MathsICSE Class 9 · Chapter 10

Pythagoras Theorem ICSE Class 9: Area-Based Proof, Converse and Applications

Updated 15 October 2026 · Fact-checked

Pythagoras theorem says that in a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides: c² = a² + b². Its converse says that if c² = a² + b², the angle opposite c is a right angle. ICSE Class 9 asks for an area-based proof and simple applications of both.

What Pythagoras Theorem is about

In a right-angled triangle, the side opposite the right angle is the hypotenuse; it is always the longest side. Pythagoras theorem links the three sides: if the legs are $a$ and $b$ and the hypotenuse is $c$ , then $c^{2} = a^{2} + b^{2}$ .

The CISCE 2028 Class IX syllabus asks for an area-based proof of the theorem and simple applications of the theorem and its converse. Area-based means the proof compares areas of squares and triangles, rather than using similar triangles.

In the school exam you will prove the theorem (4 to 5 marks), find missing sides in ladders, rectangles, rhombuses and isosceles triangles, test whether three lengths form a right triangle, and prove relations such as $A B^{2} - A C^{2} = B D^{2} - C D^{2}$ when $A D ⊥ B C$ .

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Pythagoras Theorem is in the 4. Geometry (i) Triangles (d) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Statement and the hypotenuse

Pythagoras theorem: in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

In $△ A B C$ with $∠ B = 9 0^{\circ}$ : $A C^{2} = A B^{2} + B C^{2}$ .

To find a leg, subtract: $A B^{2} = A C^{2} - B C^{2}$ . Always identify the hypotenuse first: it is opposite the right angle and is the longest side.

Worked example

The hypotenuse of a right triangle is 25 cm and one leg is 7 cm. Find the other leg.

Try it first, then show the solutionSolution

Other leg $^{2} = 2 5^{2} - 7^{2} = 625 - 49 = 576$ .
Other leg $= 24$ cm.

Answer: 24 cm

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 each corner holds a right angle and the hypotenuses form a tilted inner quadrilateral.

1. Each side of the inner figure is $c$ . Each of its angles is $18 0^{\circ} - (sum of the two acute angles) = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so it is a square of area $c^{2}$ .
2. Area of the big square $= (a + b)^{2}$ .
3. It is made of the inner square and four triangles: $(a + b)^{2} = c^{2} + 4 \times \frac{1}{2} ab$ .
4. $a^{2} + 2 ab + b^{2} = c^{2} + 2 ab$ , so $a^{2} + b^{2} = c^{2}$ .

An alternative area proof (Euclid's) draws squares on all three sides and shows each smaller square equals a rectangle inside the largest square, using congruent triangles and 'triangle = half the parallelogram on the same base and between the same parallels'.

Worked example

In the four-triangle proof with $a = 3$ and $b = 4$ , check that the areas agree.

Try it first, then show the solutionSolution

Big square: $(3 + 4)^{2} = 49$ .
Four triangles: $4 \times \frac{1}{2} \times 3 \times 4 = 24$ ; inner square $c^{2} = 25$ .
$24 + 25 = 49$ ✓, and $c = 5$ .

Answer: 49 = 24 + 25; c = 5

The converse

Converse: if the square on one side of a triangle equals the sum of the squares on the other two sides, then the angle opposite the first side is a right angle.

Proof outline: in $△ A B C$ let $A C^{2} = A B^{2} + B C^{2}$ . Construct $△ P QR$ with $∠ Q = 9 0^{\circ}$ , $P Q = A B$ and $QR = B C$ . By Pythagoras, $P R^{2} = P Q^{2} + Q R^{2} = A C^{2}$ , so $P R = A C$ . The triangles are congruent (SSS), so $∠ B = ∠ Q = 9 0^{\circ}$ .

To test three lengths, square them and check whether the largest square equals the sum of the other two.

Worked example

Is a triangle with sides 8 cm, 15 cm and 17 cm right-angled? If so, which angle is the right angle?

Try it first, then show the solutionSolution

$8^{2} + 1 5^{2} = 64 + 225 = 289 = 1 7^{2}$ .
By the converse, the triangle is right-angled.
The right angle is opposite the 17 cm side.

Answer: Yes; the right angle faces the 17 cm side.

Pythagorean triples

Whole numbers $(a, b, c)$ with $a^{2} + b^{2} = c^{2}$ are Pythagorean triples. The common ones: $(3, 4, 5)$ , $(5, 12, 13)$ , $(8, 15, 17)$ , $(7, 24, 25)$ , $(20, 21, 29)$ .

