Pythagorean Theorem Practice Problems (With Solutions)

The Pythagorean theorem is one of the most useful formulas in all of mathematics. Whether you are solving geometry homework, preparing for a standardized test, or working through real-world distance problems, this theorem comes up constantly.

This guide explains the theorem, walks through 10 practice problems with full solutions, and covers related topics like Pythagorean triples and 3D applications. For a broader look at geometry topics, visit our Geometry Guide.

The Pythagorean Theorem Explained

For any right triangle with legs a and b and hypotenuse c:

a^2 + b^2 = c^2

The hypotenuse is always the longest side, located opposite the right angle. This formula only works for right triangles.

To use it:

• Finding the hypotenuse: Plug in both legs and solve for c.
• Finding a leg: Rearrange to a^2 = c^2 - b^2, then solve for a.

Pythagorean Triples

Pythagorean triples are sets of three whole numbers that satisfy the theorem. Memorizing common triples saves time on tests.

• 3, 4, 5 (and multiples: 6-8-10, 9-12-15, 12-16-20)
• 5, 12, 13 (and multiples: 10-24-26)
• 8, 15, 17
• 7, 24, 25

If you recognize a triple in a problem, you can skip the computation entirely.

---

Want instant help on any math problem? Screenshot it with Math.Photos and get a step-by-step solution in seconds. Install the free extension.

---

Practice Problems

Problem 1: Finding the Hypotenuse (Basic)

A right triangle has legs of length 6 and 8. Find the hypotenuse.

Solution:

a^2 + b^2 = c^2

6^2 + 8^2 = c^2

36 + 64 = c^2

100 = c^2

c = 10

You might recognize this as the 3-4-5 triple scaled by 2. The hypotenuse is 10.

Problem 2: Finding the Hypotenuse (Non-Integer)

A right triangle has legs of length 5 and 7. Find the hypotenuse.

Solution:

5^2 + 7^2 = c^2

25 + 49 = c^2

74 = c^2

c = sqrt(74) ≈ 8.60

The hypotenuse is sqrt(74), or approximately 8.60.

Problem 3: Finding a Missing Leg

A right triangle has a hypotenuse of 13 and one leg of 5. Find the other leg.

Solution:

a^2 + b^2 = c^2

5^2 + b^2 = 13^2

25 + b^2 = 169

b^2 = 144

b = 12

This is the 5-12-13 triple. The missing leg is 12.

Problem 4: Finding a Missing Leg (Non-Integer)

A right triangle has a hypotenuse of 15 and one leg of 9. Find the other leg.

Solution:

9^2 + b^2 = 15^2

81 + b^2 = 225

b^2 = 144

b = 12

The missing leg is 12. (This is the 3-4-5 triple scaled by 3: 9-12-15.)

---

Stuck on a problem like these? Use our geometry solver to get help, or screenshot any problem with Math.Photos for a full walkthrough.

---

Problem 5: Word Problem (Ladder)

A 20-foot ladder leans against a wall. The base of the ladder is 8 feet from the wall. How high up the wall does the ladder reach?

Solution:

The ladder is the hypotenuse (20), the distance from the wall is one leg (8), and the height up the wall is the other leg.

8^2 + h^2 = 20^2

64 + h^2 = 400

h^2 = 336

h = sqrt(336) = 4 * sqrt(21) ≈ 18.33

The ladder reaches approximately 18.33 feet up the wall.

Problem 6: Word Problem (Distance Between Two Points)

A ship sails 9 miles north and then 12 miles east. How far is the ship from its starting point?

Solution:

The north and east directions form a right angle, so this is a right triangle problem.

9^2 + 12^2 = d^2

81 + 144 = d^2

225 = d^2

d = 15

The ship is 15 miles from its starting point. (Another 3-4-5 triple, scaled by 3.)

Problem 7: Distance Formula on the Coordinate Plane

Find the distance between points A(2, 3) and B(7, 15).

Solution:

The distance formula is derived directly from the Pythagorean theorem:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

d = sqrt((7 - 2)^2 + (15 - 3)^2)

d = sqrt(5^2 + 12^2)

d = sqrt(25 + 144)

d = sqrt(169)

d = 13

The distance is 13. (The 5-12-13 triple appears again.)

