = c²
   10 = c²
   Answer: √10 units
2. The legs of the right triangle are 1 unit and 3 units.
   1² + 3² = c²
   1 + 9 = c²
   10 = c²
   Answer: √10 units
3. The legs of the right triangle are 4 units and 3 units.
   4² + 3² = c²
   16 + 9 = c²
   25 = c²
   Answer: 5 units
4. The legs of the right triangle are 3 units and 1 unit.
   3² + 1² = c²
   9 + 1 = c²
   10 = c²
   Answer: √10 units
5. a) The horizontal distance is 7 - 2 = 5 units. The vertical distance is 13 - 1 = 12 units. The total grid line distance is 5 + 12 = 17 units.
   b) The legs of the right triangle are 5 units and 12 units.
   5² + 12² = c²
   25 + 144 = c²
   169 = c²
   c = 13. The direct path is 13 units.
   c) The direct path is 17 - 13 = 4 units shorter.
6. The horizontal distance (Δx) is 4 - (-4) = 8 units. The vertical distance (Δy) is 4 - (-2) = 6 units.
   8² + 6² = c²
   64 + 36 = c²
   100 = c²
   Answer: 10 units
7. Path 1 (W to P):
   Δx = 3 - (-3) = 6 units. Δy = 5 - (-3) = 8 units.
   6² + 8² = c₁² → 36 + 64 = 100 → c₁ = 10 units.
   Path 2 (P to S):
   Δx = 8 - 3 = 5 units. Δy = 9 - (-3) = 12 units.
   5² + 12² = c₂² → 25 + 144 = 169 → c₂ = 13 units.
   Total Distance: 10 + 13 = 23 units.
8. To find the shortest path, we reflect the lookout point L(10, 5) across the river line y=3. The new reflected point L' will be at (10, 1). The shortest path from the start to the river and then to the lookout is a straight line from the start (0, 0) to the reflected point L'(10, 1).
   We find the length of the segment from (0, 0) to (10, 1).
   Δx = 10 - 0 = 10 units. Δy = 1 - 0 = 1 unit.
   10² + 1² = c²
   100 + 1 = c²
   101 = c²
   Answer: √101 units