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Give me note on Pythagoras theorem for primary 5 with much examples and also state the law,
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Pythagoras’ theorem (for Primary 5)
The law (short):
- In a right-angled triangle, (hypotenuse)² = (one shorter side)² + (the other shorter side)².
- In symbols: c² = a² + b², where c is the hypotenuse (the side opposite the right angle), and a and b are the two other sides.
What it means (simple):
- If you know the lengths of two sides of a right triangle, you can find the third side by using c² = a² + b².
- The hypotenuse is always the longest side.
How to use it:
- To find the hypotenuse: c = √(a² + b²).
- To find a missing short side: a = √(c² − b²) (or b = √(c² − a²)).
Worked examples
1) Easy example (3, 4, ?)
- a = 3, b = 4. Find c.
- c² = 3² + 4² = 9 + 16 = 25 → c = √25 = 5.
- So the triangle sides are 3, 4, 5.
2) Another whole-number example (6, 8, ?)
- a = 6, b = 8. c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10.
3) Find a missing short side (c = 13, a = 5, find b)
- c² = a² + b² → b² = c² − a² = 13² − 5² = 169 − 25 = 144 → b = √144 = 12.
4) Rectangle diagonal (length 12 m, width 5 m)
- The diagonal forms a right triangle with sides 12 and 5, so diagonal d = √(12² + 5²) = √(144 + 25) = √169 = 13 m.
5) Word problem (ladder)
- A ladder leans against a wall. The foot of the ladder is 3 m from the wall. The ladder reaches 4 m up the wall. How long is the ladder?
- Use triangle with a = 3, b = 4 → c = 5 m (same as example 1).
6) Square diagonal (side 7 cm)
- A square’s diagonal d = √(7² + 7²) = √(49 + 49) = √98 = √(49×2) = 7√2 ≈ 9.9 cm.
- If you want a decimal: d ≈ 9.90 cm.
7) Non-perfect square example (legs 3 cm and 7 cm)
- c = √(3² + 7²) = √(9 + 49) = √58 ≈ 7.62 cm.
Short demonstration you can try (visual)
- Draw a right triangle on squared paper. Make small squares on each of the three sides (build squares with side lengths equal to each triangle side). You will see the area of the square on the hypotenuse equals the sum of the areas of the two smaller squares. This is a visual way to feel the truth of the theorem.
Tips and checks
- Always make sure the triangle is right-angled before using Pythagoras.
- The hypotenuse is the longest side — use it as c.
- If a² + b² = c² exactly, the triangle is right-angled (the converse).
Practice (try these; answers below)
1) a = 5, b = 12. Find c.
2) c = 25, a = 7. Find b.
3) Rectangle 8 m by 15 m. Find diagonal.
4) Right triangle with sides 9 and 40. Find hypotenuse.
5) Square side 10 cm. Find diagonal (exact and decimal).
6) a = 2, b = 6. Find hypotenuse (decimal).
Answers
1) c = √(5² + 12²) = √(25 + 144) = √169 = 13.
2) b = √(25² − 7²) = √(625 − 49) = √576 = 24.
3) diagonal = √(8² + 15²) = √(64 + 225) = √289 = 17 m.
4) c = √(9² + 40²) = √(81 + 1600) = √1681 = 41.
5) diagonal = √(10² + 10²) = √200 = 10√2 ≈ 14.14 cm.
6) c = √(2² + 6²) = √(4 + 36) = √40 = 2√10 ≈ 6.32.
If you want, I can make printable practice sheets, draw step-by-step diagrams, or show a simple visual proof with labelled squares. Which would you like next?
The law (short):
- In a right-angled triangle, (hypotenuse)² = (one shorter side)² + (the other shorter side)².
- In symbols: c² = a² + b², where c is the hypotenuse (the side opposite the right angle), and a and b are the two other sides.
What it means (simple):
- If you know the lengths of two sides of a right triangle, you can find the third side by using c² = a² + b².
- The hypotenuse is always the longest side.
How to use it:
- To find the hypotenuse: c = √(a² + b²).
- To find a missing short side: a = √(c² − b²) (or b = √(c² − a²)).
Worked examples
1) Easy example (3, 4, ?)
- a = 3, b = 4. Find c.
- c² = 3² + 4² = 9 + 16 = 25 → c = √25 = 5.
- So the triangle sides are 3, 4, 5.
2) Another whole-number example (6, 8, ?)
- a = 6, b = 8. c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10.
3) Find a missing short side (c = 13, a = 5, find b)
- c² = a² + b² → b² = c² − a² = 13² − 5² = 169 − 25 = 144 → b = √144 = 12.
4) Rectangle diagonal (length 12 m, width 5 m)
- The diagonal forms a right triangle with sides 12 and 5, so diagonal d = √(12² + 5²) = √(144 + 25) = √169 = 13 m.
5) Word problem (ladder)
- A ladder leans against a wall. The foot of the ladder is 3 m from the wall. The ladder reaches 4 m up the wall. How long is the ladder?
- Use triangle with a = 3, b = 4 → c = 5 m (same as example 1).
6) Square diagonal (side 7 cm)
- A square’s diagonal d = √(7² + 7²) = √(49 + 49) = √98 = √(49×2) = 7√2 ≈ 9.9 cm.
- If you want a decimal: d ≈ 9.90 cm.
7) Non-perfect square example (legs 3 cm and 7 cm)
- c = √(3² + 7²) = √(9 + 49) = √58 ≈ 7.62 cm.
Short demonstration you can try (visual)
- Draw a right triangle on squared paper. Make small squares on each of the three sides (build squares with side lengths equal to each triangle side). You will see the area of the square on the hypotenuse equals the sum of the areas of the two smaller squares. This is a visual way to feel the truth of the theorem.
Tips and checks
- Always make sure the triangle is right-angled before using Pythagoras.
- The hypotenuse is the longest side — use it as c.
- If a² + b² = c² exactly, the triangle is right-angled (the converse).
Practice (try these; answers below)
1) a = 5, b = 12. Find c.
2) c = 25, a = 7. Find b.
3) Rectangle 8 m by 15 m. Find diagonal.
4) Right triangle with sides 9 and 40. Find hypotenuse.
5) Square side 10 cm. Find diagonal (exact and decimal).
6) a = 2, b = 6. Find hypotenuse (decimal).
Answers
1) c = √(5² + 12²) = √(25 + 144) = √169 = 13.
2) b = √(25² − 7²) = √(625 − 49) = √576 = 24.
3) diagonal = √(8² + 15²) = √(64 + 225) = √289 = 17 m.
4) c = √(9² + 40²) = √(81 + 1600) = √1681 = 41.
5) diagonal = √(10² + 10²) = √200 = 10√2 ≈ 14.14 cm.
6) c = √(2² + 6²) = √(4 + 36) = √40 = 2√10 ≈ 6.32.
If you want, I can make printable practice sheets, draw step-by-step diagrams, or show a simple visual proof with labelled squares. Which would you like next?
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