MathsICSE Class 9 · Pythagoras Theorem

Pythagoras Theorem: Statement, Proof by Areas and Applications (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

In a right-angled triangle the square of the hypotenuse equals the sum of the squares of the other two sides: c² = a² + b². Use it to find an unknown side, the diagonal of a rectangle, the side of a rhombus from its diagonals, the altitude of an isosceles or equilateral triangle, and to prove relations such as AB² + CD² = AC² + BD².

Big square = four triangles + the tilted square; the 2ab terms cancel, leaving c² = a² + b².

Illustration drawn for QuestionBuddy.

Understand Pythagoras Theorem: Statement, Proof and Applications

Statement. In a right-angled triangle, $(hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}$ , i.e. $c^{2} = a^{2} + b^{2}$ . The hypotenuse is opposite the right angle and is the longest side.

Area-based proof. Take four copies of a right triangle with legs $a, b$ and hypotenuse $c$ . Arrange them inside a square of side $(a + b)$ so that the hypotenuses form a tilted square of side $c$ in the middle. Area of the big square $= (a + b)^{2} = 4 \cdot \frac{1}{2} ab + c^{2}$ . Hence $a^{2} + 2 ab + b^{2} = 2 ab + c^{2}$ and $c^{2} = a^{2} + b^{2}$ .

Standard applications.

*Ladder / pole / wire:* the wall, ground and ladder form a right triangle; height $= ladder^{2} - foot^{2}$ .
*Rectangle:* diagonal $= l^{2} + b^{2}$ ; given a side and the diagonal, the other side is $d^{2} - s^{2}$ .
*Rhombus:* diagonals bisect at right angles, so side $= (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$ .
*Isosceles triangle:* altitude to the base bisects it; $h = a^{2} - (\frac{b}{2})^{2}$ for equal sides $a$ and base $b$ .
*Equilateral triangle, side $a$ :* altitude $= \frac{3}{2} a$ and area $= \frac{3}{4} a^{2}$ .

A proof pattern. In $△ A B C$ , if $A D ⊥ B C$ then $A B^{2} = A D^{2} + B D^{2}$ and $A C^{2} = A D^{2} + C D^{2}$ . Subtracting, $A B^{2} - B D^{2} = A C^{2} - C D^{2}$ , i.e. $A B^{2} + C D^{2} = A C^{2} + B D^{2}$ . The common altitude $A D$ cancels.

Check. The hypotenuse must be longer than either leg; if your answer is shorter, you subtracted when you should have added (or conversely).

Level. Class IX is a school-exam class. Direct questions are routine; relations proved with a perpendicular are at the hardest school-exam level.

Pythagoras Theorem: Statement, Proof and Applications: formulas to remember

Pythagoras: $c^{2} = a^{2} + b^{2}$; c is the hypotenuse.
Leg from hypotenuse: $a = c^{2} - b^{2}$; Subtract.
Rhombus side: $s = (d_{1} /2)^{2} + (d_{2} /2)^{2}$; Diagonals bisect at 90°.
Equilateral altitude: $h = \frac{3}{2} a$; Area = (√3/4) a².
Perpendicular relation: $A B^{2} + C D^{2} = A C^{2} + B D^{2}$; AD ⟂ BC.

How to solve Pythagoras Theorem: Statement, Proof and Applications questions

Mark the right angle, identify the hypotenuse, then add squares for it or subtract squares for a leg.

1Locate the right angle (or draw an altitude to create one).
2Label the hypotenuse $c$ (opposite the right angle).
3Write $c^{2} = a^{2} + b^{2}$ and substitute.
4Simplify the square root; leave a surd if it is not a perfect square.
5Check that the hypotenuse is the longest side.
6Add units and finish the question (area, perimeter, cost).

Pythagoras Theorem: Statement, Proof and Applications 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.

A ladder on a wall

Suppose a 17 m ladder rests against a wall with its foot 8 m from the base of the wall.

How high does it reach?

Try it first, then show the solutionSolution

$h = 1 7^{2} - 8^{2} = 225 = 15$ m.

Answer: 15 m.

Memory hook: 8-15-17 triple.

Wire between two poles

Imagine poles 12 m and 7 m tall standing 12 m apart on level ground with a wire joining their tops, costing ₹30 per metre.

Find the wire length and its cost.

Try it first, then show the solutionSolution

Height difference $= 5$ m; length $= 1 2^{2} + 5^{2} = 13$ m.
Cost $= 13 \times 30 = ₹390$ .

Answer: 13 m; ₹390.

Memory hook: Horizontal gap and height difference are the legs.

Tricks to remember Pythagoras Theorem: Statement, Proof and Applications

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

Hypotenuse is opposite the right angle
It is always the longest side.
Use it when: Choosing which side to square and add.
Add for the long side, subtract for a short one
Hypotenuse: add squares. Leg: subtract.
Use it when: Avoiding sign errors.
Common altitude cancels
In proofs with a perpendicular, AD² appears in both equations and cancels.
Use it when: Relation proofs.

Pythagoras Theorem: Statement, Proof and Applications: solved examples

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

Example 1: Hypotenuse · 1 mark · MCQBoard level

The legs of a right triangle are $9$ cm and $12$ cm. The hypotenuse is:
(a) 13 cm
(b) 15 cm
(c) 18 cm
(d) 21 cm

Try it first, then show the solutionSolution

$81 + 144 = 225 = 15$ .

Answer: (b) 15 cm

Example 2: Ladder · 3 marks · Short answerBoard level

A $17$ m ladder leans against a wall with its foot $8$ m from the wall. How high up the wall does it reach?

