Pythagoras Theorem | ICSE Class 9 Maths Notes

This note covers right-angled triangles, Pythagoras Theorem, its area-based proof, the converse and its proof, finding missing sides, testing for right angles, square diagonals, and applications involving the heights, areas and perimeters of triangles.

What does Pythagoras Theorem state?

Theorem: Pythagoras Theorem

Definition: A triangle is a closed plane shape with three straight sides. A right-angled triangle has one angle equal to 90°, where ° denotes degrees, a unit of angle measurement. Its hypotenuse is the side opposite the right angle. The two sides meeting at the right angle are its legs.

A theorem is a mathematical statement established by proof. Pythagoras Theorem states that, in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. A square of a number means that number multiplied by itself.

Let a and b denote the positive lengths of the legs, and let c denote the positive length of the hypotenuse, all measured in the same unit. The theorem gives c² = a² + b². Here, a² means a × a, and similarly for b² and c².

The right-angle condition is essential. The formula cannot be applied to an arbitrary triangle merely because three side lengths have been labelled a, b and c. Identify the right angle first, then identify the side opposite it.

How do the labels affect the equation?

In a triangle with vertices, or corner points, A, B and C, the notation ∠ABC means the angle at B. The notation AB means the segment joining A and B, or its length when used in an equation.

If ∠ABC = 90°, AC is the hypotenuse, so AC² = AB² + BC². The middle letter in the angle name locates the right angle. Changing the triangle's orientation does not change which side lies opposite that angle.

The hypotenuse is longer than either leg. Indeed, c² exceeds a² by the positive quantity b², and exceeds b² by the positive quantity a². This gives a useful check on a calculated answer before it is accepted.

How can areas prove Pythagoras Theorem?

What areas are needed?

Area measures the space enclosed by a plane shape. A square has four equal sides and four right angles. A square of side c has area c² square units. A right-angled triangle with perpendicular legs a and b has area ab/2 square units. Here ab means a × b, and division by 2 takes half the product.

Perpendicular lines meet at a right angle. Thus, either leg of a right-angled triangle can serve as its base, while the other serves as its perpendicular height. The sloping hypotenuse is not the height corresponding to either leg.

Draw a large square of side a + b. Fit four identical right-angled triangles, each with legs a and b and hypotenuse c, into its corners. Arrange the legs along the outer boundary so that the four hypotenuses enclose a central square.

Why is the central region a square?

Congruent shapes have the same shape and size. The four triangles are congruent copies, so each side of the central region is a hypotenuse of length c. Equal sides alone would not be enough to establish that the central region is a square.

The two acute angles, meaning angles smaller than 90°, in each right triangle add to 90°. At each central corner, these two angles and the angle inside the central region lie along a straight angle of 180°. The central angle is therefore 90°.

Consequently, the central region has four equal sides and four right angles. It is a square of side c. Each outer side consists of one leg of length a and one of length b, giving total length a + b.

How does the area equation simplify?

1. The area of the outer square is (a + b)² square units.
2. The four corner triangles together have area 4 × ab/2 = 2ab square units.
3. The central square has area c² square units. The pieces fill the outer square without gaps or overlaps.
4. Equating the whole area to the sum of its parts gives (a + b)² = 2ab + c².
5. Expanding the left side gives a² + 2ab + b² = 2ab + c². Subtracting 2ab from both sides gives a² + b² = c².

This is an area-based proof: two expressions measure the same region, so they are equal. It works for arbitrary positive leg lengths. Checking a particular numerical triangle illustrates the theorem, but does not replace this general argument.

What is the converse, and how is it proved?

Theorem: Converse of Pythagoras Theorem

A converse reverses the condition and conclusion of a statement. Pythagoras Theorem starts with a right angle and concludes a relation between squared lengths. Its converse starts with that relation and concludes that the triangle has a right angle.

Definition: If the square of one side of a triangle equals the sum of the squares of the other two sides, the angle opposite that side is a right angle.

Let a triangle ABC have side lengths BC = a, CA = b and AB = c, with a² + b² = c². The angle to be proved right is ∠ACB, opposite AB. Do not assume this right angle before proving it.

