Two Squares on the Legs Fill the Square on the Hypotenuse
Prove the Pythagoras theorem by comparing areas, find any missing side of a right-angled triangle, use the converse to test for a right angle, and solve ladder, pole, rhombus and diagonal problems.
What does the Pythagoras theorem actually say about areas?
Most students meet this theorem as an equation about lengths. It began as a statement about areas, and that version is worth seeing because it is what makes the proof possible.
Take a right-angled triangle with legs cm and cm and hypotenuse cm. Build a square on each side. Their areas are
and . The two small squares together have exactly the area of the big one. Cut them up and they would fit inside it with nothing left over and no gaps.
That is the theorem: the square on the hypotenuse equals the sum of the squares on the other two sides. Written with letters, with the hypotenuse,
**The word square is doing double duty**, and keeping both meanings in view is useful. is a number when you are calculating and a region when you are proving.
This page covers the ICSE Class 9 Mathematics chapter on the Pythagoras theorem: an area-based proof, calculating a missing side, the converse as a test for a right angle, and applications to ladders, poles, diagonals and distances.
Take a right-angled triangle with legs cm and cm and hypotenuse cm. Build a square on each side. Their areas are
and . The two small squares together have exactly the area of the big one. Cut them up and they would fit inside it with nothing left over and no gaps.
That is the theorem: the square on the hypotenuse equals the sum of the squares on the other two sides. Written with letters, with the hypotenuse,
**The word square is doing double duty**, and keeping both meanings in view is useful. is a number when you are calculating and a region when you are proving.
This page covers the ICSE Class 9 Mathematics chapter on the Pythagoras theorem: an area-based proof, calculating a missing side, the converse as a test for a right angle, and applications to ladders, poles, diagonals and distances.
Formula
How do you prove the Pythagoras theorem by comparing areas?
Build one big square two different ways and set the two expressions for its area equal.
The construction. Take four copies of a right-angled triangle with legs and and hypotenuse . Arrange them inside a square of side , one in each corner, with the legs along the edges. The four hypotenuses then enclose a tilted square of side in the middle.
Now count the area twice.
First, directly from the outer square's side:
Second, as the four triangles plus the tilted square. Each triangle has area , so
The two counts describe the same region, so they are equal:
**The cancels, and that cancellation is the whole proof.** Notice which chapter supplied it: is the identity from expansions, and the four triangles are exactly the it produces.
One step in the construction needs justification, and examiners ask for it. Why is the inner figure a square rather than any rhombus? Because at each point where two triangles meet along an edge of the outer square, the two acute angles of the triangles sit together on a straight line with the corner angle of the inner figure. The acute angles of a right-angled triangle add to , so the inner angle must be . **All four inner angles are right angles and all four sides are , so it is a square.
Verify the whole thing numerically.** With and : the outer square has side and area ; the four triangles have total area ; so the inner square is , giving , as required.
The construction. Take four copies of a right-angled triangle with legs and and hypotenuse . Arrange them inside a square of side , one in each corner, with the legs along the edges. The four hypotenuses then enclose a tilted square of side in the middle.
Now count the area twice.
First, directly from the outer square's side:
Second, as the four triangles plus the tilted square. Each triangle has area , so
The two counts describe the same region, so they are equal:
**The cancels, and that cancellation is the whole proof.** Notice which chapter supplied it: is the identity from expansions, and the four triangles are exactly the it produces.
One step in the construction needs justification, and examiners ask for it. Why is the inner figure a square rather than any rhombus? Because at each point where two triangles meet along an edge of the outer square, the two acute angles of the triangles sit together on a straight line with the corner angle of the inner figure. The acute angles of a right-angled triangle add to , so the inner angle must be . **All four inner angles are right angles and all four sides are , so it is a square.
Verify the whole thing numerically.** With and : the outer square has side and area ; the four triangles have total area ; so the inner square is , giving , as required.
How do you calculate a missing side of a right-angled triangle?
Identify the hypotenuse first — it is the side opposite the right angle — then add or subtract. Adding finds the hypotenuse; subtracting finds a leg.
Worked example 1 — finding the hypotenuse. A right-angled triangle has legs cm and cm.
Worked example 2 — finding a leg. The hypotenuse is cm and one leg is cm.
The subtraction must have the hypotenuse first. Writing gives a negative number, which is the arithmetic telling you that you have named the sides wrongly — no leg can exceed the hypotenuse.
