Pythagoras Theorem Exercises: Practice Questions, Solutions and Real-Life Applications

The Pythagoras Theorem is a fundamental principle in geometry that explains the relationship between the sides of a right-angled triangle. According to the theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Understanding this concept is essential for solving geometrical problems and developing analytical reasoning skills in mathematics. This section presents a structured set of Pythagorean theorem exercises designed to help learners understand and apply these concepts effectively through step-by-step solutions.

Table of Contents

• What Is the Pythagoras Theorem?
• Common Pythagorean Triples
• Exercises Set 1: Finding the Hypotenuse
• Exercises Set 2: Finding a Missing Leg
• Exercises Set 3: Checking If a Triangle Is Right-Angled
• Exercises Set 4: Word Problems & Real-Life Applications
• Exercises Set 5: Applied Geometry
• Common Mistakes to Avoid

What Is the Pythagoras Theorem?

The Pythagoras Theorem states:

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Mathematically:

a² + b² = c²

Where:

c is the hypotenuse (the longest side, opposite the right angle)

a and b are the two legs (the other two sides)

Common Pythagorean Triples

Pythagorean triples are sets of three positive integers that perfectly satisfy a² + b² = c². Memorising a few of these saves a lot of calculation time.

 

Any multiple of these triples also works: so 6, 8, 10 is valid (double of 3-4-5), as is 9, 12, 15, and so on.

Formula to generate Pythagorean triples: For any integer m > 1, the triple (2m, m²−1, m²+1) is a Pythagorean triple.

Exercises Set 1: Finding the Hypotenuse

Exercise 1.1: Find the hypotenuse of a right-angled triangle with legs 6 cm and 11 cm.

Solution: Given a = 6 cm and b = 11cm

Using a² + b² = c²

⇒ 6² + 11² = c²

⇒ 36 + 121 = c²

⇒ c² = 157

⇒ c = √157 ≈ 12.53 cm

Therefore, hypotenuse = 12.53 cm.

Exercise 1.2: The perpendicular sides of a right triangle are √3 cm and √6 cm. Find the hypotenuse.

Solution: Given a = √3  cm and b =  √6 cm

⇒ (√3)² + (√6)² = c²

⇒ 3 + 6 = c²

⇒ c² = 9

⇒ c = 3 cm

Therefore, hypotenuse = 3 cm.

Exercise 1.3: A right-angled triangle has legs of length 3 cm and 4 cm. Find the hypotenuse.

Solution: Given a = 3  cm and b =  4 cm

Using a² + b² = c²

⇒ 3² + 4² = c²

⇒ 9 + 16 = c²

⇒ c² = 25

⇒ c = 5 cm

Exercise 1.4: The two legs of a right triangle measure 5 cm and 12 cm. Find the length of the hypotenuse.

Solution: Given a = 5 cm and b =  12 cm

⇒ 5² + 12² = c²

⇒ 25 + 144 = c²

⇒ c² = 169

⇒ c = 13 cm

Therefore, hypotenuse = 13 cm.

Exercises Set 2: Finding a Missing Leg

Here, one leg is unknown. You rearrange the formula:

a² = c² − b² or b² = c² − a²

Exercise 2.1: The hypotenuse of a right triangle is 17 cm and one leg is 8 cm. Find the other leg.

Solution: Given c = 17 cm and b =  8 cm

Using a² + b² = c²

⇒ a² + 8² = 17²

⇒ a² + 64 = 289

⇒  a² = 289 − 64 = 225

⇒  a = 15 cm

This is the 8-15-17 Pythagorean triple.

Exercise 2.2: A right triangle has a hypotenuse of 16 cm and one leg of 12 cm. Find the missing leg.

Solution: Given c = 16 cm and a =  12 cm

Using a² + b² = c²

⇒ 12² + b² = 16²

⇒ 144 + b² = 256

⇒ b² = 256 − 144 = 112

⇒ b = √112 = 4√7 cm

Exercise 2.3: The hypotenuse of a right triangle is 4√2 cm. If the longest leg is half the hypotenuse (i.e., 2√2 cm), find the shortest leg.

Solution: Given c = 4√2 cm and b =  2√2 cm

Using a² + b² = c²

⇒ a² + (2√2)² = (4√2)²

⇒ a² + 8 = 32

⇒ a² = 24

⇒ a = 2√6 cm

 Therefore, the shortest leg is 2√6 cm

Exercise 2.4:  Find the width of a rectangle whose length is 144 cm and diagonal is 145 cm.

Solution: The diagonal of a rectangle divides it into two right-angled triangles. Using the Pythagoras theorem: width² + length² = diagonal²

⇒ width² + 144² = 145²

⇒ width² = 21025 − 20736

⇒ width² = 289

⇒ width = 17 cm

Therefore, width of the rectangle = 17 cm.

 

Exercises Set 3: Checking If a Triangle Is Right-Angled

The converse of the Pythagoras theorem says, 'If a² + b² = c², then the triangle is right-angled.'