How does a second triangle establish the result?

1. Construct a second triangle XYZ with a right angle at Z, and choose YZ = a and XZ = b. The letters X, Y and Z name its vertices.
2. Apply Pythagoras Theorem to this constructed right triangle. Its hypotenuse satisfies XY² = a² + b².
3. The given equation is AB² = a² + b², so XY² = AB². Since side lengths are positive, XY = AB = c.
4. The triangles have BC = YZ, CA = ZX and AB = YX. They are congruent by the side-side-side criterion, abbreviated SSS: equality of three corresponding sides establishes congruence.
5. Corresponding angles of congruent triangles are equal. Therefore ∠ACB equals the right angle at Z, and triangle ABC is right-angled at C.

The vertex correspondence is A to X, B to Y, and C to Z. Writing the matching sides prevents an incorrect conclusion about which angle is right. The largest side becomes the hypotenuse after the converse has established the right angle.

The proof uses the original theorem on a triangle constructed to be right-angled. It then transfers that angle to the given triangle through congruence. This avoids assuming the very fact that the converse is intended to prove.

How do you find an unknown hypotenuse?

Why are the squared legs added?

When both legs are known, their squares are added because the theorem gives the hypotenuse squared as their sum. A final square root is needed to obtain the length itself. A squared length and a length are different quantities.

The symbol √ denotes the non-negative square root. For a positive side length, use the positive value. Thus the formula for the hypotenuse is c = √(a² + b²), with a and b the perpendicular leg lengths and c the opposite side.

Worked example 1. Triangle ABC is right-angled at B. If AB = 6 cm and BC = 8 cm, find AC. The abbreviation cm means centimetres.

Answer: AC is opposite the right angle, so it is the hypotenuse. By Pythagoras Theorem, AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100 cm². Therefore AC = √100 = 10 cm.

Here cm² means square centimetres. The equation first finds AC² in square units. Taking its square root returns a length measured in centimetres. The answer 10 cm is longer than both 6 cm and 8 cm, as required.

How can substitution check the answer?

Substitute the final length back into the original equation: 10² = 6² + 8². Both sides equal 100. This check confirms the arithmetic and that the calculated side occupies the hypotenuse position in the equation.

Keep the square-root sign over the whole sum. Adding the two leg lengths would measure a broken path along the legs. Pythagoras Theorem concerns the straight side joining their other endpoints, so it requires squared lengths followed by a square root.

How do you find an unknown leg?

Why does the calculation use subtraction?

When the hypotenuse and one leg are known, write the complete theorem before rearranging. If c is the hypotenuse, a is the known leg and b is the unknown leg, then b² = c² − a². The symbol − denotes subtraction.

Subtract the square of the known leg from the square of the hypotenuse. Then take the positive square root: b = √(c² − a²). The subtraction occurs between squares, before taking the root. It is not a subtraction between the original lengths.

Worked example 2. A right-angled triangle has hypotenuse 5 cm and one leg 3 cm. Find the other leg, whose length is denoted by x.

Answer: Pythagoras Theorem gives 5² = x² + 3². Hence x² = 25 − 9 = 16 cm². The positive length is x = √16 = 4 cm. Substitution confirms that 4² + 3² = 5².

Worked example 3. Triangle ABC is right-angled at B. If AC = 13 cm and BC = 5 cm, find AB.

Answer: AC is the hypotenuse because it is opposite B. Therefore AB² = AC² − BC² = 13² − 5² = 169 − 25 = 144 cm². Taking the positive square root gives AB = 12 cm.

What checks catch a wrong rearrangement?

In the second calculation, 12 cm is less than the hypotenuse length 13 cm. Also, 12² + 5² = 144 + 25 = 169 = 13². Both the size check and the substitution check support the answer.

Note: A negative value for the square of a side length cannot be correct. Recheck the identification of the hypotenuse, the order of subtraction and the given measurements before proceeding.

These examples use the same theorem as the hypotenuse calculation. Addition and subtraction are not competing rules. They arise from rearranging one equation according to which side is unknown.