Worked example 3 — an answer in surd form. A right-angled isosceles triangle has both legs cm.
Leave surds in exact form unless a decimal is asked for, and simplify them as you learned in the surds chapter: , not carried through later working.
Worked example 4 — two steps. In , , cm and cm. is the perpendicular from to . Find , then the area of the triangle two ways, and hence .
The area using the two legs as base and height is cm. Using as the base and as the height gives the same area:
Check that this is sensible: must be shorter than both legs, since the perpendicular is the shortest distance from to the line — and , as the inequality chapter promised.
The triples worth recognising on sight save time in every later chapter: , , , and , together with all their multiples such as and . **If three numbers in a question are a multiple of , you do not need to square anything.**
Worked example 1 — finding the hypotenuse. A right-angled triangle has legs cm and cm.
Worked example 2 — finding a leg. The hypotenuse is cm and one leg is cm.
The subtraction must have the hypotenuse first. Writing gives a negative number, which is the arithmetic telling you that you have named the sides wrongly — no leg can exceed the hypotenuse.
Worked example 3 — an answer in surd form. A right-angled isosceles triangle has both legs cm.
Leave surds in exact form unless a decimal is asked for, and simplify them as you learned in the surds chapter: , not carried through later working.
Worked example 4 — two steps. In , , cm and cm. is the perpendicular from to . Find , then the area of the triangle two ways, and hence .
The area using the two legs as base and height is cm. Using as the base and as the height gives the same area:
Check that this is sensible: must be shorter than both legs, since the perpendicular is the shortest distance from to the line — and , as the inequality chapter promised.
The triples worth recognising on sight save time in every later chapter: , , , and , together with all their multiples such as and . **If three numbers in a question are a multiple of , you do not need to square anything.**
How does the converse tell you whether a triangle is right-angled?
If the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled — and the right angle is opposite the longest side.
This is a genuine test, not a restatement. The theorem starts from a right angle and produces an equation; the converse starts from the equation and produces the right angle.
Worked example 1. Is a triangle with sides cm, cm and cm right-angled?
The longest side is cm, so test it against the other two:
They agree, so the triangle is right-angled, with the right angle opposite the cm side.
Worked example 2. Is a triangle with sides cm, cm and cm right-angled?
, so it is not right-angled. And the comparison tells you more: since , the angle opposite the cm side is slightly more than , so the triangle is obtuse.
Worked example 3. Is a triangle with sides cm, cm and cm right-angled?
Yes — and this is one of the triples worth memorising.
The single most common error is testing the wrong side. For sides , and , checking against gives against and the wrong conclusion. Order the three lengths first and square the largest on its own side of the equation.
A boundary case that connects to the previous chapter. Try the lengths , and . The converse test gives , so not right-angled — but in fact no triangle exists at all, since . Run the triangle inequality first: the converse test is only meaningful once you know the three lengths actually close up into a triangle.
Where the converse is genuinely useful. It is how you check a right angle without a set square. A mason marking out the corner of a room measures units along one wall, along the other, and adjusts until the diagonal reads exactly — the converse of the theorem, used as a tool.
This is a genuine test, not a restatement. The theorem starts from a right angle and produces an equation; the converse starts from the equation and produces the right angle.
Worked example 1. Is a triangle with sides cm, cm and cm right-angled?
The longest side is cm, so test it against the other two:
They agree, so the triangle is right-angled, with the right angle opposite the cm side.
Worked example 2. Is a triangle with sides cm, cm and cm right-angled?
, so it is not right-angled. And the comparison tells you more: since , the angle opposite the cm side is slightly more than , so the triangle is obtuse.
Worked example 3. Is a triangle with sides cm, cm and cm right-angled?
Yes — and this is one of the triples worth memorising.
The single most common error is testing the wrong side. For sides , and , checking against gives against and the wrong conclusion. Order the three lengths first and square the largest on its own side of the equation.
A boundary case that connects to the previous chapter. Try the lengths , and . The converse test gives , so not right-angled — but in fact no triangle exists at all, since . Run the triangle inequality first: the converse test is only meaningful once you know the three lengths actually close up into a triangle.
Where the converse is genuinely useful. It is how you check a right angle without a set square. A mason marking out the corner of a room measures units along one wall, along the other, and adjusts until the diagonal reads exactly — the converse of the theorem, used as a tool.
