Practice set
Pythagorean Theorem Word Problems
Eight real-world problems solved step by step, from ladders and navigation to construction and screen sizes. Each solution shows exactly how to turn a word problem into a right triangle calculation.
Problems in This Set
Method
How to Solve Any Pythagorean Theorem Word Problem
Every word problem that involves the Pythagorean theorem follows the same pattern. Before diving into the problems, learn this four-step approach. It works for all eight problems on this page and any others you encounter.
Read and visualize
Sketch a rough diagram. Look for two perpendicular measurements. Those are your legs. The direct distance you need is usually the hypotenuse.
Label the triangle
Assign a, b to the legs and c to the hypotenuse. Write down
the values you know and circle what you are solving for.
Choose the right formula
Finding hypotenuse: c = √(a² + b²). Finding a leg:
a = √(c² − b²). Verifying a right angle: check whether
a² + b² = c².
Calculate and check
Substitute the numbers, compute step by step, and verify the answer makes sense. The hypotenuse must be the longest side.
The most common mistake in word problems is misidentifying the hypotenuse. The hypotenuse is always the side you travel directly: the shortcut, the ladder length, the diagonal, the signal path. The two perpendicular measurements are always the legs.
Practice
Word Problems with Full Solutions
Basic Problems
These problems involve a single right triangle with straightforward labeling. Focus on identifying which side is the hypotenuse.
Ladder Against a Wall
A ladder leans against a vertical wall. The base of the ladder is 1.5 m from the wall, and the ladder reaches 2 m up the wall. How long is the ladder?
| Identify the triangle | The wall, 2 m, and the ground, 1.5 m, are the two legs. The ladder is the hypotenuse. |
|---|---|
| Assign values | a = 2 m, b = 1.5 m, find c |
| Apply formula | c = √(a² + b²) = √(2² + 1.5²) |
| Calculate | = √(4 + 2.25) = √6.25 |
| Result | c = 2.5 m |
Check: 2² + 1.5² = 4 + 2.25 = 6.25 = 2.5² ✓
Shortcut Across a Field
A rectangular field is 60 m long and 80 m wide. A student walks diagonally across the field from one corner to the opposite corner. How far does the student walk?
| Identify the triangle | The length, 80 m, and width, 60 m, are the two legs. The diagonal path is the hypotenuse. |
|---|---|
| Assign values | a = 80 m, b = 60 m, find c |
| Apply formula | c = √(80² + 60²) |
| Calculate | = √(6400 + 3600) = √10000 |
| Result | c = 100 m |
60-80-100 is a multiple of the 3-4-5 Pythagorean triple, scaled by 20.
Try in Calculator →TV Screen Diagonal
A television screen is 96 cm wide and 54 cm tall. What is the diagonal size of the screen in centimeters?
| Identify the triangle | The width, 96 cm, and height, 54 cm, are the legs. The diagonal is the hypotenuse. |
|---|---|
| Assign values | a = 96 cm, b = 54 cm, find c |
| Apply formula | c = √(96² + 54²) |
| Calculate | = √(9216 + 2916) = √12132 |
| Result | c ≈ 110.15 cm |
That is about 43.4 inches.
Try in Calculator →Intermediate Problems
These problems require an extra modeling step, either finding a diagonal from a real object or recognizing a square or route as a right triangle.
Roof Rafter Length
A roof has a horizontal span of 8 m from the center ridge to the outer wall. The ridge is 3 m above the level of the outer wall. What is the length of the roof rafter?
| Identify the triangle | The vertical rise, 3 m, and horizontal run, 8 m, are the legs. The rafter is the hypotenuse. |
|---|---|
| Assign values | a = 3 m, b = 8 m, find c |
| Apply formula | c = √(3² + 8²) |
| Calculate | = √(9 + 64) = √73 |
| Result | c ≈ 8.544 m |
Baseball Diamond
A baseball diamond is a square with 27.4 m between each base. A catcher at home plate throws directly to second base. How far is that throw?
| Identify the triangle | The throw from home to second is the diagonal of the square. Each leg is 27.4 m. |
|---|---|
| Assign values | a = 27.4 m, b = 27.4 m, find c |
| Apply formula | c = √(27.4² + 27.4²) = 27.4 × √2 |
| Calculate | = 27.4 × 1.414 ≈ 38.74 m |
| Result | c ≈ 38.74 m |
When both legs are equal, c = a√2. See the 45-45-90 Triangle Calculator.
Boat Navigation
A boat travels 12 km due north, then 9 km due east. How far is the boat from its starting point in a straight line?
| Identify the triangle | North travel, 12 km, and east travel, 9 km, are perpendicular. The straight-line return is the hypotenuse. |
|---|---|
| Assign values | a = 12 km, b = 9 km, find c |
| Apply formula | c = √(12² + 9²) |
| Calculate | = √(144 + 81) = √225 |
| Result | c = 15 km |
9-12-15 is a multiple of the 3-4-5 triple, scaled by 3.
Try in Calculator →Advanced Problems
These problems either find a missing leg instead of the hypotenuse, or require slightly less obvious setup from the wording.
How High Does the Ladder Reach?
A 5 m ladder leans against a wall. The base of the ladder is placed 2 m from the wall. How high up the wall does the top of the ladder reach?
| Identify the triangle | The ladder, 5 m, is the hypotenuse. The base distance, 2 m, is one leg. The wall height is the missing leg. |
|---|---|
| Assign values | c = 5 m, b = 2 m, find a |
| Apply formula | a = √(c² − b²) = √(5² − 2²) |
| Calculate | = √(25 − 4) = √21 |
| Result | a ≈ 4.583 m |
This problem finds a missing leg, so the formula subtracts instead of adds.
Try in Calculator →Staircase Design
A staircase must cover a horizontal distance of 4.2 m and rise to a height of 3.15 m. What is the total length of the staircase stringer?
| Identify the triangle | The horizontal run, 4.2 m, and vertical rise, 3.15 m, are the legs. The stringer is the hypotenuse. |
|---|---|
| Assign values | a = 3.15 m, b = 4.2 m, find c |
| Apply formula | c = √(3.15² + 4.2²) |
| Calculate | = √(9.9225 + 17.64) = √27.5625 |
| Result | c = 5.25 m |
3.15-4.2-5.25 is a scaled version of the 3-4-5 triple, scaled by 1.05.
Try in Calculator →Calculator
Check Your Own Word Problem
Have a word problem with different numbers? Enter your values into the calculator and get the answer with full step-by-step working.
FAQ
Frequently Asked Questions
Start by drawing a sketch and identifying the right angle in the scenario. Label the two
perpendicular sides as legs, a and b, and the direct distance as the
hypotenuse, c. Then decide what you are solving for and apply the correct form of
the formula.
The hypotenuse is always the direct path, the ladder length, diagonal distance, or straight-line route. The two legs are the perpendicular measurements that define the right angle, such as height and base, north and east, or width and height.
Many word problems hide the triangle in plain language. Any time you see two perpendicular directions and need a direct distance, a right triangle is implied. Draw those perpendicular paths as legs and the direct distance as the hypotenuse.
Yes, with an extension. For a box with dimensions l, w, and h,
the space diagonal is d = √(l² + w² + h²). This effectively applies the theorem
twice.
Ladder problems and diagonal distance problems are the most common in school settings. Both types follow the same four-step method described above, even though one may ask for a hypotenuse and another for a missing leg.
Next steps