Exercise 3.1: Check whether the sides 21 cm, 20 cm, and 29 cm form a right-angled triangle.

Solution: The largest side is 29 cm. Check if 21² + 20² = 29²

⇒  441 + 400 = 841

And 841 = 29² 

Yes, this is a right-angled triangle. (And 20, 21, 29 is a Pythagorean triple!)

Exercise 3.2: Do the sides 8 cm, 15 cm, and 16 cm form a right-angled triangle?

Solution: The largest side is 16 cm. Check if 8² + 15² = 16²

⇒ 64 + 225 = 289

289 ≠ 256

No, this is not a right-angled triangle.

Exercise 3.3: Are the sides 10 cm, 24 cm, and 26 cm a valid right-angled triangle?

Solution: The largest side is 26 cm. Check if 10² + 24² = 26²

10² + 24² = 100 + 576 = 676

26² = 676

Yes. The sides 10 cm, 24 cm, and 26 cm form a valid right-angled triangle. This is a multiple of the 5-12-13 triple, scaled by 2.

Exercises Set 4: Word Problems & Real-Life Applications

Exercise 4.1: A ladder leans against a wall. The foot of the ladder is 15 cm away from the base of the wall, and the top of the ladder reaches 8 cm up the wall. How long is the ladder?

Solution: The wall, the ground, and the ladder form a right-angled triangle.

⇒  Ladder² = 8² + 15²

⇒  Ladder² = 64 + 225 = 289

⇒  Ladder = 17 cm

Therefore, the ladder is 17 cm long.

Exercise 4.2: The diagonal of a square is 6 cm. Find the side of the square.

Solution: A diagonal cuts a square into two right-angled isosceles triangles. If the side is s, then:

⇒  s² + s² = 6²

⇒  2s² = 36

⇒  s² = 18

⇒  s = 3√2 cm

Therefore, the side of the square is 3√2 cm long

Exercise 4.3: A boy walks 24 km due east from his house, then turns left and walks 10 km due north. How far is he from his starting point (displacement)?

Solution: The path forms a right-angled triangle with legs 24 km and 10 km.

⇒  Distance² = 24² + 10²

⇒  Distance² = 576 + 100 = 676

⇒  Distance = 26 km

(This is the 5-12-13 triple scaled by 2.)

Therefore, the boy is 26 km away from the starting point.

Exercise 4.4: A television screen is described as 32 inches. This refers to the diagonal. If the screen width is 28 inches, find the height of the screen.

Solution: Given width = 28 inches and diagonal = 32 inches

⇒  Height² + 28² = 32²

⇒ Height² = 1024 − 784 = 240

⇒  Height = 4√15 ≈ 15.49 inches

Therefore the height of the screen is 15.49 inches

Exercise 4.5: A rectangular field is 40 m long and 30 m wide. A student walks along the diagonal instead of around two sides. How much distance does the student save?

Solution: Length of the field = 40 m

Width of the field = 30 m.

Diagonal = √(40² + 30²) = √(1600 + 900) = √2500 = 50 m

Distance around two sides = 40 + 30 = 70 m

Distance saved = 70 − 50 = 20 m

Exercises Set 5: Applied Geometry

Exercise 5.1: In a right isosceles triangle, the hypotenuse is 2√3 cm. Find the equal legs.

Solution: Let each equal leg = a.

⇒  a² + a² = (2√3)²

⇒  2a² = 12

⇒  a² = 6

⇒  a = √6 cm

Exercise 5.2: Find the height of an equilateral triangle with side 10 cm.

Solution:  The height bisects the base, creating two right-angled triangles with hypotenuse 10 cm and one leg = 5 cm.

⇒  h² + 5² = 10²

⇒  h² = 100 − 25 = 75

⇒  h = 5√3 cm ≈ 8.66 cm

Exercise 5.3: A cuboid has a base of 6 cm × 8 cm and a height of 10 cm. Find the length of the space diagonal (from one corner to the opposite corner).

Solution: Given, b = 6 cm, l = 8 cm and h = 10 cm.

Step 1: Find the base diagonal.

⇒  d_base = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

Step 2: Find the space diagonal.

⇒  d_space = √(d_base² + height²) = √(10² + 10²) = √200 = 10√2 cm

Exercise 5.4: Find the Pythagorean triple where one of the numbers is 6.

Solution: Using the formula for m = 3, the triple = (2m, m²−1, m²+1) = (6, 8, 10).

Check: 6² + 8² = 36 + 64 = 100 = 10² 

The Pythagorean triple is (6, 8, 10).

Common Mistakes to Avoid

1. Confusing the hypotenuse with a leg. Always identify the hypotenuse (side opposite 90°) before setting up the equation.

2. Forgetting to take the square root. Once you solve for c², you still need c = √c².

3. Applying the theorem to non-right triangles. The Pythagorean theorem only works when there is a 90° angle.

4. Rounding mid-calculation. Always keep values in exact form (like √157) until the final step unless a decimal answer is explicitly asked for.

5. Setting up the wrong equation in word problems. Sketch a diagram first; this helps eliminate most errors.