How do you solve ladder, pole and diagonal problems?
Draw the figure, mark the right angle, and name the hypotenuse before writing any equation. Almost every application question hides a right-angled triangle whose right angle is at the ground, at a wall, or at a corner.
Worked example 1 — a ladder. A ladder m long rests against a vertical wall with its foot m from the wall. How high does it reach? If the foot is pulled out to m from the wall, how far does the top slide down?
The wall is vertical and the ground horizontal, so the right angle is at the base. The ladder is the hypotenuse:
After the foot is moved:
So the top slides down m.
Notice how unequal the two movements are. The foot moved out m and the top fell m here, but that is a coincidence of these numbers. The ladder's length never changes, and that constant hypotenuse is the whole of the physics in the question.
Worked example 2 — two poles. Two vertical poles of heights m and m stand on level ground m apart. Find the distance between their tops.
Join the tops and drop a horizontal from the shorter pole's top to the taller one. That creates a right-angled triangle with a horizontal leg of m and a vertical leg equal to the difference in heights, m:
The vertical leg is the difference, not the sum. Adding the heights is the standard error here, and it gives an answer longer than either pole is tall — which a glance at the sketch rules out.
Worked example 3 — a room diagonal. A rectangular hall measures m by m. Find the length of its diagonal.
And a square of side cm has diagonal cm. **The diagonal of any square is times its side, which is worth carrying as a fact.
Worked example 4 — a rhombus.** A rhombus has side cm and one diagonal cm. Find the other diagonal.
The diagonals of a rhombus bisect each other at right angles, so they cut it into four congruent right-angled triangles whose legs are the half-diagonals and whose hypotenuse is the side:
So the other diagonal is cm.
Worked example 5 — a journey. A cyclist rides km due south, then km due east. How far is she from her starting point?
South and east are perpendicular, so
**She travelled km but ended up km away** — and the gap between those two numbers is the triangle inequality from the previous chapter, measured exactly.
Worked example 1 — a ladder. A ladder m long rests against a vertical wall with its foot m from the wall. How high does it reach? If the foot is pulled out to m from the wall, how far does the top slide down?
The wall is vertical and the ground horizontal, so the right angle is at the base. The ladder is the hypotenuse:
After the foot is moved:
So the top slides down m.
Notice how unequal the two movements are. The foot moved out m and the top fell m here, but that is a coincidence of these numbers. The ladder's length never changes, and that constant hypotenuse is the whole of the physics in the question.
Worked example 2 — two poles. Two vertical poles of heights m and m stand on level ground m apart. Find the distance between their tops.
Join the tops and drop a horizontal from the shorter pole's top to the taller one. That creates a right-angled triangle with a horizontal leg of m and a vertical leg equal to the difference in heights, m:
The vertical leg is the difference, not the sum. Adding the heights is the standard error here, and it gives an answer longer than either pole is tall — which a glance at the sketch rules out.
Worked example 3 — a room diagonal. A rectangular hall measures m by m. Find the length of its diagonal.
And a square of side cm has diagonal cm. **The diagonal of any square is times its side, which is worth carrying as a fact.
Worked example 4 — a rhombus.** A rhombus has side cm and one diagonal cm. Find the other diagonal.
The diagonals of a rhombus bisect each other at right angles, so they cut it into four congruent right-angled triangles whose legs are the half-diagonals and whose hypotenuse is the side:
So the other diagonal is cm.
Worked example 5 — a journey. A cyclist rides km due south, then km due east. How far is she from her starting point?
South and east are perpendicular, so
**She travelled km but ended up km away** — and the gap between those two numbers is the triangle inequality from the previous chapter, measured exactly.
Exam tip
What is the safest way to write out a Pythagoras calculation?
Sketch, mark the right angle, label the hypotenuse, and only then write the equation. Almost every lost mark in this chapter comes from a mislabelled sketch rather than a wrong calculation.
