Geometry and Measurement Sec 2 E-Mathematics

Pythagoras' Theorem

Pythagoras’ Theorem

Key Concepts

  • Pythagoras’ theorem applies only to a right-angled triangle.

    • A right-angled triangle is a triangle that has one angle of 90°.
    • The theorem states:
    a2+b2=c2 a^2 + b^2 = c^2
  • In the formula:

    • aa and bb are the lengths of the two shorter sides that form the right angle.
    • cc is the length of the hypotenuse.
    • The hypotenuse is always:
      • the longest side of a right-angled triangle
      • the side opposite the 90° angle
  • What the formula means:

    • If you square the lengths of the two shorter sides and add them together, you get the square of the hypotenuse.

    • Example:

      32+42=52 3^2 + 4^2 = 5^2
      9+16=25 9 + 16 = 25
  • Finding the hypotenuse:

    • If the two shorter sides are known, use:

      c=a2+b2 c = \sqrt{a^2 + b^2}
    • Steps:

      1. Square each shorter side.
      2. Add the squares.
      3. Take the square root.
  • Finding a shorter side:

    • If the hypotenuse and one shorter side are known, rearrange the theorem:

      a2=c2b2 a^2 = c^2 - b^2

      or

      b2=c2a2 b^2 = c^2 - a^2
    • Then take the square root.

    • Steps:

      1. Square the hypotenuse.
      2. Square the known shorter side.
      3. Subtract.
      4. Take the square root.
  • Pythagorean triples are sets of 3 whole numbers that satisfy Pythagoras’ theorem exactly.

    • Common examples:
      • 3,4,53, 4, 5
      • 5,12,135, 12, 13
      • 8,15,178, 15, 17
      • 7,24,257, 24, 25
    • These are useful because they allow quick answers without using a calculator much.
  • Applications in real-world contexts:

    • Pythagoras’ theorem is used to find distances that cannot be measured directly.
    • Examples include:
      • the length of a ladder leaning against a wall
      • the diagonal of a rectangle
      • the shortest straight-line distance between two points
      • the height of an object using a sloping side
      • distances on maps or floor plans
  • Converse of Pythagoras’ theorem:

    • The converse is used to check whether a triangle is right-angled.

    • If the side lengths of a triangle satisfy:

      a2+b2=c2 a^2 + b^2 = c^2

      where cc is the longest side, then the triangle is a right-angled triangle.

    • If the equation is not true, the triangle is not right-angled.

  • Important note about units:

    • All side lengths must be in the same unit before using the formula.
    • Example: convert cm to m, or m to cm, before calculating.
  • Important note about square roots:

    • A length cannot be negative.
    • When finding a side length, use the positive square root only.

Important Definitions

  • Pythagoras’ theorem: the rule that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, a2+b2=c2\,a^2+b^2=c^2.

  • Right-angled triangle: a triangle with one angle equal to 90°.

  • Hypotenuse: the longest side of a right-angled triangle, opposite the right angle.

  • Shorter sides: the two sides that meet to form the right angle in a right-angled triangle.

  • Square of a number: the result of multiplying a number by itself, for example 62=366^2 = 36.

  • Square root: a number that, when multiplied by itself, gives the original number, for example 49=7\sqrt{49}=7.

  • Pythagorean triple: a set of three whole numbers that satisfy Pythagoras’ theorem exactly.

  • Converse of Pythagoras’ theorem: the rule used to test whether a triangle is right-angled by checking whether the side lengths satisfy a2+b2=c2\,a^2+b^2=c^2.

  • Diagonal: a straight line joining two opposite corners of a shape such as a rectangle or square.

  • Perpendicular: meeting at an angle of 90°.


Worked Examples

Example 1: Finding the hypotenuse

A right-angled triangle has shorter sides of 6 cm and 8 cm. Find the hypotenuse.

Step 1: Write the formula

a2+b2=c2 a^2 + b^2 = c^2

Step 2: Substitute the values

62+82=c2 6^2 + 8^2 = c^2

Step 3: Square the numbers

36+64=c2 36 + 64 = c^2

Step 4: Add

100=c2 100 = c^2

Step 5: Take the square root

c=100=10 c = \sqrt{100} = 10

Answer: The hypotenuse is 10 cm.


