MathsICSE Class 9 · Pythagoras Theorem

Converse of Pythagoras, Pythagorean Triples and Proving Relations (ICSE Class 9)

Q: Is the converse proved in Class IX?

It is used without proof; you must apply it correctly.

Q: Can a triangle with integer sides but not a triple be right-angled?

No; a right triangle with integer sides is a triple.

Q: Do triples need to be in lowest terms?

No; multiples of a triple are also triples.

Updated 17 October 2026 · Fact-checked

If the square on the longest side of a triangle equals the sum of the squares on the other two sides, the triangle is right-angled, with the right angle opposite the longest side. Pythagorean triples such as (3, 4, 5), (5, 12, 13), (8, 15, 17) and (7, 24, 25) follow from (m² − n², 2mn, m² + n²).

Compare the two smaller squares with the largest one: 9 + 16 = 25 gives a right angle opposite the longest side; c is the longest side.

Illustration drawn for QuestionBuddy.

Understand Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations

Converse. If in $△ A B C$ we have $A C^{2} = A B^{2} + B C^{2}$ , then $∠ B = 9 0^{\circ}$ . To test a triangle: find the longest side, square all three sides, and see whether the largest square equals the sum of the other two. If it does, the right angle is opposite the longest side. (If the largest square is smaller than the sum, all angles are acute; if larger, one angle is obtuse.)

*Example.* $9, 40, 41$ : $81 + 1600 = 1681 = 4 1^{2}$ , so right-angled.

Pythagorean triples. Positive integers $(a, b, c)$ with $a^{2} + b^{2} = c^{2}$ . For integers $m > n > 0$ :
$ $a = m^{2} - n^{2}, b = 2 mn, c = m^{2} + n^{2} .$ $
With $m = 5$ , $n = 2$ : $(21, 20, 29)$ . Multiples of a triple are triples. Learn $(3, 4, 5)$ , $(5, 12, 13)$ , $(8, 15, 17)$ , $(7, 24, 25)$ , $(20, 21, 29)$ , $(9, 40, 41)$ .

Median and Pythagoras. In right $△ A B C$ with right angle at $C$ and $M$ the mid-point of $B C$ : $A M^{2} = A C^{2} + C M^{2} = A C^{2} + \frac{1}{4} B C^{2}$ .

Point inside a rectangle. For any point $O$ inside rectangle $A B C D$ : $O A^{2} + O C^{2} = O B^{2} + O D^{2}$ . Proof: through $O$ draw $P Q ∥ A B$ (meeting $A D$ at $P$ , $B C$ at $Q$ ) and $R S ∥ A D$ (meeting $A B$ at $R$ , $C D$ at $S$ ). Apply Pythagoras in the four right triangles at $O$ : $O A^{2} = A R^{2} + A P^{2}$ , $O C^{2} = C S^{2} + C Q^{2}$ , $O B^{2} = B R^{2} + B Q^{2}$ , $O D^{2} = D S^{2} + D P^{2}$ . Since $A R = D S$ , $A P = B Q$ , $B R = C S$ and $C Q = D P$ (opposite sides of rectangles), sums of the pairs match.

Level. Class IX is a school-exam class. The converse is routine; the point-in-rectangle proof is at the hardest school-exam level.

Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations: formulas to remember

Converse: $c^{2} = a^{2} + b^{2} \Rightarrow right angle opposite c$; c is the longest side.
Triples: $(m^{2} - n^{2}, 2 mn, m^{2} + n^{2})$; m > n > 0.
Median to a leg: $A M^{2} = A C^{2} + \frac{1}{4} B C^{2}$; Right angle at C.
Point in a rectangle: $O A^{2} + O C^{2} = O B^{2} + O D^{2}$; O inside ABCD.

How to solve Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations questions

Sort the sides, compare the largest square with the sum of the others, then state the right angle.

1Identify the longest side; it is the only possible hypotenuse.
2Compute the squares of all three sides.
3Compare: equal means right-angled; state where the right angle is.
4For triples, substitute $m, n$ into the three formulas and verify.
5For a proof, draw the perpendiculars needed and apply Pythagoras in each right triangle.

Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Checking a square corner

Suppose a carpenter measures a frame with sides 15 cm, 20 cm and a diagonal of 25 cm.