Problem 8: Diagonal of a Rectangle

A rectangle has a length of 16 and a width of 12. Find the length of its diagonal.

Solution:

The diagonal of a rectangle forms a right triangle with the length and width.

12^2 + 16^2 = d^2

144 + 256 = d^2

400 = d^2

d = 20

The diagonal is 20. (The 3-4-5 triple scaled by 4.)

---

These problems getting harder? Math.Photos handles everything from basic right triangles to advanced geometry. Just screenshot and solve. Try it free.

---

Problem 9: 3D Application (Space Diagonal of a Box)

A rectangular box has dimensions 3 x 4 x 12. Find the length of the space diagonal (the diagonal running from one corner to the opposite corner through the interior).

Solution:

For a 3D space diagonal, extend the Pythagorean theorem:

d = sqrt(l^2 + w^2 + h^2)

d = sqrt(3^2 + 4^2 + 12^2)

d = sqrt(9 + 16 + 144)

d = sqrt(169)

d = 13

The space diagonal is 13.

You can also think of this in two steps. First find the diagonal of the base: sqrt(3^2 + 4^2) = 5. Then use that base diagonal with the height: sqrt(5^2 + 12^2) = 13.

Problem 10: 3D Application (Distance in Space)

Find the distance between points P(1, 2, 3) and Q(4, 6, 15) in three-dimensional space.

Solution:

d = sqrt((4-1)^2 + (6-2)^2 + (15-3)^2)

d = sqrt(9 + 16 + 144)

d = sqrt(169)

d = 13

The distance is 13.

When the Pythagorean Theorem Does Not Apply

Remember these limitations:

• The theorem only works for right triangles. For non-right triangles, you need the Law of Cosines.
• If a problem does not involve a 90-degree angle, check whether you can construct one. For example, dropping an altitude from a vertex often creates right triangles within a larger non-right triangle.
• In proofs, you may need to justify why a triangle is a right triangle before applying the theorem. See our guide on geometry proof strategies for techniques on structuring these arguments.

Tips for Standardized Tests

1. Memorize common triples. Recognizing 3-4-5, 5-12-13, and 8-15-17 (plus their multiples) will save significant time.
2. Check for right triangles first. Many word problems and coordinate geometry questions are Pythagorean theorem problems in disguise.
3. Draw a diagram. If the problem is described in words, sketch the triangle and label the sides before computing.
4. Estimate before calculating. The hypotenuse must be longer than either leg but shorter than the sum of both legs. Use this to check your answer.

---

Want to ace your next geometry test? Math.Photos is like having a tutor in your browser. Screenshot any problem — right triangles, proofs, coordinate geometry — and get clear, step-by-step solutions. Install the free Chrome extension.

---

Frequently Asked Questions

Can the Pythagorean theorem be used on any triangle?

No. It only applies to right triangles (triangles with one 90-degree angle). For other triangles, use the Law of Cosines: c^2 = a^2 + b^2 - 2ab * cos(C). The Pythagorean theorem is actually a special case of this formula where angle C is 90 degrees and cos(90) = 0.

What is the converse of the Pythagorean theorem?

The converse states: if a^2 + b^2 = c^2 for the three sides of a triangle (where c is the longest side), then the triangle is a right triangle. This is useful for proving that an angle is 90 degrees without measuring it directly.

How does the distance formula relate to the Pythagorean theorem?

The distance formula is the Pythagorean theorem applied to the coordinate plane. The horizontal distance between two points is one leg, the vertical distance is the other leg, and the straight-line distance is the hypotenuse. Problems 7 and 10 above demonstrate this directly.

What are some real-world applications of the Pythagorean theorem?

Construction workers use it to verify right angles (the 3-4-5 method). Architects calculate diagonal measurements. Navigation and GPS systems compute distances. Screen sizes are measured diagonally using this theorem. Any time you need a straight-line distance and you know two perpendicular measurements, the Pythagorean theorem applies.