Finding Shortest Distance with Pythagorean Theorem

Geometry & Distance 9th-10th Grade
PROBLEM
Mark is on his way home from work. He drives 35 miles due North and then 42 miles due West. Find the shortest distance he can cover to reach home early. use pythag therom

What This Problem Teaches

  • Recognizing when perpendicular movements form a right triangle
  • Applying the Pythagorean theorem to find direct distances
  • Simplifying square roots by factoring out perfect squares
  • Understanding the difference between path distance and straight-line distance
  • Converting between exact radical form and decimal approximations

Picture This

Mark is on his way home from work. He drives 35 miles due North and then 42 miles due West. Find the shortest...

Mark's path creates a right triangle. The shortest distance home is the hypotenuse.

Solution: Method 1 — The Direct Pythagorean Approach

Step 1 — Identify the right triangle

Mark's path forms a perfect right triangle. He drives 35 miles North, then 42 miles West. Since North and West are perpendicular directions, these two segments are the legs of a right triangle, and the shortest distance home is the hypotenuse.

Step 2 — Set up the Pythagorean theorem

For any right triangle, the Pythagorean theorem states that c² = a² + b², where c is the hypotenuse and a and b are the legs.

c² = 35² + 42²

Step 3 — Calculate the squares

Let's compute each term:

  • 35² = 35 × 35 = 1,225
  • 42² = 42 × 42 = 1,764

Step 4 — Add and take the square root

c² = 1,225 + 1,764 = 2,989 c = √2,989

Step 5 — Simplify the square root

To simplify √2,989, we need to factor it. Let's check if it has any perfect square factors:

  • 2,989 ÷ 49 = 61 (since 49 = 7²)
  • So 2,989 = 49 × 61 = 7² × 61
c = √(7² × 61) = 7√61

For a decimal approximation: √61 ≈ 7.81, so c ≈ 7 × 7.81 = 54.67 miles.

Solution: Method 2 — Recognition of the 7-Factor Pattern

Step 1 — Notice the common factor

Both legs are multiples of 7:

  • 35 = 7 × 5
  • 42 = 7 × 6

Step 2 — Factor out the 7

We can rewrite the Pythagorean theorem as:

c² = (7 × 5)² + (7 × 6)² c² = 7²(5² + 6²) c² = 49(25 + 36) = 49 × 61

Step 3 — Take the square root directly

c = √(49 × 61) = √49 × √61 = 7√61

This method is faster when you recognize that both legs share a common factor, allowing you to factor before calculating large squares.

The Answer: The shortest distance Mark can cover to reach home is 7√61 miles, which is approximately 54.7 miles.

Verification

Let's verify our answer by substituting back into the Pythagorean theorem:

(7√61)² = 35² + 42² 49 × 61 = 1,225 + 1,764 2,989 = 2,989 ✓

We can also check reasonableness: Mark's total path distance is 35 + 42 = 77 miles. The straight-line distance should be shorter, and indeed 54.7 < 77. Additionally, the hypotenuse should be longer than either leg, and 54.7 > 42 > 35 ✓.

Common Pitfalls

✗ Mistake 1: Adding the distances
Some students calculate 35 + 42 = 77 miles. This gives the total path distance, not the straight-line distance. The problem asks for the shortest distance, which requires the Pythagorean theorem.
✗ Mistake 2: Forgetting to take the square root
Calculating 35² + 42² = 2,989 and stopping there. Remember that c² = 2,989 means c = √2,989, not c = 2,989.
✗ Mistake 3: Incorrect radical simplification
Writing √2,989 = √(49 × 61) = 49√61 instead of 7√61. Only perfect squares can be taken out of the radical as their square root: √49 = 7.

The Math Beneath the Surface

This problem demonstrates the fundamental principle that the shortest distance between two points is a straight line. When you have perpendicular movements (North-South and East-West), they form the legs of a right triangle, and the Pythagorean theorem gives you that straight-line distance.

The general formula for this type of problem is:

Shortest distance = √(northward distance² + eastward distance²)

This same mathematical structure appears in:

  • GPS navigation systems calculating "as the crow flies" distances
  • Vector addition in physics (combining perpendicular forces)
  • Distance formula in coordinate geometry: d = √[(x₂-x₁)² + (y₂-y₁)²]

Beyond the Textbook

This calculation appears constantly in real-world navigation:

  • Aviation: Pilots calculate direct flight paths when air traffic control allows deviation from standard routes
  • Emergency services: Fire departments and ambulances use straight-line distances to estimate response times
  • Delivery logistics: Companies like Amazon use these calculations to optimize drone delivery routes

How to Spot This Problem Type

Look for these key phrases and setups:

  • "Due North," "due South," "due East," or "due West" — these signal perpendicular movements
  • "Shortest distance" or "direct distance" — this asks for the hypotenuse, not the path length
  • Any movement in two perpendicular directions followed by a question about returning to the starting point
  • "As the crow flies" — another way of asking for straight-line distance

If you see a problem involving two perpendicular movements and a question about distance, you're likely looking at a Pythagorean theorem problem.