Any multiple of a triple is a triple: $(6, 8, 10)$ , $(9, 12, 15)$ , $(15, 20, 25)$ .

A general family: for a whole number $m > 1$ , $(2 m, m^{2} - 1, m^{2} + 1)$ is a triple, since $(2 m)^{2} + (m^{2} - 1)^{2} = m^{4} + 2 m^{2} + 1 = (m^{2} + 1)^{2}$ .

Worked example

Use $(2 m, m^{2} - 1, m^{2} + 1)$ with $m = 6$ to write a triple, and check it.

Try it first, then show the solutionSolution

$2 m = 12$ , $m^{2} - 1 = 35$ , $m^{2} + 1 = 37$ .
$144 + 1225 = 1369 = 3 7^{2}$ ✓.

Answer: (12, 35, 37)

Standard applications

Look for the hidden right triangle:

Rectangle: diagonal $^{2} = l^{2} + b^{2}$ .
Rhombus: diagonals bisect each other at right angles, so side $^{2} = (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$ .
Isosceles triangle: the altitude to the base bisects it.
Equilateral triangle of side $a$ : height $= \frac{3}{2} a$ .
Ladder, poles, travel north then east: draw the right angle between the wall and the ground, or between two directions at $9 0^{\circ}$ .

Worked example

The diagonals of a rhombus are 16 cm and 30 cm. Find its side and perimeter.

Try it first, then show the solutionSolution

Half-diagonals: 8 cm and 15 cm, meeting at $9 0^{\circ}$ .
Side $= 64 + 225 = 289 = 17$ cm.
Perimeter $= 68$ cm.

Answer: Side 17 cm, perimeter 68 cm

Proving relations with Pythagoras

Many proofs use two right triangles that share a side.

Example result: if $A D ⊥ B C$ in $△ A B C$ , then $A B^{2} - A C^{2} = B D^{2} - C D^{2}$ , because $A B^{2} = A D^{2} + B D^{2}$ and $A C^{2} = A D^{2} + C D^{2}$ ; subtracting removes $A D^{2}$ .

Perpendicular diagonals: if the diagonals of quadrilateral $A B C D$ meet at right angles at $O$ , then $A B^{2} + C D^{2} = B C^{2} + D A^{2}$ (each side is a hypotenuse in one of four right triangles at $O$ ).

Worked example

In $△ A B C$ , $A D ⊥ B C$ , $A B = 13$ cm, $A C = 15$ cm and $A D = 12$ cm. Find $B C$ (with $D$ between $B$ and $C$ ).

Try it first, then show the solutionSolution

$B D = 169 - 144 = 5$ cm.
$D C = 225 - 144 = 9$ cm.
$B C = 5 + 9 = 14$ cm.

Answer: 14 cm

Maths in real life: Pythagoras Theorem

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.

Ladder against a wall

Suppose a 13 m ladder leans against a vertical wall with its foot 5 m from the wall on level ground.

How high up the wall does it reach? If the top slides down by 7 m, how far does the foot move out?

Try it first, then show the solutionSolution

Height $= 169 - 25 = 12$ m.
New height $= 5$ m; new foot distance $= 169 - 25 = 12$ m.
Foot moves out $12 - 5 = 7$ m.

Answer: 12 m; the foot moves 7 m.

Memory hook: The ladder is always the hypotenuse.

Distance between two pole tops

Imagine two vertical poles, 6 m and 11 m tall, standing 12 m apart on level ground, and a wire is to be stretched between their tops.

How long must the wire be?

Try it first, then show the solutionSolution

Difference in heights $= 5$ m; horizontal gap $= 12$ m.
Wire $= 25 + 144 = 13$ m.

Answer: 13 m.

Memory hook: Slice off the shorter pole to see the right triangle.

Travelling at right angles

Suppose a small plane flies 300 km due north and then 400 km due east.

How far is it from its starting point?

Try it first, then show the solutionSolution

North and east are at $9 0^{\circ}$ .
Distance $= 30 0^{2} + 40 0^{2} = 500$ km.

Answer: 500 km.

Memory hook: 3-4-5 scaled up by 100.

Topics in this chapter

Pythagoras Theorem formula sheet

Pythagoras theorem: $c^{2} = a^{2} + b^{2}$; c is the hypotenuse.
Finding a leg: $a = c^{2} - b^{2}$; Subtract from the hypotenuse.
Rectangle diagonal: $d = l^{2} + b^{2}$; Square: d = a√2.
Rhombus side: $s^{2} = (\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$; Diagonals bisect at 90°.
Equilateral height: $h = \frac{3}{2} a$; From a² = h² + (a/2)².
Triple family: $(2 m, m^{2} - 1, m^{2} + 1)$; m > 1 a whole number.

