Pythagorean Theorem Word Problems with Answers

Grade: 8-9 | Topic: Geometry

What You Will Learn

After completing this page you will be able to translate real-world scenarios into right-triangle diagrams, decide which sides are the legs and which is the hypotenuse, and solve for the unknown measurement. You will see problems involving ladders, sports fields, construction, navigation, and everyday situations — all powered by the same formula: $a^2 + b^2 = c^2$.

Theory

From words to triangles

Every Pythagorean theorem word problem follows three steps:

1. Sketch a right triangle. Identify where the 90-degree angle is hiding in the scenario.
2. Label the sides. Assign the two known measurements to legs or hypotenuse.
3. Solve. Use $a^2 + b^2 = c^2$ (or its rearranged forms) and take the square root.

Recognizing right angles in the real world

Right angles appear whenever two directions are perpendicular:

Scenario	Where the right angle is
Ladder against a wall	Wall meets the ground
Walking north then east	The turn from north to east
Diagonal of a rectangle	Corner of the rectangle
Ramp to a raised platform	Ground meets the vertical rise
TV/monitor screen	Corner of the rectangular screen

Once you spot the right angle, everything else follows from the formula.

Which side is the hypotenuse?

The hypotenuse is always:

• The side opposite the right angle
• The longest side of the triangle
• Often the "direct" or "straight-line" distance (e.g., the ladder itself, the diagonal, the distance "as the crow flies")

Worked Examples

Example 1: Ladder against a wall (easy)

Problem: A 13-foot ladder leans against a building. The base of the ladder is 5 feet from the building. How high up the wall does the ladder reach?

Step 1: Sketch the right triangle. The wall and the ground form the right angle. The ladder is the hypotenuse.

• Hypotenuse: $c = 13$ ft (the ladder)
• One leg: $b = 5$ ft (distance from wall)
• Unknown leg: $a = \;?$ (height on wall)

Step 2: Apply the formula for a missing leg. $a = \sqrt{c^2 - b^2} = \sqrt{13^2 - 5^2}$

Step 3: Calculate. $a = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ ft}$

Answer: The ladder reaches 12 feet up the wall.

Example 2: Walking two directions (easy)

Problem: Maya walks 6 blocks north and then 8 blocks east. How far is she from her starting point in a straight line?

Step 1: North and east are perpendicular, forming a right angle. The straight-line distance back is the hypotenuse.

• Leg $a = 6$ blocks (north)
• Leg $b = 8$ blocks (east)

Step 2: Solve for the hypotenuse. $c = \sqrt{a^2 + b^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$

Step 3: Simplify. $c = 10 \text{ blocks}$

Answer: Maya is 10 blocks from her starting point. (This is the 3-4-5 triple scaled by 2.)

Example 3: Diagonal of a TV screen (medium)

Problem: A TV screen is 48 inches wide and 27 inches tall. The screen size is measured along the diagonal. What is the screen size?

Step 1: The screen is a rectangle. The diagonal cuts it into two right triangles.

• Leg $a = 48$ in (width)
• Leg $b = 27$ in (height)

Step 2: Find the diagonal (hypotenuse). $d = \sqrt{48^2 + 27^2} = \sqrt{2304 + 729} = \sqrt{3033}$

Step 3: Approximate. $d \approx 55.1 \text{ inches}$

Answer: The screen is approximately a 55-inch TV.

Example 4: Building a wheelchair ramp (medium)

Problem: A wheelchair ramp must reach a doorway that is 2 feet above the ground. The ramp extends 24 feet horizontally from the building. How long is the ramp surface?

Step 1: The ground is horizontal, the rise is vertical, and the ramp is the hypotenuse.

• Leg $a = 2$ ft (vertical rise)
• Leg $b = 24$ ft (horizontal run)

Step 2: Calculate the ramp length. $c = \sqrt{a^2 + b^2} = \sqrt{2^2 + 24^2} = \sqrt{4 + 576} = \sqrt{580}$

Step 3: Simplify. $\sqrt{580} = \sqrt{4 \times 145} = 2\sqrt{145} \approx 24.08 \text{ ft}$

Answer: The ramp is approximately 24.08 feet long. (Notice the ramp is only slightly longer than the horizontal distance because the rise is small.)