Is the corner a right angle? What is the area of the triangle?

Try it first, then show the solutionSolution

$1 5^{2} + 2 0^{2} = 225 + 400 = 625 = 2 5^{2}$ , so yes.
Area $= \frac{1}{2} \times 15 \times 20 = 150$ cm².

Answer: Yes; 150 cm².

Memory hook: A 3-4-5 triple scaled by 5.

Shortcut across a field

Imagine a rectangular field 40 m by 30 m, and a person walking across the diagonal instead of along two sides.

How much shorter is the diagonal path?

Try it first, then show the solutionSolution

Diagonal $= 4 0^{2} + 3 0^{2} = 50$ m.
Along the sides: $70$ m. Saving $= 20$ m.

Answer: 20 m shorter.

Memory hook: Diagonal < sum of the two sides.

Tricks to remember Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Longest first
Test only the longest side as the hypotenuse.
Use it when: Converse questions.
m and n
Triples come from m² − n², 2mn, m² + n².
Use it when: Generating triples.
Opposite pairs match
In the rectangle proof, opposite segments are equal in pairs.
Use it when: Pairing the squares.

Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Which is right-angled? · 1 mark · MCQBoard level

Which set of numbers can be the sides of a right-angled triangle?
(a) 6, 7, 8
(b) 7, 24, 25
(c) 5, 6, 10
(d) 8, 9, 10

Try it first, then show the solutionSolution

$7^{2} + 2 4^{2} = 49 + 576 = 625 = 2 5^{2}$ .

Answer: (b)

Example 2: Using the converse · 3 marks · Short answerBoard level

Show that a triangle with sides $11$ cm, $60$ cm and $61$ cm is right-angled and state where the right angle is.

Try it first, then show the solutionSolution

$1 1^{2} + 6 0^{2} = 121 + 3600 = 3721 = 6 1^{2}$ .
By the converse, the triangle is right-angled, with the right angle opposite the $61$ cm side (between the $11$ cm and $60$ cm sides).

Answer: Right-angled; the right angle is between the sides 11 cm and 60 cm.

Example 3: Generating a triple · 3 marks · Short answerHarder board level

Using $m = 5$ and $n = 2$ in $(m^{2} - n^{2}, 2 mn, m^{2} + n^{2})$ , find a Pythagorean triple and verify it.

Try it first, then show the solutionSolution

$m^{2} - n^{2} = 21$ , $2 mn = 20$ , $m^{2} + n^{2} = 29$ .
Check: $2 1^{2} + 2 0^{2} = 441 + 400 = 841 = 2 9^{2}$ . ✓

Answer: (21, 20, 29)

Example 4: A median in a right triangle · 4 marks · Short answerHarder board level

In $△ A B C$ , $∠ C = 9 0^{\circ}$ , $A C = 12$ cm and $B C = 10$ cm. $M$ is the mid-point of $B C$ . Prove that $A M^{2} = A C^{2} + \frac{1}{4} B C^{2}$ and find $A M$ .

Try it first, then show the solutionSolution

In right $△ A C M$ , $A M^{2} = A C^{2} + C M^{2}$ and $C M = \frac{1}{2} B C$ .
So $A M^{2} = A C^{2} + \frac{1}{4} B C^{2}$ .
$A M^{2} = 144 + 25 = 169$ , so $A M = 13$ cm.

Answer: Proved; AM = 13 cm

Example 5: Point inside a rectangle · 5 marks · Long answerHOTS / competency

$O$ is any point inside rectangle $A B C D$ . Prove that $O A^{2} + O C^{2} = O B^{2} + O D^{2}$ . Verify for a $8 \times 6$ rectangle with $O$ at $3$ units from $A D$ and $2$ units from $A B$ .