How can side lengths establish a right angle?

How do you apply the converse numerically?

If a triangle is specified by its three sides without a right-angle condition, use the converse to test whether it is right-angled. Identify the largest side, square it, and compare this with the sum of the squares of the other two sides.

Use exact values in the comparison. An equality establishes a right angle opposite the largest side. If equality fails, the triangle is not right-angled. Naming the angle is part of the conclusion when the triangle's vertices have been supplied.

Worked example 4. A triangle ABC has AB = 4 units, BC = 3 units and AC = 5 units. Show that it is right-angled and find its area.

Answer: AC is the largest side. AB² + BC² = 4² + 3² = 16 + 9 = 25 = AC². By the converse, ∠ABC = 90°. Therefore its area is half the product of AB and BC: 4 × 3/2 = 6 square units.

The triangle has A above B, C to the right of B, AB labelled 4, BC labelled 3 and the sloping side AC labelled 5.

See Fig. 6.25 in your NCERT textbook

Worked example 5. A triangle has sides 7 cm, 24 cm and 25 cm. Use Pythagoras Theorem or its converse to find its area.

Answer: The largest side is 25 cm. Since 7² + 24² = 49 + 576 = 625 = 25², the converse establishes a right angle between the 7 cm and 24 cm sides. Its area is 7 × 24/2 = 84 cm².

A Pythagorean triplet consists of three positive integers, meaning whole numbers greater than zero, whose largest square equals the sum of the other two squares. The numbers 3, 4, 5 and the numbers 6, 8, 10 are examples. Recognising them is useful, but writing the equality explains the conclusion.

How does the theorem give the diagonal of a square?

Result: Equal perpendicular legs

An isosceles triangle has two equal sides. In an isosceles right-angled triangle, the two legs are equal. Let each leg have length a and the hypotenuse have length c. Then c² = 2a², so c = a√2.

A diagonal joins two non-adjacent vertices of a polygon, a closed plane shape bounded by straight sides. A square has equal sides and right angles. Drawing a diagonal produces a right-angled triangle whose legs are adjacent sides of the square and whose hypotenuse is the diagonal.

Thus the same result applies to a square: if its side length is a and its diagonal length is d, then d = a√2. Here d is a length, not an area. Its square, d², is twice the square of the side length.

The square is labelled A at the upper left, B at the lower left, C at the lower right and D at the upper right. The segment AC joins opposite corners, and a right angle is marked at B.

See Fig. 5.1 in your NCERT textbook

Worked example 6. Find the length of a diagonal of a square whose side is 8 cm.

Answer: Let d be the diagonal length. The two perpendicular sides are each 8 cm, so d² = 8² + 8² = 128 cm². Therefore d = √128 = √(64 × 2) = 8√2 cm.

The answer 8√2 cm is exact. It need not be replaced by a decimal approximation. Squaring it gives 64 × 2 = 128 cm², matching the sum of the squared sides. The diagonal is longer than a side, consistent with its role as hypotenuse.

How does a perpendicular help with isosceles triangles?

Why is half the base used?

An altitude is a perpendicular segment from a triangle's vertex to the line containing its opposite side. In an isosceles triangle, the altitude from the vertex between the equal sides bisects the base, meaning it divides the base into two equal parts.

To justify this, compare the two right triangles formed by the altitude. Their hypotenuses are the original equal sides and they share the altitude. The right-angle-hypotenuse-side criterion, abbreviated RHS, makes them congruent, so their base segments are equal.

In triangle ABC, let AB = AC = a, BC = 2b and AD = h, where D is the foot of the altitude on BC. Here a is an equal-side length, b is half the base and h is the height.

Each smaller right triangle has hypotenuse a and legs b and h. Consequently, h² = a² − b² and h = √(a² − b²). The whole triangle's area is half its full base times its height, giving b√(a² − b²) square units.

A is the top vertex, B and C are the base endpoints, and D lies on BC. AB and AC are labelled a, BD and DC are labelled b, and the dashed perpendicular AD is labelled h.

See Fig. 6.24 in your NCERT textbook

Worked example 7. An isosceles triangle has perimeter 40 cm and equal sides of 15 cm each. Find its area. Perimeter means the total length around the boundary.