Try it first, then show the solutionSolution

$h^{2} = 1 7^{2} - 8^{2} = 289 - 64 = 225$ .
$h = 15$ m.

Answer: 15 m

Example 3: Rhombus · 3 marks · Short answerHarder board level

The two diagonals of a rhombus measure $18$ cm and $80$ cm. Find the length of its side and its perimeter.

Try it first, then show the solutionSolution

The diagonals bisect at right angles: half-diagonals $9$ and $40$ .
Side $= 9^{2} + 4 0^{2} = 1681 = 41$ cm.
Perimeter $= 4 \times 41 = 164$ cm.

Answer: Side 41 cm; perimeter 164 cm

Example 4: Rectangle · 4 marks · Short answerHarder board level

A diagonal of a rectangle is $25$ cm and one side is $20$ cm. Find the other side, the perimeter and the area.

Try it first, then show the solutionSolution

Other side $= 2 5^{2} - 2 0^{2} = 225 = 15$ cm.
Perimeter $= 2 (20 + 15) = 70$ cm.
Area $= 20 \times 15 = 300$ cm².

Answer: 15 cm; 70 cm; 300 cm²

Example 5: A relation with a perpendicular · 5 marks · Long answerHOTS / competency

In $△ A B C$ , $A D ⊥ B C$ with $D$ on $B C$ . Prove that $A B^{2} + C D^{2} = A C^{2} + B D^{2}$ . Verify for $A D = 12$ , $B D = 5$ , $C D = 9$ .

Try it first, then show the solutionSolution

In right $△ A D B$ : $A B^{2} = A D^{2} + B D^{2}$ ... (1).
In right $△ A D C$ : $A C^{2} = A D^{2} + C D^{2}$ ... (2).
Subtract: $A B^{2} - A C^{2} = B D^{2} - C D^{2}$ , so $A B^{2} + C D^{2} = A C^{2} + B D^{2}$ . Hence proved.
Numbers: $A B = 144 + 25 = 13$ , $A C = 144 + 81 = 15$ .
$A B^{2} + C D^{2} = 169 + 81 = 250$ and $A C^{2} + B D^{2} = 225 + 25 = 250$ . ✓

Answer: Proved; both sides equal 250.

Where students lose marks in Pythagoras Theorem: Statement, Proof and Applications

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

Mistake: Adding when a leg is wanted
To find a leg, subtract.
Wrong working
$1 7^{2} + 8^{2}$ for the height.
Right working
$1 7^{2} - 8^{2} = 15$ .
Do this: Check which side is the hypotenuse.
Usually costs: Whole answer.
Mistake: Not squaring the half-diagonals
A rhombus uses half the diagonals.
Wrong working
Side $= 1 6^{2} + 3 0^{2}$ .
Right working
$8^{2} + 1 5^{2} = 17$ .
Do this: Halve first.
Usually costs: Whole answer.
Mistake: Ignoring units
Unit lost.
Wrong working
15
Right working
15 m
Do this: Write units.
Usually costs: One mark.

Exam tips

Class IX is examined in school papers; the area-based proof is a standard 4 to 5 mark theorem.
Mark the right angle with a small square and the hypotenuse in a different colour.
In relation proofs, label the equations (1) and (2) and show the subtraction.
Know the triples (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25).

Practice questions on Pythagoras Theorem: Statement, Proof and Applications

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
In a right triangle the longest side is $25$ cm and another side is $7$ cm. The third side is:
(a) 18 cm
(b) 20 cm
(c) 24 cm
(d) 32 cm
Hint
$625 - 49$ .
Answer
(c) 24 cm
Q2 · 3 marks · Short answerHarder board level
Find the altitude and the area of an equilateral triangle of side $12$ cm.
Hint
The altitude bisects the base.
Answer
$h = 144 - 36 = 108 = 63 \approx 10.39$ cm; area $= \frac{1}{2} \times 12 \times 63 = 363 \approx 62.35$ cm².
Q3 · 3 marks · Short answerHarder board level
An isosceles triangle has sides $17$ cm, $17$ cm and $16$ cm. Find the altitude to the base and the area.
Hint
Half the base is 8.
Answer
$h = 289 - 64 = 15$ cm; area $= \frac{1}{2} \times 16 \times 15 = 120$ cm².
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a right-angled triangle the hypotenuse is the longest side.
Reason (R): The square of the hypotenuse equals the sum of the squares of the other two sides, so it exceeds each of them.
(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
$c^{2} = a^{2} + b^{2} > a^{2}$ .
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Two poles, $12$ m and $7$ m tall, stand $12$ m apart on level ground. A wire joins their tops and costs ₹30 per metre.
(a) Find the difference in the heights.
(b) Find the length of the wire.
(c) Find the cost of the wire.
Hint
Draw the horizontal from the lower top.
Answer
(a) 5 m (b) 13 m (c) ₹390

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 Pythagoras theorem.Show answer
In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
2.How do you find the side of a rhombus from its diagonals?Show answer
Square half of each diagonal, add, and take the root.
3.Altitude of an equilateral triangle of side a?Show answer
(√3/2)a.
4.Which side is the hypotenuse?Show answer
The one opposite the right angle.

Pythagoras Theorem: Statement, Proof and Applications: frequently asked questions

Do I have to prove the theorem?

The area-based proof is part of the ICSE Class IX syllabus.

Is the surd answer acceptable?

Yes; leave it simplified, with a decimal if asked.

Can the triangle be non-right?

No; the theorem needs a right angle.