Pythagoras Theorem: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Figure for the area-based proof

Prove Pythagoras theorem using areas; 4 to 5 marks.

Draw

Draw a large square and mark each side divided into lengths a and b, going round in the same order.
Join the division points in order to form a tilted inner quadrilateral.
Label the four corner triangles: legs a and b, hypotenuse c.
Mark the right angles at the corners of the big square.

Label

a and b on each side of the big square
c on each side of the inner square
right angles at the four corners
angles x and y at a corner of the inner square

Watch out: Prove that the inner figure is a square (sides c, angles 90°) before using its area c²; examiners look for this step.

How the school exam asks Pythagoras Theorem

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find a side	Two sides of a right triangle given; find the third.	Identify the hypotenuse; add squares for it, subtract for a leg.
Is it right-angled?	Three lengths given; test with the converse and name the right angle.	Compare the largest square with the sum of the other two.
Real-life setting	Ladder against a wall, two poles, a journey north then east, a diagonal path.	Sketch, mark the right angle, label known lengths, then apply the theorem.
Rhombus, rectangle, isosceles triangle	Find a side, a diagonal or an altitude.	Use the property that creates the right angle (diagonals, altitude to base).
Prove the theorem	Prove Pythagoras theorem using areas, or prove the converse.	Figure, given, to prove, construction, proof with reasons.
Prove a relation	AB² − AC² = BD² − CD², AB² = 4AD² − 3AC², sums of squares in a quadrilateral.	Write Pythagoras for each right triangle; add or subtract to eliminate the shared side.

Tricks to remember Pythagoras Theorem

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

Hypotenuse stands alone
The hypotenuse goes alone on one side of the equals sign: c² = a² + b².
Use it when: Setting up the equation correctly.
Add for the long side, subtract for a short side
Want the hypotenuse? Add squares. Want a leg? Subtract.
Use it when: Finding the missing side.
Five triples
3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29.
Use it when: Spotting answers and checking.
2m, m² ± 1
Pick m, double it, square it and add or subtract 1: a new triple.
Use it when: Making or checking triples.

Pythagoras Theorem: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Which is a right triangle? · 1 mark · MCQExam level

Which set of lengths forms a right-angled triangle?
(a) 4, 5, 6
(b) 6, 8, 11
(c) 9, 40, 41
(d) 2, 3, 4

Try it first, then show the solutionSolution

$9^{2} + 4 0^{2} = 81 + 1600 = 1681 = 4 1^{2}$ .

Answer: (c)

Example 2: Altitude of an isosceles triangle · 3 marks · Short answerExam level

In $△ A B C$ , $A B = A C = 10$ cm and $B C = 12$ cm. Find the altitude from $A$ and the area.

Try it first, then show the solutionSolution

The altitude $A D$ bisects $B C$ : $B D = 6$ cm.
$A D = 100 - 36 = 8$ cm.
Area $= \frac{1}{2} \times 12 \times 8 = 48$ cm².

Answer: 8 cm; 48 cm²

Example 3: Equilateral triangle · 2 marks · Short answerExam level

Find the height of an equilateral triangle of side 8 cm, correct to two decimal places ( $3 = 1.732$ ).

Try it first, then show the solutionSolution

$h^{2} = 8^{2} - 4^{2} = 48$ , so $h = 43$ .
$h = 4 \times 1.732 = 6.928 \approx 6.93$ cm.

Answer: 4√3 ≈ 6.93 cm

Example 4: Unknown side in algebra · 3 marks · Short answerHarder exam level

The sides of a right triangle are $x$ , $x + 7$ and $x + 8$ cm, the last being the hypotenuse. Find $x$ .

Try it first, then show the solutionSolution

$(x + 8)^{2} = x^{2} + (x + 7)^{2}$ .
$x^{2} + 16 x + 64 = 2 x^{2} + 14 x + 49 \Rightarrow x^{2} - 2 x - 15 = 0$ .
$(x - 5) (x + 3) = 0$ , so $x = 5$ (length is positive). Sides 5, 12, 13.

Answer: x = 5

Example 5: Perpendicular diagonals · 4 marks · Long answerHarder exam level

The diagonals of quadrilateral $A B C D$ cut at right angles at $O$ . Prove that $A B^{2} + C D^{2} = B C^{2} + D A^{2}$ .