Answer: The base is 40 − 15 − 15 = 10 cm, so half the base is 5 cm. The altitude h satisfies h² = 15² − 5² = 225 − 25 = 200 cm². Hence h = 10√2 cm.

Answer: The area is half the full base times the height: 10 × 10√2/2 = 50√2 cm².

Worked example 8. An isosceles triangle has base 10 cm and area 60 cm². Find the lengths of its equal sides.

Answer: Let h be its altitude. From 10h/2 = 60, h = 12 cm. The altitude bisects the base into lengths of 5 cm. Each equal side has square 12² + 5² = 144 + 25 = 169 cm², so each equals 13 cm.

In the first example, the equal side is known and the altitude is found by subtraction. In the second, the height is found from area first, and the equal side is found by addition. The right triangle determines which operation is appropriate.

How do area and perimeter combine with Pythagoras Theorem?

Which quantity should be found first?

A problem may provide area rather than two side lengths. For a right triangle, the legs are perpendicular, so its area is half their product. Use that relationship to find a missing leg before calculating the hypotenuse.

Let K denote the area of a right triangle, a and b its leg lengths, and c its hypotenuse. Then K = ab/2. If K and a are known, rearranging gives b = 2K/a. These letters distinguish area from side lengths.

Worked example 9. A right-angled triangle has area 54 cm² and one leg 12 cm. Find its perimeter.

Answer: Let b be the other leg. Since 12b/2 = 54, b = 9 cm. The hypotenuse c satisfies c² = 12² + 9² = 144 + 81 = 225 cm², so c = 15 cm.

Answer: Its perimeter is the sum of all three sides: 12 + 9 + 15 = 36 cm.

Why must area and perimeter be kept separate?

Area uses square units, whereas perimeter uses length units. The given 54 cm² measures the enclosed region. The answer 36 cm measures the boundary. They come from different operations, even though both refer to the same triangle.

The order of work matters: first extract a leg from the area formula, then obtain the hypotenuse using the theorem, and finally add the three lengths. Attempting to calculate the perimeter before finding the hypotenuse leaves a required boundary length unknown.

A final check can recover the original area: 12 × 9/2 = 54 cm². Checking the hypotenuse separately gives 15² = 12² + 9². Together these checks use both conditions supplied in the question.

How can you choose and present the correct method?

What does the given information establish?

Begin by separating a given right angle from a right angle that still needs proof. A drawing that looks square at one corner is not a stated measurement. Use the wording, marked angle or established property of the shape to justify the condition.

What makes a solution complete?

1. Name the triangle being used and justify its right angle, or state that the converse will test for one.
2. Identify the hypotenuse and define any letter introduced for an unknown length.
3. Write the equation with side names before substituting the supplied measurements.
4. Show the squared calculation, take the positive square root when finding a length, and attach the correct unit.
5. State the requested result, including the location of a proved right angle or the complete area or perimeter calculation.

For a proof, identify the given condition and the conclusion separately. In the area proof, explain why the central region is a square. In the converse proof, explain why the constructed and given triangles have three matching sides.

For a numerical answer, distinguish a check from a justification. Substitution checks arithmetic, while the right-angle condition justifies applying the theorem. A complete solution needs both a valid geometric starting point and calculations consistent with it.

Glossary

• Right-angled triangle — A triangle containing one right angle, with its opposite side called the hypotenuse.
• Hypotenuse — The side opposite the right angle, longer than either leg of a right-angled triangle.
• Legs — The two sides of a right-angled triangle that meet at its right angle.
• Perpendicular — Describes lines or segments that meet to form a right angle.
• Square of a number — The product obtained by multiplying a number by that same number.
• Square root — A number which, when multiplied by itself, produces the specified number.
• Converse — A statement formed by exchanging the condition and conclusion of an original statement.
• Congruent triangles — Triangles with the same shape and size, having equal corresponding sides and angles.
• Altitude — A perpendicular segment from a triangle's vertex to the line containing its opposite side.
• Isosceles triangle — A triangle with two equal sides; the altitude between them bisects the opposite base.
• Diagonal — A line segment joining two vertices of a polygon that are not adjacent.
• Pythagorean triplet — Three positive integers whose largest square equals the sum of the other two squares.
• Area — The measure of the region enclosed by a plane shape, expressed in square units.
• Perimeter — The total length around a shape's boundary, expressed in units of length.