Example 2: Finding a shorter side

A right-angled triangle has hypotenuse 13 m and one shorter side 5 m. Find the other shorter side.

Step 1: Write the formula

a2+b2=c2 a^2 + b^2 = c^2

Let the unknown side be aa.

a2+52=132 a^2 + 5^2 = 13^2

Step 2: Square the known numbers

a2+25=169 a^2 + 25 = 169

Step 3: Rearrange

a2=16925 a^2 = 169 - 25
a2=144 a^2 = 144

Step 4: Take the square root

a=144=12 a = \sqrt{144} = 12

Answer: The other shorter side is 12 m.


Example 3: Using the converse of Pythagoras’ theorem

Check whether a triangle with side lengths 9 cm, 12 cm, and 15 cm is right-angled.

Step 1: Identify the longest side

  • The longest side is 15 cm.
  • So let c=15c = 15, a=9a = 9, b=12b = 12.

Step 2: Use the converse Check whether:

a2+b2=c2 a^2 + b^2 = c^2

Step 3: Square the side lengths

92+122=152 9^2 + 12^2 = 15^2
81+144=225 81 + 144 = 225
225=225 225 = 225

Step 4: Conclude Since both sides are equal, the triangle satisfies Pythagoras’ theorem.

Answer: Yes, the triangle is right-angled.


Common Mistakes to Avoid

  • Using Pythagoras’ theorem on a triangle that is not right-angled.
  • Choosing the wrong side as the hypotenuse.
    • Remember: the hypotenuse is always opposite the 90° angle and is the longest side.
  • Forgetting to square the numbers.
    • Example: writing 3+4=53 + 4 = 5 instead of 32+42=523^2 + 4^2 = 5^2.
  • Forgetting to take the square root at the end when finding a side.
    • Example: stopping at c2=81c^2 = 81 and saying the answer is 81 instead of 9.
  • Mixing up the formula when finding a shorter side.
    • Correct method:

      shorter side2=hypotenuse2other shorter side2 \text{shorter side}^2 = \text{hypotenuse}^2 - \text{other shorter side}^2
  • Subtracting in the wrong order.
    • Always do:

      c2shorter side2 c^2 - \text{shorter side}^2
    • Not the other way round.

  • Forgetting to write units in the final answer.
  • Using different units without converting first.
    • Example: one side in cm and another in m.
  • For the converse, not using the longest side as cc.
  • Rounding too early during working, which may cause small errors.

Exam Tips

  • First check whether the triangle is right-angled before using Pythagoras’ theorem.

  • In your working, clearly state:

    • “Using Pythagoras’ theorem”
    • or “Using the converse of Pythagoras’ theorem”
  • Label the sides carefully:

    • cc = hypotenuse
    • aa and bb = shorter sides
  • Show substitution clearly:

    a2+b2=c2 a^2 + b^2 = c^2
    52+122=c2 5^2 + 12^2 = c^2
  • If checking whether a triangle is right-angled, write a full conclusion such as:

    • “Since a2+b2=c2a^2 + b^2 = c^2, the triangle is right-angled.”
    • “Since a2+b2c2a^2 + b^2 \ne c^2, the triangle is not right-angled.”
  • For word problems:

    • Identify the horizontal, vertical, and sloping sides.
    • Draw a neat diagram before calculating.
  • Write the final answer with:

    • correct units
    • correct degree of accuracy if required
  • If the answer is a decimal, keep enough figures during working and round only at the end.

  • Memorise common Pythagorean triples:

    • 3,4,53,4,5
    • 5,12,135,12,13
    • 8,15,178,15,17
    • 7,24,257,24,25

Quick Summary

  • Pythagoras’ theorem applies only to a right-angled triangle.

  • The formula is:

    a2+b2=c2 a^2 + b^2 = c^2
  • The hypotenuse is the side opposite the 90° angle and is the longest side.

  • To find the hypotenuse:

    c=a2+b2 c = \sqrt{a^2+b^2}
  • To find a shorter side:

    a=c2b2 a = \sqrt{c^2-b^2}

    or

    b=c2a2 b = \sqrt{c^2-a^2}
  • Common Pythagorean triples include:

    • 3,4,53,4,5
    • 5,12,135,12,13
    • 8,15,178,15,17
    • 7,24,257,24,25
  • Real-life applications include ladders, diagonals, map distances, and heights.

