Pythagoras Theorem and Its Converse
There is a fixed relationship among the lengths of the three sides of every right-angled triangle. This relationship is called the Pythagoras Theorem. It helps us find an unknown side of a right triangle. Its converse also helps us decide whether a triangle is right-angled when its three side lengths are known.
📚 Understanding the relation through areas
Suppose the two perpendicular sides of a right triangle have lengths a and b, and the hypotenuse has length c.
If squares are drawn on these three sides, their areas are respectively
a², b² and c².
The Pythagorean relation is
c² = a² + b².
In words,
Area of the square on the hypotenuse = sum of the areas of the squares on the other two sides.
📚 A simple example
If the perpendicular sides of a right triangle are 3 units and 4 units, and the hypotenuse is 5 units, then
5² = 3² + 4²25 = 9 + 16.
So the areas of the three squares satisfy the same relation.
📚 An area-based verification
Take two congruent right triangles whose perpendicular sides are a and b, and whose hypotenuse is c. Arrange them to form a trapezium.
The parallel sides of the trapezium have lengths a and b, and the distance between them is a + b. Therefore,
Area = 1/2 × (a + b) × (a + b).
The same region can also be divided into two right triangles and the middle triangle. Therefore,
1/2(a + b)² = 1/2 ab + 1/2 ab + 1/2 c².
Multiplying both sides by 2,
(a + b)² = 2ab + c².
So,
a² + 2ab + b² = 2ab + c².
Hence,
a² + b² = c².
This area interpretation makes the basic Pythagorean relation easy to visualise.
⭐ Theorem 49: Pythagoras Theorem
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
| Given: | △ABC is a right triangle and ∠A = 90°. Therefore, BC is the hypotenuse. |
| To prove: | BC² = AB² + AC². |
| Construction: | From A, draw AD ⟂ BC, meeting the hypotenuse BC at D. |
| Proof: | In △ABD and △CBA, ∠ADB = ∠CAB = 90° and ∠ABD = ∠CBA. Therefore, △ABD ~ △CBA. |
| ∴ | AB/BC = BD/AB. |
| ∴ | AB² = BC × BD. ...(i) |
| Again, | In △CAD and △CBA, ∠CDA = ∠CAB = 90° and ∠ACD = ∠BCA. Therefore, △CAD ~ △CBA. |
| ∴ | AC/BC = DC/AC. |
| ∴ | AC² = BC × DC. ...(ii) |
| Now, | Adding (i) and (ii), AB² + AC² = BC × BD + BC × DC. |
| ∴ | AB² + AC² = BC(BD + DC) = BC × BC = BC². |
| Therefore, | BC² = AB² + AC². [Proved] |
📚 General form of the theorem
If the hypotenuse is c and the other two sides are a and b, then
c² = a² + b².
Therefore,
c = √(a² + b²).
To find an unknown perpendicular side,
a = √(c² - b²)or
b = √(c² - a²).
📚 Historical note: Baudhāyana
Ancient Indian mathematical texts associated with Baudhāyana contain a statement equivalent to the Pythagorean relation for the diagonal of a rectangle. This work predates Pythagoras by several centuries and is an important part of the history of geometry in India.
⭐ Theorem 50: Converse of Pythagoras Theorem
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
| Given: | △ABC satisfies AB² = AC² + BC². |
| To prove: | ∠ACB = 90°. |
| Construction: | Take FE = CB. At F, draw a perpendicular to FE and mark FD = AC on it. Join D and E. |
| Proof: | Given AB² = BC² + AC². Also, EF = BC and DF = AC. Hence, |
| ∴ | AB² = EF² + DF². |
| But, | △DFE is right-angled because DF ⟂ FE. By Pythagoras theorem, DE² = DF² + EF². |
| ∴ | AB² = DE², so AB = DE. |
| Now, | In △ABC and △DEF, AB = DE, BC = EF and AC = DF. |
| ∴ | △ABC ≅ △DEF. [SSS congruence criterion] |
| ∴ | ∠ACB = ∠DFE. |
| But, | ∠DFE = 90°. |
| Therefore, | ∠ACB = 90°. [Proved] |
📚 Identifying a right triangle using the converse
Let the largest side of a triangle be c and the other two sides be a and b. Check whether
c² = a² + b².