- Write the right angle on the figure with a small square, and put the hypotenuse label opposite it. The hypotenuse is always the longest side, so a labelled sketch is self-checking
- State the triangle you are using: *In , *. Application questions contain several triangles and the examiner needs to know which one
- **Keep the equation in the form ** and substitute into it, rather than rearranging in your head. For a leg, write as its own line
- Carry units throughout and give the answer with them. A length in metres squared during the working must come back to metres
- Simplify surds rather than decimalising early. is exact; is not, and rounding twice compounds the error
- For differences in height, subtract before squaring. The vertical leg between two pole tops is , and doing it in the wrong order is the classic slip
- Sanity-check against the inequalities chapter: the hypotenuse must be longer than either leg, and any perpendicular must be shorter than the slanting segments
The trap that produces a negative answer. If a subtraction gives you a negative square, you have used a leg as the hypotenuse. Do not take the root and change the sign — go back to the sketch and find the right angle again. The arithmetic is telling you the figure is wrong, and that is worth more than the arithmetic.
- Write the right angle on the figure with a small square, and put the hypotenuse label opposite it. The hypotenuse is always the longest side, so a labelled sketch is self-checking
- State the triangle you are using: *In , *. Application questions contain several triangles and the examiner needs to know which one
- **Keep the equation in the form ** and substitute into it, rather than rearranging in your head. For a leg, write as its own line
- Carry units throughout and give the answer with them. A length in metres squared during the working must come back to metres
- Simplify surds rather than decimalising early. is exact; is not, and rounding twice compounds the error
- For differences in height, subtract before squaring. The vertical leg between two pole tops is , and doing it in the wrong order is the classic slip
- Sanity-check against the inequalities chapter: the hypotenuse must be longer than either leg, and any perpendicular must be shorter than the slanting segments
The trap that produces a negative answer. If a subtraction gives you a negative square, you have used a leg as the hypotenuse. Do not take the root and change the sign — go back to the sketch and find the right angle again. The arithmetic is telling you the figure is wrong, and that is worth more than the arithmetic.
Did you know
How can you produce a right-angled triangle from any two numbers?
The triples and look like lucky accidents. They are not — there is a recipe, and you can derive it with the identity from the expansions chapter.
Take any two whole numbers and with , and form these three:
These always satisfy . Check the algebra. Expanding both sides,
and that is exactly . The middle terms are what cancel and reassemble — the same mechanism as in the area proof above.
Now generate a few. With , : . With , : . With , : . With , : . Every triple you were told to memorise comes out of two small numbers, so you can rebuild the list instead of remembering it.
And the recipe explains a practical trick still used on building sites. A mason with a knotted rope divided into equal parts can form a -- triangle and get an exact right angle with no instrument at all — the converse of the theorem turned into a tool, and the reason the triple is the one everyone knows.
One thing the recipe does not produce, and this is the curious part: there are no such triples for cubes. No three positive whole numbers satisfy , nor any higher power. The square is the only exponent for which this whole family of solutions exists — a fact that turns out to be far harder to prove than the theorem this chapter is about.
Take any two whole numbers and with , and form these three:
These always satisfy . Check the algebra. Expanding both sides,
and that is exactly . The middle terms are what cancel and reassemble — the same mechanism as in the area proof above.
Now generate a few. With , : . With , : . With , : . With , : . Every triple you were told to memorise comes out of two small numbers, so you can rebuild the list instead of remembering it.
And the recipe explains a practical trick still used on building sites. A mason with a knotted rope divided into equal parts can form a -- triangle and get an exact right angle with no instrument at all — the converse of the theorem turned into a tool, and the reason the triple is the one everyone knows.
One thing the recipe does not produce, and this is the curious part: there are no such triples for cubes. No three positive whole numbers satisfy , nor any higher power. The square is the only exponent for which this whole family of solutions exists — a fact that turns out to be far harder to prove than the theorem this chapter is about.
Exam relevance
Why is the Pythagoras theorem the most reused result in JEE and NEET?
Of everything in Class 9 geometry, this is the result with the widest reach — it is foundation work that never stops being used.
Where it leads in Mathematics. In Class 10 it is proved again using similarity, and its converse becomes a standard classification tool. In Class 11 it becomes the distance formula , which is the theorem in coordinates, and then the modulus of a complex number and the magnitude of a vector. In trigonometry, is the Pythagoras theorem on a triangle with hypotenuse — which is why it is called a Pythagorean identity.
Where it leads in Physics. Every resultant of two perpendicular quantities uses it: velocity components, forces at right angles, the amplitude of two perpendicular oscillations, and the relationship between the components of an electric or magnetic field. Both JEE Main and NEET ask questions whose only geometry is this one step, repeated.