  • The converse of Pythagoras’ theorem is used to test whether a triangle is right-angled.

  • When using the converse, always treat the longest side as cc.

  • Use the same units for all sides before calculating.

  • A side length must be positive, so use the positive square root only.

  • Always include clear working, units, and a proper final statement in exam answers.

✏️ 27 practice questions available

30 questions from school exam papers

Q1

In the figure PQRS, PS = 3.75 cm, PQ = 9 cm, QS = 9.75 cm, QR = 18 cm and angle RSQ = 90°. Show that PQS is a right-angled triangle.

A quadrilateral PQRS with point S having a right angle marked. PS = 3.75 cm, PQ = 9 cm, QS = 9.75 cm, QR = 18 cm. Point S is connected to P and Q forming triangle PQS at the top. Point R is below S, with SR marked as vertical and QR = 18 cm connecting Q to R.
📊 Diagram: A quadrilateral PQRS with point S having a right angle marked. PS = 3.75 cm, PQ = 9 cm, QS = 9.75 cm, QR = 18 cm. Point S is connected to P and Q forming triangle PQS at the top. Point R is below S, with SR marked as vertical and QR = 18 cm connecting Q to R.
2 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q2

Find the length of RS.

Diagram for question 7b
1 mark
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q3

ABC is a triangle in which AB = 17 cm, BC = 15 cm and AC = 8 cm. Show that triangle ABC is a right-angled triangle.

A right-angled triangle ABC with the right angle at C. Side AB (hypotenuse) is labeled 17, side BC (base) is labeled 15, and side AC (height) is labeled 8.
📊 Diagram: A right-angled triangle ABC with the right angle at C. Side AB (hypotenuse) is labeled 17, side BC (base) is labeled 15, and side AC (height) is labeled 8.
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q4

ABC is a straight line. AB = 5 cm, DB = 12 cm, DC = 16 cm and angle CBD = 90°. Find the length of AD.

Triangle with vertices A, B, C on a straight line (with A on the left, B in the middle, C on the right). Point D is above the line ABC, forming a triangle. A perpendicular line is drawn from D to B on line AC, labeled 12 cm. Side DC is labeled 16 cm. AB is labeled 5 cm. The diagram shows a right angle at B where DB meets AC.
📊 Diagram: Triangle with vertices A, B, C on a straight line (with A on the left, B in the middle, C on the right). Point D is above the line ABC, forming a triangle. A perpendicular line is drawn from D to B on line AC, labeled 12 cm. Side DC is labeled 16 cm. AB is labeled 5 cm. The diagram shows a right angle at B where DB meets AC.
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q5

Find the number of students who scored at least 40 marks.

Same triangle as in 11a.
📊 Diagram: Same triangle as in 11a.
2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q6

In triangle PQR, PQ = 20 cm, QR = 25 cm and PR = 32 cm. Show that triangle PQR is not a right-angled triangle.

Diagram for question 8
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q7

Find the length of AC.

Same diagrams as in part (a)
📊 Diagram: Same diagrams as in part (a)
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q8

Hence, show that the triangle is a right-angled triangle.

Diagram for question 4b
3 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q9

ABC is a right-angled triangle. Find the length of AC.

A right-angled triangle ABC with the right angle at C. Side AB (hypotenuse) is labeled 41 cm. Side BC is labeled 9 cm. Side AC is unknown and needs to be found.
📊 Diagram: A right-angled triangle ABC with the right angle at C. Side AB (hypotenuse) is labeled 41 cm. Side BC is labeled 9 cm. Side AC is unknown and needs to be found.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q10

Another triangle PQR has the dimensions shown below. PQ = 20 cm, QR = 10 cm, PR = 18 cm. Stating your reasons clearly, determine whether triangle PQR is a right-angled triangle.