Try it first, then show the solutionSolution

In right triangles at $O$ : $A B^{2} = O A^{2} + O B^{2}$ , $C D^{2} = O C^{2} + O D^{2}$ .
$B C^{2} = O B^{2} + O C^{2}$ , $D A^{2} = O D^{2} + O A^{2}$ .
So $A B^{2} + C D^{2} = O A^{2} + O B^{2} + O C^{2} + O D^{2} = B C^{2} + D A^{2}$ .

Answer: Proved.

Example 6: Median in a right triangle · 4 marks · Long answerHarder exam level

In $△ A B C$ , $∠ C = 9 0^{\circ}$ and $D$ is the mid-point of $B C$ . Prove that $A B^{2} = 4 A D^{2} - 3 A C^{2}$ .

Try it first, then show the solutionSolution

In $△ A C D$ : $A D^{2} = A C^{2} + C D^{2} = A C^{2} + \frac{B C ^{2}}{4}$ .
So $4 A D^{2} = 4 A C^{2} + B C^{2}$ , i.e. $B C^{2} = 4 A D^{2} - 4 A C^{2}$ .
In $△ A C B$ : $A B^{2} = A C^{2} + B C^{2} = A C^{2} + 4 A D^{2} - 4 A C^{2} = 4 A D^{2} - 3 A C^{2}$ .

Answer: Proved.

Example 7: Two medians of a right triangle · 5 marks · Long answerHOTS / competency

In $△ A B C$ , $∠ A = 9 0^{\circ}$ and $B E$ , $C F$ are medians. Prove that $4 (B E^{2} + C F^{2}) = 5 B C^{2}$ .

Try it first, then show the solutionSolution

$E$ is the mid-point of $A C$ : $B E^{2} = A B^{2} + \frac{A C ^{2}}{4}$ , so $4 B E^{2} = 4 A B^{2} + A C^{2}$ .
$F$ is the mid-point of $A B$ : $C F^{2} = A C^{2} + \frac{A B ^{2}}{4}$ , so $4 C F^{2} = 4 A C^{2} + A B^{2}$ .
Add: $4 (B E^{2} + C F^{2}) = 5 A B^{2} + 5 A C^{2}$ .
$= 5 (A B^{2} + A C^{2}) = 5 B C^{2}$ .

Answer: Proved.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

A park with a diagonal path

A rectangular park is 80 m long and 60 m wide. A straight path runs along one diagonal. A lamp post stands at the mid-point of the diagonal, and a children's play area is a rhombus whose diagonals are 24 m and 10 m.

(i) · 3 marks
Find the length of the diagonal path, and how much shorter it is than walking along two sides.
Answer
Answer: 100 m; 40 m shorter
$8 0^{2} + 6 0^{2} = 100$ m. Two sides: $140$ m. Saving $= 40$ m.
(ii) · 3 marks
How far is the lamp post from each corner of the park? Give a reason.
Answer
Answer: 50 m
The diagonals of a rectangle are equal and bisect each other, so the mid-point is 50 m from all four corners.
(iii) · 4 marks
Find the fencing needed for the play area and check that its sides with the half-diagonals form a Pythagorean triple.
Answer
Answer: 52 m
Half-diagonals 12 m and 5 m: side $= 144 + 25 = 13$ m. Fencing $= 4 \times 13 = 52$ m. $(5, 12, 13)$ is a triple.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A rectangle is 24 cm long and its diagonal is 25 cm. Find its breadth and perimeter. (3 marks)

Step	Marks
$b^{2} = 2 5^{2} - 2 4^{2} = 49$	1
$b = 7$ cm	1
Perimeter $= 2 (24 + 7) = 62$ cm	1

The sides of a triangle are 9 cm, 12 cm and 15 cm. Show that it is right-angled, and find the length of the perpendicular from the right-angled vertex to the longest side. (4 marks)

Step	Marks
$9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$	1
By the converse, the angle opposite 15 cm is $9 0^{\circ}$	1
Area $= \frac{1}{2} \times 9 \times 12 = 54$ cm² $= \frac{1}{2} \times 15 \times h$	1
$h = 7.2$ cm	1

Prove Pythagoras theorem using areas. (5 marks)

Step	Marks
Figure: four right triangles (legs $a$ , $b$ , hypotenuse $c$ ) inside a square of side $a + b$	1
Inner figure has sides $c$ and each angle $18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ , so it is a square	1
Area of big square $= (a + b)^{2}$	1
Also $= c^{2} + 4 \times \frac{1}{2} ab$	1
$a^{2} + 2 ab + b^{2} = c^{2} + 2 ab$ , so $a^{2} + b^{2} = c^{2}$	1