Common errors and misconceptions

• Misconception: The sloping side in any drawing is the hypotenuse. Correct: The hypotenuse is identified by its position opposite the right angle, regardless of the drawing's orientation.
• Misconception: Pythagoras Theorem applies directly to every triangle. Correct: It requires a right angle; when only three sides are given, test them using the converse.
• Misconception: Finding any unknown side requires adding squares. Correct: Add the squared legs for the hypotenuse; subtract the known leg's square from the hypotenuse's square for an unknown leg.
• Misconception: A calculated squared side is already the required length. Correct: Take the positive square root and express the resulting length in length units.
• Misconception: The full base of an isosceles triangle belongs in either smaller right triangle. Correct: The altitude from the vertex between equal sides bisects the base, so use half of it.
• Misconception: Three sides satisfying the squared relation make any chosen angle right. Correct: The right angle is opposite the side whose square equals the sum.
• Misconception: Equal sides alone make the central region of the area proof a square. Correct: Also establish that its interior angles are right angles.

Exam-style questions with model answers

Q1. State Pythagoras Theorem for a right triangle whose legs have lengths a and b and whose hypotenuse has length c. Explain which side is the hypotenuse. [2 marks]

1. In this right-angled triangle, the squared hypotenuse equals the sum of the squared legs: c² = a² + b².
2. The hypotenuse is the side opposite the right angle; it is longer than either leg.

Q2. Triangle ABC is right-angled at B, with AB = 6 cm and BC = 8 cm. Find AC. [3 marks]

1. The side AC is opposite the right angle at B, so it is the hypotenuse. Pythagoras Theorem gives AC² = AB² + BC².
2. Substituting the supplied lengths gives AC² = 6² + 8² = 36 + 64 = 100 cm².
3. Taking the positive square root, AC = 10 cm. A length is positive, and this answer is longer than either given leg.

Q3. Triangle ABC is right-angled at B, with AC = 13 cm and BC = 5 cm. Find AB. [3 marks]

1. AC lies opposite the right angle, so it is the hypotenuse. The theorem gives AC² = AB² + BC².
2. Rearranging before substitution gives AB² = AC² − BC² = 13² − 5² = 169 − 25 = 144 cm².
3. Therefore AB = √144 = 12 cm, using the positive root for a length. Checking gives 12² + 5² = 169 = 13².

Q4. A triangle has sides 7 cm, 24 cm and 25 cm. Prove it is right-angled and find its area. [4 marks]

1. The largest side is 25 cm, so compare its square with the sum of the squares of the other two sides.
2. The comparison gives 7² + 24² = 49 + 576 = 625, while 25² = 625.
3. By the converse of Pythagoras Theorem, the angle opposite the 25 cm side is a right angle.
4. The perpendicular legs are consequently 7 cm and 24 cm, giving area 7 × 24/2 = 84 cm².

Q5. Prove Pythagoras Theorem by comparing areas. Take a right triangle with positive leg lengths a and b and hypotenuse c. Arrange four congruent copies at the corners of a square of side a + b, with their hypotenuses enclosing the central region. [5 marks]

1. The outer square has area (a + b)². Each of its sides consists of one leg of length a and one leg of length b.
2. The central region has four sides of length c. At each corner, the adjacent acute angles of the right triangles sum to 90°, leaving a central angle of 90° on the straight line.
3. The central region is therefore a square of area c². Each corner triangle has area ab/2, so the four triangles together have area 2ab.
4. The pieces cover the outer square without gaps or overlaps. Hence (a + b)² = 2ab + c².
5. Expand and cancel the common term: a² + 2ab + b² = 2ab + c², giving a² + b² = c², as required.