Triangle PQR with side PQ (base) = 20 cm, side QR = 10 cm, and side PR = 18 cm. The triangle appears to be scalene with no right angle symbol indicated.
📊 Diagram: Triangle PQR with side PQ (base) = 20 cm, side QR = 10 cm, and side PR = 18 cm. The triangle appears to be scalene with no right angle symbol indicated.
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q11

A ladder BC is leaning against the wall AB and touching the top of the wall at B. The height of the wall is 6 m and the distance from the foot of the ladder to the foot of the wall, AC, is 9.6 m. Find the length of the ladder BC.

A diagram showing a triangular configuration with a vertical wall AB of height 6 m, a horizontal ground line with point A at the base of the wall and point C at distance 9.6 m from A. A ladder BC leans against the wall, touching the top at B. The ladder forms the hypotenuse of a right triangle. There is also a point D on the ground to the left of A, and BD is marked as 8.2 m, indicating the shadow cast by the wall.
📊 Diagram: A diagram showing a triangular configuration with a vertical wall AB of height 6 m, a horizontal ground line with point A at the base of the wall and point C at distance 9.6 m from A. A ladder BC leans against the wall, touching the top at B. The ladder forms the hypotenuse of a right triangle. There is also a point D on the ground to the left of A, and BD is marked as 8.2 m, indicating the shadow cast by the wall.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q12

The tanker delivers the water to a factory. The factory uses the water to fill 500 cm³ bottles. How many full bottles can it fill from the tanker?

Same diagram as 7a, with emphasis on the shadow measurement. D is positioned on the ground to the left of A, with BD = 8.2 m marked as the distance from the top of the wall B to point D on the ground.
📊 Diagram: Same diagram as 7a, with emphasis on the shadow measurement. D is positioned on the ground to the left of A, with BD = 8.2 m marked as the distance from the top of the wall B to point D on the ground.
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q13

Measure and write down the angle opposite the longest side of the triangle.

Diagram for question 5b
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q14

Find the length of AB.

Diagram for question 6b
2 marks
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q15

Find the perpendicular distance from B to AC.

Triangle ABC with a dashed line from B perpendicular to AC.
📊 Diagram: Triangle ABC with a dashed line from B perpendicular to AC.
1 mark
Math_Sec2NA_SA2_2023_Chung_Cheng_Yishun 2023
Q16

In the diagram, Alice takes the route AN while Bala takes the route BN to travel to school from their houses. AOB is a straight line. AN = 10 km, BN = 8 km and ON = 6 km. Find the distance AOB.

A triangle with vertices at A (Alice's house, left), B (Bala's house, right), and N (School, top). The base AB is a straight line passing through point O. ON is perpendicular to AB with length 6 km marked. AN is marked as 10 km, BN is marked as 8 km. O lies on the line segment AB between A and B.
📊 Diagram: A triangle with vertices at A (Alice's house, left), B (Bala's house, right), and N (School, top). The base AB is a straight line passing through point O. ON is perpendicular to AB with length 6 km marked. AN is marked as 10 km, BN is marked as 8 km. O lies on the line segment AB between A and B.
3 marks
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q17

Show that ABN is a right-angled triangle.

Triangle ABC with point N on side BC. AB = 15 cm, BN = 12 cm, AN = 9 cm. The diagram shows vertex A at the top, vertex B at the bottom right, vertex C at the left, and point N on segment BC between B and C. Lines are drawn connecting A to B (15 cm), A to N (9 cm), and N to B (12 cm).
📊 Diagram: Triangle ABC with point N on side BC. AB = 15 cm, BN = 12 cm, AN = 9 cm. The diagram shows vertex A at the top, vertex B at the bottom right, vertex C at the left, and point N on segment BC between B and C. Lines are drawn connecting A to B (15 cm), A to N (9 cm), and N to B (12 cm).
Math_Sec2NA_SA2_2023_Springfield_Sec 2023
Q18

Calculate the total surface area of the tent, including the base.

Diagram for question 7c
2 marks
Math_Sec2_SA2_2023_Bedok_South_Sec 2023
Q19

A triangle ABC has sides AB = 36 cm, BC = 39 cm and AC = 15 cm. Prove that triangle ABC is a right-angled triangle.