If the equality is true, the angle opposite the largest side c is 90°, so the triangle is right-angled.
Example: 5, 12, 13
13² = 169and
5² + 12² = 25 + 144 = 169.
Therefore, 13² = 5² + 12². Hence the triangle is right-angled, and the side of length 13 is the hypotenuse.
📚 Generating Pythagorean triples
The algebraic identity
(m² - n²)² + (2mn)² = (m² + n²)²shows that, for m > n,
m² - n², 2mn and m² + n²
can be used as the side lengths of a right triangle. The hypotenuse is
m² + n².
Example 1
For m = 2, n = 1,
m² - n² = 3, 2mn = 4, m² + n² = 5.
So 3, 4, 5 is a Pythagorean triple.
Example 2
For m = 3, n = 2,
m² - n² = 5, 2mn = 12, m² + n² = 13.
So 5, 12, 13 is another Pythagorean triple.
📚 Solved examples
Example 3: Find the hypotenuse
The perpendicular sides of a right triangle are 6 cm and 8 cm. Find its hypotenuse.
c² = 6² + 8² = 36 + 64 = 100Therefore,
c = 10 cm.
Example 4: Find an unknown side
The hypotenuse of a right triangle is 17 cm and one perpendicular side is 8 cm. Find the other side.
Let the unknown side be x.
x² + 8² = 17² x² = 289 - 64 = 225Therefore,
x = 15 cm.
Example 5: Is the triangle right-angled?
Check a triangle with sides 7 cm, 24 cm and 25 cm.
The largest side is 25.
25² = 625and
7² + 24² = 49 + 576 = 625.
Therefore, the triangle is right-angled.
✍️ Practice problems
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A right triangle has perpendicular sides 9 cm and 12 cm. Find its hypotenuse.
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The hypotenuse of a right triangle is 13 cm and one perpendicular side is 5 cm. Find the other side.
-
Check whether a triangle with sides 8 cm, 15 cm and 17 cm is right-angled.
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Check whether a triangle with sides 6 cm, 8 cm and 11 cm is right-angled.
-
A rectangular field is 24 m long and 10 m wide. Find the length of its diagonal.
-
Use m = 4 and n = 1 to generate a Pythagorean triple.
Answers
- 15 cm
- 12 cm
- Yes, because 17² = 8² + 15².
- No, because 11² ≠ 6² + 8².
- 26 m
- 15, 8, 17
⚠️ Common mistakes
👉 Mistake: Using a² + b² = c² for every triangle.
Correct: Pythagoras theorem applies directly only to a right-angled triangle.
👉 Mistake: Taking any side as c.
Correct: c must be the side opposite the right angle, that is, the hypotenuse.
👉 Mistake: Adding squares while finding an unknown perpendicular side when the hypotenuse is known.
Correct: Square of the unknown perpendicular side = square of the hypotenuse − square of the other perpendicular side.
👉 Mistake: Testing the square of the smallest side when three side lengths are given.
Correct: In the converse theorem, compare the square of the largest side with the sum of the squares of the other two sides.
👉 Mistake: Treating 5² + 12² = 13² as only a numerical coincidence.
Correct: The equality allows us to conclude, by the converse theorem, that the triangle is right-angled.
📚 Important points to remember
👉 In a right triangle, the side opposite the right angle is the hypotenuse, and it is the longest side.
👉 If ∠A = 90°, then BC² = AB² + AC².
👉 In general, c² = a² + b².
👉 Unknown hypotenuse: c = √(a² + b²).
👉 Unknown perpendicular side: a = √(c² - b²) or b = √(c² - a²).
👉 Converse theorem: if c² = a² + b², then the angle opposite c is a right angle.
👉 3, 4, 5, 5, 12, 13 and 8, 15, 17 are common Pythagorean triples.
👉 (m² - n²), 2mn, (m² + n²) generates a Pythagorean triple when m > n.