Q6. Triangle ABC has BC = a, CA = b and AB = c, where the positive lengths satisfy a² + b² = c². Prove that ∠ACB = 90° by constructing a right triangle and using congruence. [5 marks]

1. Construct triangle XYZ with a right angle at Z and perpendicular sides YZ = a and XZ = b. These match two given side lengths.
2. Apply Pythagoras Theorem to this constructed triangle: XY² = a² + b². The given relation therefore gives XY² = c².
3. Since lengths are positive, XY = c = AB. Also YZ = BC = a and XZ = CA = b.
4. All three pairs of corresponding sides are equal. By side-side-side congruence, triangle ABC is congruent to triangle XYZ, with C corresponding to Z.
5. Corresponding angles are equal, so ∠ACB equals the right angle at Z. Hence ∠ACB = 90°, proving the converse.

Q7. An isosceles triangle has base 10 cm and area 60 cm². Find the length of each equal side, explaining how you use its altitude. [4 marks]

1. Let h denote the altitude from the vertex between the equal sides to the base. The area formula gives 10h/2 = 60.
2. Solving gives h = 12 cm. In an isosceles triangle this altitude bisects the base, giving two segments of 5 cm each.
3. Each equal side is the hypotenuse of a right triangle with legs 12 cm and 5 cm. Its square is 12² + 5² = 169 cm².
4. Taking the positive square root gives √169 = 13 cm. Both equal sides therefore measure 13 cm.

Q8. A right-angled triangle has area 54 cm² and one leg of length 12 cm. Find its perimeter. [4 marks]

1. Let b be the other leg length. The two legs are perpendicular, so the area equation is 12b/2 = 54.
2. Solving this equation gives b = 9 cm. Thus both perpendicular sides are now known.
3. Let c be the hypotenuse length. Pythagoras Theorem gives c² = 12² + 9² = 225 cm², so c = 15 cm.
4. The perimeter is the sum of the three side lengths: 12 + 9 + 15 = 36 cm.

Key takeaways

• Pythagoras Theorem relates squared side lengths in a right-angled triangle: the hypotenuse square equals the sum of the leg squares.
• Identify the hypotenuse by locating the side opposite the right angle before writing any equation.
• The area proof compares a large square with four congruent right triangles and the smaller square between them.
• The converse establishes a right angle from the squared side relation, and congruence proves that conclusion.
• Add squared legs to find the squared hypotenuse; subtract a squared leg from the squared hypotenuse to find another squared leg.
• Take the positive square root to obtain a side length, and distinguish length units from square units.
• A square's diagonal forms a right triangle with two equal sides, so its length is the side multiplied by √2.
• The altitude from the vertex between an isosceles triangle's equal sides bisects its base and produces two right triangles.

Test yourself

In triangle ABC, ∠ABC = 90°. Which side is the hypotenuse, and what equation connects the sides?

AC is opposite the right angle, so it is the hypotenuse. The equation is AC² = AB² + BC².

Why does the central region in the four-triangle area proof have right angles?

At each corner, the two adjacent acute triangle angles total 90°. Subtracting them from the straight angle of 180° leaves a central right angle.

How does the starting information differ between Pythagoras Theorem and its converse?

The theorem starts with a right angle and gives a squared side relation. The converse starts with that relation and establishes the right angle.

A right triangle has hypotenuse 5 cm and one leg 3 cm. What is the other leg?

Its square is 5² − 3² = 16 cm². Taking the positive square root gives a length of 4 cm.

A square has side 8 cm. What is its exact diagonal length?

The diagonal square is 8² + 8² = 128 cm², so its exact length is 8√2 cm.

A triangle has sides 7 cm, 24 cm and 25 cm. Which angle is right?

Since 7² + 24² = 25², the converse makes the angle opposite the 25 cm side a right angle.

An isosceles triangle has base 10 cm. What base length enters either right triangle made by its altitude from the vertex between equal sides?

Use 5 cm, because that altitude bisects the 10 cm base into two equal segments.

A right triangle has area 54 cm² and one leg 12 cm. What is the other leg?

Twice the area divided by the known leg gives the other leg: 2 × 54/12 = 9 cm.