Triangle ABC with base CB = 39 cm, side AB = 36 cm, and side AC = 15 cm. A perpendicular line is drawn from vertex A to base CB, meeting at point D. The perpendicular AD is labeled as 15 cm, and the base segment DB is labeled as 39 cm.
📊 Diagram: Triangle ABC with base CB = 39 cm, side AB = 36 cm, and side AC = 15 cm. A perpendicular line is drawn from vertex A to base CB, meeting at point D. The perpendicular AD is labeled as 15 cm, and the base segment DB is labeled as 39 cm.
2 marks
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q20

A bag contains 5 blue balls, 8 green balls and 3 yellow balls. A ball is drawn at random. Find the probability of getting a green ball.

Triangle ABC with point D on BC such that AD is perpendicular to BC. AB = 14.8 cm, AC = 17 cm, angle ACD = 41°. AD is drawn perpendicular to BC, forming a right angle at D.
📊 Diagram: Triangle ABC with point D on BC such that AD is perpendicular to BC. AB = 14.8 cm, AC = 17 cm, angle ACD = 41°. AD is drawn perpendicular to BC, forming a right angle at D.
2 marks
Math_Sec2_SA2_2023_Broadrick_Sec 2023
Q21

Triangle ABD is similar to triangle BCD. Given that angle BAD = 28°, angle BDA = 35°, AD = 13.8 cm and CD = 5 cm. Find angle ABC.

Two triangles are shown. Triangle ABC is congruent to triangle DAE. Points are labeled A, B, C, D, E, F where ACE is a straight line. AD = 8.7 cm, CE = 2.2 cm, DE = 6.8 cm, angle ADE = 70°, and angle ACB = 65°. Point F appears to be on line segment from A to another point, with various line segments connecting the vertices.
📊 Diagram: Two triangles are shown. Triangle ABC is congruent to triangle DAE. Points are labeled A, B, C, D, E, F where ACE is a straight line. AD = 8.7 cm, CE = 2.2 cm, DE = 6.8 cm, angle ADE = 70°, and angle ACB = 65°. Point F appears to be on line segment from A to another point, with various line segments connecting the vertices.
1 mark
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q22

The total surface area of the cuboid is 325 cm². Form an equation, in terms of x, to represent this information and show that it simplifies to 18x² + 65x - 275 = 0.

Same diagram as part (a)
📊 Diagram: Same diagram as part (a)
2 marks
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q23

Show that triangle ABD is a right-angled triangle.

A diagram showing triangle ABD with point C on line BD. AB = 7.5 cm, BD = 18 cm, AD = 19.5 cm, and AC = 18 cm. BCD is a straight line.
📊 Diagram: A diagram showing triangle ABD with point C on line BD. AB = 7.5 cm, BD = 18 cm, AD = 19.5 cm, and AC = 18 cm. BCD is a straight line.
2 marks
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q24

(b) Calculate the total amount, including credit card fee, Jade is charged for fuel. Give your answer in Singapore dollars correct to the nearest cent.

Diagram for question 9b
2 marks
Math_Sec2_SA2_2023_Bukit_Merah_Sec 2023
Q25

A triangle has sides AB = 8 cm, BC = 15 cm and AC = 17 cm.

Math_Sec2_SA2_2023_Canberra_Sec 2023
Q26

Hence, calculate the length of XZ.

Diagram for question 10b
3 marks
Math_Sec2_SA2_2023_Juying_Sec 2023
Q27

Calculate angle ADB.

A diagram showing a wooden pole AB of height 45m built on horizontal ground. Point B is at the base of the pole on the ground. A semi-elastic steel wire AC is attached at point A (top of pole) and makes an angle of 36° to the horizontal. Point C is on the ground. Point D is 22m from C on the ground, with BCD forming a straight line. The distances shown are: AB = 45m (vertical), AC at 36° angle, CD = 22m.
📊 Diagram: A diagram showing a wooden pole AB of height 45m built on horizontal ground. Point B is at the base of the pole on the ground. A semi-elastic steel wire AC is attached at point A (top of pole) and makes an angle of 36° to the horizontal. Point C is on the ground. Point D is 22m from C on the ground, with BCD forming a straight line. The distances shown are: AB = 45m (vertical), AC at 36° angle, CD = 22m.
4 marks
Math_Sec2_SA2_2023_Juying_Sec 2023

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