Try it first, then show the solutionSolution

Through $O$ draw $P Q ∥ A B$ and $R S ∥ A D$ , meeting the sides at $P, Q, R, S$ . All the angles at $O$ are right angles.
$O A^{2} = A R^{2} + A P^{2}$ , $O C^{2} = C S^{2} + C Q^{2}$ , $O B^{2} = B R^{2} + B Q^{2}$ , $O D^{2} = D S^{2} + D P^{2}$ .
With $A R = D S$ , $B R = C S$ , $A P = B Q$ , $D P = C Q$ : $O A^{2} + O C^{2} = A R^{2} + A P^{2} + C S^{2} + C Q^{2} = O B^{2} + O D^{2}$ after pairing $A R^{2} = D S^{2}$ , $C S^{2} = B R^{2}$ , $A P^{2} = B Q^{2}$ , $C Q^{2} = D P^{2}$ . Hence proved.
Numbers: with $A (0, 0)$ , $B (8, 0)$ , $C (8, 6)$ , $D (0, 6)$ , $O (3, 2)$ : $O A^{2} = 13$ , $O C^{2} = 41$ , $O B^{2} = 29$ , $O D^{2} = 25$ .
$13 + 41 = 54$ and $29 + 25 = 54$ . ✓

Answer: Proved; both sums are 54.

Where students lose marks in Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations

The slips examiners see most often, and what to write instead.

Mistake: Testing the wrong side as hypotenuse
Only the longest side can be the hypotenuse.
Wrong working
$9^{2} + 4 1^{2} = 4 0^{2}$ ?
Right working
$9^{2} + 4 0^{2} = 4 1^{2}$ .
Do this: Sort the sides.
Usually costs: Whole answer.
Mistake: Right angle at the wrong vertex
The right angle is opposite the longest side.
Wrong working
Right angle between 40 and 41.
Right working
Right angle between 9 and 40.
Do this: Name the vertex.
Usually costs: One mark.
Mistake: Assuming a triple from two numbers
Always verify the third.
Wrong working
(5, 12, 14) because 5 and 12.
Right working
(5, 12, 13).
Do this: Check the sum of squares.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; the converse is used without proof.
State the converse by name when you use it.
Show all three squares in a column before comparing.
In the rectangle proof, label the four feet of perpendiculars clearly.

Practice questions on Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these is a Pythagorean triple?
(a) 8, 15, 17
(b) 8, 15, 16
(c) 9, 15, 18
(d) 7, 15, 17
Hint
Check the largest square.
Answer
(a)
Q2 · 3 marks · Short answerHarder board level
The sides of a triangle are $(x - 1)$ , $x$ and $(x + 1)$ cm. For what $x$ is it right-angled? Name the triangle.
Hint
$(x - 1)^{2} + x^{2} = (x + 1)^{2}$ .
Answer
$x^{2} - 4 x = 0$ , so $x = 4$ (reject $0$ ): sides $3, 4, 5$ .
Q3 · 3 marks · Short answerHarder board level
Show that a triangle with sides $6.5$ cm, $6$ cm and $2.5$ cm is right-angled and find its area.
Hint
$6. 5^{2} = 42.25$ .
Answer
$6^{2} + 2. 5^{2} = 36 + 6.25 = 42.25 = 6. 5^{2}$ ; right-angled. Area $= \frac{1}{2} \times 6 \times 2.5 = 7.5$ cm².
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A triangle with sides $4$ cm, $5$ cm and $6$ cm is right-angled.
Reason (R): In a right-angled triangle the square of the longest side equals the sum of the squares of the other two sides.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
$16 + 25 = 41 \neq = 36$ .
Answer
(d)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A rectangular field is $40$ m long and $30$ m wide. A path runs along a diagonal. A second, triangular plot has sides $15$ m, $20$ m and $25$ m.
(a) Find the length of the diagonal path.
(b) How much shorter is it than walking along two sides?
(c) Show that the triangular plot is right-angled and find its area.
Hint
Both are 3-4-5 multiples.
Answer
(a) 50 m (b) 70 − 50 = 20 m (c) 225 + 400 = 625 = 25², right-angled; area 150 m².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the converse of Pythagoras.Show answer
If the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled.
2.Where is the right angle?Show answer
Opposite the longest side.
3.Triple from m = 3, n = 2?Show answer
(5, 12, 13).
4.What is OA² + OC² for a point in a rectangle?Show answer
It equals OB² + OD².

Converse of Pythagoras Theorem, Pythagorean Triples and Proving Relations: frequently asked questions

Is the converse proved in Class IX?

It is used without proof; you must apply it correctly.

Can a triangle with integer sides but not a triple be right-angled?

No; a right triangle with integer sides is a triple.

Do triples need to be in lowest terms?

No; multiples of a triple are also triples.