Example 5: Baseball diamond (challenging)

Problem: A baseball diamond is a square with 90-foot sides. How far does a catcher at home plate need to throw the ball to reach second base?

Step 1: The throw from home plate to second base is the diagonal of the square. A square's diagonal creates a right triangle using two sides of the square as legs.

• Leg $a = 90$ ft (home to first)
• Leg $b = 90$ ft (first to second)

Step 2: Calculate the diagonal. $d = \sqrt{90^2 + 90^2} = \sqrt{8100 + 8100} = \sqrt{16{,}200}$

Step 3: Simplify. $\sqrt{16{,}200} = \sqrt{8100 \times 2} = 90\sqrt{2} \approx 127.3 \text{ ft}$

Answer: The catcher must throw approximately 127.3 feet (or exactly $90\sqrt{2}$ feet).

Common Mistakes

Mistake 1: Treating the "direct distance" as a leg instead of the hypotenuse

❌ A 20-ft rope stretches from the top of a pole to the ground 12 ft away. Student solves: $c = \sqrt{20^2 + 12^2} = \sqrt{544} \approx 23.3$ ft.

✅ The rope (direct distance) is the hypotenuse, not a leg. The height is: $a = \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16$ ft.

Why this matters: The straight-line distance between two points (the rope, the ladder, the diagonal) is almost always the hypotenuse. Misidentifying it means you add when you should subtract.

Mistake 2: Forgetting to look for the right angle

❌ "Two cars start from the same point. One drives 30 miles and the other drives 40 miles. How far apart are they?" Student assumes the answer is 50 miles.

✅ This is only true if they drive at right angles to each other (e.g., one north, one east). If they drive in the same direction, they could be 10 miles apart. Always confirm the right angle exists.

Why this matters: Without a 90-degree angle the Pythagorean theorem does not apply. The problem must specify perpendicular directions.

Mistake 3: Mixing up units or rounding too early

❌ Converting 3 feet and 8 inches to "3.8 feet" and plugging that into the formula.

✅ Convert properly: 8 inches = $\frac{8}{12} = 0.667$ feet, so the total is $3.667$ feet.

Why this matters: Rounding or converting incorrectly introduces errors that compound through the calculation. Keep full precision until the final answer.

Practice Problems

Try these on your own before checking the answers:

1. A kite is flying 40 meters above the ground. The string makes a straight line from your hand (at ground level) to the kite. If you are standing 30 meters from the point directly below the kite, how long is the string?
2. A ship sails 5 km east and then 12 km north. How far is it from the starting port?
3. A rectangular swimming pool is 25 m long and 10 m wide. A swimmer goes from one corner to the opposite corner diagonally. How far does the swimmer travel?
4. A tree casts a shadow 18 feet long. A wire from the top of the tree to the tip of the shadow is 30 feet. How tall is the tree?
5. Two hikers start from the same campsite. One hikes 9 km due west and the other hikes 12 km due south. How far apart are the two hikers?

Click to see answers

1. The string is the hypotenuse. $c = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$ m.
2. $c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ km.
3. $d = \sqrt{25^2 + 10^2} = \sqrt{625 + 100} = \sqrt{725} = 5\sqrt{29} \approx 26.93$ m.
4. The tree height is a leg: $a = \sqrt{30^2 - 18^2} = \sqrt{900 - 324} = \sqrt{576} = 24$ ft.
5. West and south are perpendicular: $d = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ km.

Summary

• Every word problem starts by sketching a right triangle and identifying the 90-degree angle.
• The direct or diagonal distance is usually the hypotenuse.
• Add the squares of the legs to find the hypotenuse; subtract to find a missing leg.
• Always confirm that a right angle exists before applying $a^2 + b^2 = c^2$.
• Keep units consistent and avoid rounding until the final step.

Related Topics

• Pythagorean Theorem -- Formula, Proof, and Examples
• Pythagorean Theorem Examples with Step-by-Step Solutions
• Distance Formula -- How to Find Distance Between Two Points

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