Where students lose marks in Pythagoras Theorem

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

Mistake: Adding squares to find a leg
Students add even when the unknown is a leg, giving a 'leg' longer than the hypotenuse.
Wrong working
Leg $= 2 5^{2} + 7^{2}$
Right working
Leg $= 2 5^{2} - 7^{2} = 24$
Do this: Mark the hypotenuse first; subtract when it is known.
Usually costs: The answer.
Mistake: Taking the square root of each term separately
$a^{2} + b^{2}$ is not $a + b$ .
Wrong working
$9^{2} + 1 2^{2} = 9 + 12 = 21$
Right working
$81 + 144 = 15$
Do this: Add the squares first, then take one square root.
Usually costs: The answer.
Mistake: Not proving the inner figure is a square
In the area-based proof the inner quadrilateral has equal sides, but a rhombus is not necessarily a square; the right angles must be shown.
Do this: Use the angle sum: the two acute angles of each triangle add to $9 0^{\circ}$ .
Usually costs: 1 mark.
Mistake: Using the converse on the wrong side
Students check $a^{2} + c^{2} = b^{2}$ with the largest side on the right-hand side missing.
Do this: Always compare the square of the longest side with the sum of the other two.
Usually costs: 1 mark.
Mistake: Units of area for a length
Writing '5 cm²' for a side length.
Do this: Squares are an intermediate step; the final side is in cm, area in cm².
Usually costs: Accuracy mark.

Practice questions on Pythagoras Theorem

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 · MCQExam level
The diagonal of a square is $102$ cm. Its side is:
(a) 5 cm
(b) 10 cm
(c) 20 cm
(d) $52$ cm
Hint
d = a√2.
Answer
(b)
Q2 · 2 marks · Short answerExam level
Find the hypotenuse of a right triangle whose legs are 9 cm and 12 cm.
Hint
Add squares.
Answer
$81 + 144 = 15$ cm.
Q3 · 2 marks · Short answerExam level
Check whether 8 cm, 15 cm and 18 cm form a right triangle.
Hint
Compare 18² with 8² + 15².
Answer
$64 + 225 = 289 \neq = 324$ . Not right-angled.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): A triangle with sides 20, 21 and 29 is right-angled.
Reason (R): $20 + 21 > 29$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
R only shows a triangle exists.
Answer
(d). $400 + 441 = 841 = 2 9^{2}$ , so (A) is true; (R) is true but only shows the triangle exists.
Q5 · 3 marks · Short answerHarder exam level
In $△ A B C$ , $A D ⊥ B C$ . Prove that $A B^{2} - A C^{2} = B D^{2} - C D^{2}$ .
Hint
Write Pythagoras in ΔABD and ΔACD.
Answer
$A B^{2} = A D^{2} + B D^{2}$ and $A C^{2} = A D^{2} + C D^{2}$ . Subtracting: $A B^{2} - A C^{2} = B D^{2} - C D^{2}$ .
Q6 · 3 marks · Short answerHarder exam level
A 10 m ladder reaches a window 8 m above the ground on one side of a street. Turning it about its foot, it reaches a window 6 m high on the other side. Find the width of the street.
Hint
Two right triangles sharing the foot.
Answer
Distances from the foot: $100 - 64 = 6$ m and $100 - 36 = 8$ m. Width $= 14$ m.
Q7 · 3 marks · Short answerHarder exam level
In $△ P QR$ , $∠ Q = 9 0^{\circ}$ , $P Q = 2 x$ cm, $QR = (x + 1)$ cm and $P R = (2 x + 1)$ cm. Find $x$ and the three sides.
Hint
(2x + 1)² = (2x)² + (x + 1)².
Answer
$4 x^{2} + 4 x + 1 = 4 x^{2} + x^{2} + 2 x + 1 \Rightarrow x^{2} - 2 x = 0 \Rightarrow x = 2$ . Sides 4, 3, 5 cm.
Q8 · 4 marks · Long answerHarder exam level
In $△ A B C$ , $∠ B = 9 0^{\circ}$ and $D$ is a point on $B C$ . Prove that $A C^{2} + B D^{2} = A D^{2} + B C^{2}$ .
Hint
Write AC² and AD² using ΔABC and ΔABD.
Answer
$A C^{2} = A B^{2} + B C^{2}$ , $A D^{2} = A B^{2} + B D^{2}$ . So $A C^{2} + B D^{2} = A B^{2} + B C^{2} + B D^{2} = A D^{2} + B C^{2}$ .
Q9 · 4 marks · Long answerHarder exam level
$A B C D$ is a rhombus. Prove that $A B^{2} + B C^{2} + C D^{2} + D A^{2} = A C^{2} + B D^{2}$ .
Hint
Each side² = (AC/2)² + (BD/2)².
Answer
Diagonals bisect at right angles at $O$ : $A B^{2} = \frac{A C ^{2}}{4} + \frac{B D ^{2}}{4}$ , and the same for each side. Sum of four sides squared $= A C^{2} + B D^{2}$ .
Q10 · 3 marks · Short answerHarder exam level
In an equilateral triangle $A B C$ , $A D ⊥ B C$ . Prove that $3 A B^{2} = 4 A D^{2}$ .
Hint
BD = AB/2.
Answer
$A D^{2} = A B^{2} - \frac{A B ^{2}}{4} = \frac{3 A B ^{2}}{4}$ , so $4 A D^{2} = 3 A B^{2}$ .
Q11 · 4 marks · Long answerHOTS / competency
A point $O$ inside rectangle $A B C D$ is joined to the four corners. Prove that $O A^{2} + O C^{2} = O B^{2} + O D^{2}$ .
Hint
Draw a line through O parallel to AB, meeting AD at P and BC at Q.
Answer
With $P$ on $A D$ and $Q$ on $B C$ : $O A^{2} = O P^{2} + A P^{2}$ , $O C^{2} = O Q^{2} + Q C^{2}$ , $O B^{2} = O Q^{2} + B Q^{2}$ , $O D^{2} = O P^{2} + P D^{2}$ . Since $A P = B Q$ and $P D = QC$ (ABQP and PQCD are rectangles), $O A^{2} + O C^{2} = O P^{2} + O Q^{2} + B Q^{2} + P D^{2} = O B^{2} + O D^{2}$ .
Q12 · 4 marks · Long answerHOTS / competency
A right triangle has perimeter 30 cm and hypotenuse 13 cm. Find the other two sides.
Hint
a + b = 17 and a² + b² = 169.
Answer
$(a + b)^{2} = 289 = 169 + 2 ab$ , so $ab = 60$ . $a + b = 17$ , $ab = 60$ : $a = 12$ , $b = 5$ . Sides 5 cm and 12 cm.