Question types to expect. At this level: find a side, test the converse, solve an application. In competitive papers the theorem is almost never the question — it is one line inside a longer one, which means a slip here costs a whole question elsewhere. Diagram-based items test whether you can spot the right-angled triangle in a figure that does not advertise it.
The single trap that costs marks. Assuming the largest number given is the hypotenuse when the right angle is somewhere else. If a question says , the hypotenuse is whatever the numbers look like, and in coordinate or vector questions this becomes using the wrong pair of components. Read the right angle off the given information, never off the figure.
Board versus competitive emphasis. ICSE asks for the proof and the full application layout, so the construction and the area comparison are worth learning properly. A competitive paper wants the value in one line, so the premium is on recognising the triples instantly and on spotting the hidden right angle. Both reward the same first move: find the right angle, then name the hypotenuse.
Where it leads in Mathematics. In Class 10 it is proved again using similarity, and its converse becomes a standard classification tool. In Class 11 it becomes the distance formula , which is the theorem in coordinates, and then the modulus of a complex number and the magnitude of a vector. In trigonometry, is the Pythagoras theorem on a triangle with hypotenuse — which is why it is called a Pythagorean identity.
Where it leads in Physics. Every resultant of two perpendicular quantities uses it: velocity components, forces at right angles, the amplitude of two perpendicular oscillations, and the relationship between the components of an electric or magnetic field. Both JEE Main and NEET ask questions whose only geometry is this one step, repeated.
Question types to expect. At this level: find a side, test the converse, solve an application. In competitive papers the theorem is almost never the question — it is one line inside a longer one, which means a slip here costs a whole question elsewhere. Diagram-based items test whether you can spot the right-angled triangle in a figure that does not advertise it.
The single trap that costs marks. Assuming the largest number given is the hypotenuse when the right angle is somewhere else. If a question says , the hypotenuse is whatever the numbers look like, and in coordinate or vector questions this becomes using the wrong pair of components. Read the right angle off the given information, never off the figure.
Board versus competitive emphasis. ICSE asks for the proof and the full application layout, so the construction and the area comparison are worth learning properly. A competitive paper wants the value in one line, so the premium is on recognising the triples instantly and on spotting the hidden right angle. Both reward the same first move: find the right angle, then name the hypotenuse.
Key takeaways
What should you be able to do with the Pythagoras theorem before moving on?
One equation, one converse, and an enormous number of applications.
- **** with the hypotenuse — the side opposite the right angle
- The area proof: four copies of the triangle inside a square of side , counted two ways, with the cancelling
- The inner figure is a square because the two acute angles of a right-angled triangle add to — state this when asked to prove
- Add to find the hypotenuse, subtract to find a leg, with the hypotenuse written first in the subtraction
- The converse: if the square of the longest side equals the sum of the other two squares, the triangle is right-angled — and check the triangle inequality first
- Learn the triples , , , and their multiples, and generate more from , ,
- Applications: a ladder is a hypotenuse, the gap between two pole tops uses the difference in heights, the diagonal of a square is times its side, and a rhombus splits into four right-angled triangles on its half-diagonals
- Keep surds exact and check every answer against the size rules for sides
The fastest self-test is the double-area trick from worked example 4 in the calculation section: find in a - triangle, then get the perpendicular from to by equating two expressions for the area. If that comes out at cm without prompting, this chapter is secure.
- **** with the hypotenuse — the side opposite the right angle
- The area proof: four copies of the triangle inside a square of side , counted two ways, with the cancelling
- The inner figure is a square because the two acute angles of a right-angled triangle add to — state this when asked to prove
- Add to find the hypotenuse, subtract to find a leg, with the hypotenuse written first in the subtraction
- The converse: if the square of the longest side equals the sum of the other two squares, the triangle is right-angled — and check the triangle inequality first
- Learn the triples , , , and their multiples, and generate more from , ,
- Applications: a ladder is a hypotenuse, the gap between two pole tops uses the difference in heights, the diagonal of a square is times its side, and a rhombus splits into four right-angled triangles on its half-diagonals
- Keep surds exact and check every answer against the size rules for sides
The fastest self-test is the double-area trick from worked example 4 in the calculation section: find in a - triangle, then get the perpendicular from to by equating two expressions for the area. If that comes out at cm without prompting, this chapter is secure.