Try These Variations

1Reverse the Unknown
Mark drives 35 miles due North, then some distance due West, and the straight-line distance home is 65 miles. How far West did he drive?
Step 1 — Set up the equation

We know the hypotenuse (65 miles) and one leg (35 miles). Using the Pythagorean theorem: 65² = 35² + w² where w is the westward distance.

Step 2 — Calculate the squares

4225 = 1225 + w²

Step 3 — Solve for w²

w² = 4225 - 1225 = 3000

Step 4 — Take the square root

w = √3000 = √(100 × 30) = 10√30 ≈ 54.8 miles

Step 5 — Verify

√(35² + (10√30)²) = √(1225 + 3000) = √4225 = 65

Answer: Mark drove 10√30 ≈ 54.8 miles West.

2Add a Third Leg
Mark drives 35 miles North, then 42 miles West, then 20 miles South. What is the shortest distance back to his starting point?
Step 1 — Find net displacement

Net North movement: 35 - 20 = 15 miles North
Net West movement: 42 miles West

Step 2 — Apply Pythagorean theorem

The net displacement forms a right triangle with legs of 15 and 42 miles.

Step 3 — Calculate

c² = 15² + 42² = 225 + 1764 = 1989

Step 4 — Simplify

1989 = 9 × 221 = 9 × 13 × 17
c = √1989 = 3√221 ≈ 44.6 miles

Step 5 — Verify

(3√221)² = 9 × 221 = 1989 = 15² + 42²

Answer: 3√221 ≈ 44.6 miles

3Different Distances
Sarah drives 24 miles due East and then 32 miles due North. Find the shortest distance back to her starting point.
Step 1 — Identify the triangle

Sarah's path forms a right triangle with legs of 24 miles (East) and 32 miles (North).

Step 2 — Notice the pattern

Both distances are multiples of 8: 24 = 8 × 3 and 32 = 8 × 4. This suggests a 3-4-5 triangle scaled by 8.

Step 3 — Apply Pythagorean theorem

c² = 24² + 32² = 576 + 1024 = 1600

Step 4 — Take the square root

c = √1600 = 40 miles

Step 5 — Verify the pattern

Indeed, this is a 3-4-5 triangle scaled by 8: 8 × 5 = 40

Answer: 40 miles

4Time and Speed Comparison
Mark drives 35 miles North at 50 mph, then 42 miles West at 60 mph. If he takes the straight-line path home at 55 mph, how much time does he save?
Step 1 — Calculate original route time

North leg: 35 ÷ 50 = 0.7 hours
West leg: 42 ÷ 60 = 0.7 hours
Total: 0.7 + 0.7 = 1.4 hours

Step 2 — Find straight-line distance

From the original problem: d = 7√61 ≈ 54.67 miles

Step 3 — Calculate direct route time

Time = 54.67 ÷ 55 ≈ 0.994 hours

Step 4 — Find time saved

Time saved = 1.4 - 0.994 = 0.406 hours

Step 5 — Convert to minutes

0.406 × 60 ≈ 24.4 minutes

Answer: Mark saves approximately 24.4 minutes

Frequently Asked Questions

How do you find the shortest distance between two points when you know the legs of a right triangle?+
Use the Pythagorean theorem: c² = a² + b². The shortest distance is the hypotenuse of the right triangle. In this problem, with legs of 35 miles and 42 miles, we get c = √(35² + 42²) = √(1225 + 1764) = √2989 = 7√61 ≈ 54.7 miles.
When should you use the Pythagorean theorem instead of adding distances?+
Use the Pythagorean theorem when you can travel in a straight line between two points that form a right triangle with your current position. Adding distances gives you the path length; the Pythagorean theorem gives you the direct distance, which is always shorter.
How do you simplify square roots in Pythagorean theorem problems?+
Factor the number under the square root to find perfect squares. Here, 2989 = 49 × 61 = 7² × 61, so √2989 = √(7² × 61) = 7√61. This gives both an exact answer (7√61) and allows easy decimal approximation.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-07-28