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 Pythagoras theorem.Show answer
In a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.
2.State the converse.Show answer
If c² = a² + b² for the sides of a triangle, the angle opposite c is a right angle.
3.Which side is the hypotenuse?Show answer
The side opposite the right angle; it is the longest.
4.Diagonal of a 6 × 8 rectangle?Show answer
10.
5.Height of an equilateral triangle of side a?Show answer
(√3/2)a.
6.Name three Pythagorean triples.Show answer
3-4-5, 5-12-13, 8-15-17.
7.Is 5, 6, 8 a right triangle?Show answer
No; 25 + 36 = 61 ≠ 64.
8.Side of a rhombus with diagonals 6 and 8?Show answer
5.

Best order to study it

Learn the statement and practise finding hypotenuses and legs.
Learn the area-based proof with its figure.
Learn the converse and its proof; practise testing triples.
Practise rectangles, rhombuses, isosceles and equilateral triangles, and ladder problems.
Finish with proofs of relations using two right triangles.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Do you identify the hypotenuse before writing the equation?
Can you write the area-based proof with a labelled figure?
Can you prove the converse using a constructed triangle and SSS?
Do you know the five common triples?
Can you find a rhombus side from its diagonals?
Can you prove AB² − AC² = BD² − CD²?
Can you set up a ladder or two-pole problem from words?

Pythagoras Theorem: frequently asked questions

Which proof should I learn?

The syllabus asks for an area-based proof. The four-triangle arrangement in a square of side a + b is short and clear; Euclid's proof with squares on all three sides is also area-based.

Does Pythagoras theorem work for all triangles?

No, only right-angled triangles. For other triangles, c² is greater or less than a² + b².

What if the answer is not a whole number?

Leave it as a surd (such as 4√3) or use the value given for √2 or √3, as the question asks.

Are Pythagorean triples examined?

Not as a separate topic, but recognising them saves time and is a quick check on answers.

How is the converse used?

To prove an angle is 90° from side lengths alone, for example to show a triangle with sides 9, 12, 15 is right-angled.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028