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IB MYP 4-5 Maths-Pythagoras’ theorem- Study Notes

IB MYP 4-5 Maths- Pythagoras’ theorem- Study Notes - New Syllabus

IB MYP 4-5 Maths- Pythagoras’ theorem – Study Notes

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  • Pythagoras’ theorem

IB MYP 4-5 Maths- Pythagoras’ theorem – Study Notes – All topics

Pythagoras’ Theorem

Pythagoras’ Theorem

Pythagoras’ Theorem applies to right-angled triangles and states that:

“The square of the hypotenuse is equal to the sum of the squares of the other two sides.”

Formula: \( a^2 + b^2 = c^2 \)

  • \( c \) = hypotenuse (longest side, opposite the 90° angle)
  • \( a, b \) = other two sides (legs)

Important Properties and Notes:

  • Only applies to right-angled triangles.
  • The hypotenuse is always the longest side and opposite the right angle.
  • The theorem can be used to:
    • Find the length of a missing side.
    • Check if a triangle is right-angled by verifying \( a^2 + b^2 = c^2 \).
  • The converse of Pythagoras’ theorem: If \( a^2 + b^2 = c^2 \), then the triangle is right-angled.
  • Extended to 3D: \( d = \sqrt{x^2 + y^2 + z^2} \) for the distance in 3D space.

Steps to Use Pythagoras’ Theorem:

  1. Check if the triangle is right-angled (or if given).
  2. Identify the hypotenuse (longest side).
  3. Apply the formula: \( a^2 + b^2 = c^2 \).
  4. Solve for the unknown side (take square root if needed).

Common Uses in Real Life:

  • Finding the shortest distance between two points (straight line).
  • Construction and architecture for ensuring right angles.
  • Navigation and mapping (finding distances using coordinates).

Example : 

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

▶️ Answer/Explanation

\( c^2 = a^2 + b^2 = 6^2 + 8^2 = 36 + 64 = 100 \)

\( c = \sqrt{100} = \boxed{10 \text{ cm}} \)

Example :

The hypotenuse of a right-angled triangle is 13 cm, and one side is 5 cm. Find the other side.

▶️ Answer/Explanation

\( c^2 = a^2 + b^2 \)

\( 13^2 = 5^2 + b^2 \Rightarrow 169 = 25 + b^2 \Rightarrow b^2 = 144 \Rightarrow b = 12 \)

Answer: \( \boxed{12 \text{ cm}} \)

Example : 

Check whether a triangle with sides 7 cm, 24 cm, and 25 cm is right-angled.

▶️ Answer/Explanation

Longest side = 25 cm, so check \( a^2 + b^2 = c^2 \):

\( 7^2 + 24^2 = 49 + 576 = 625 \)

\( c^2 = 25^2 = 625 \), so the triangle is right-angled.

Example : 

A ladder 10 m long is leaning against a wall. The bottom of the ladder is 6 m away from the wall. Find how high the ladder reaches on the wall.

▶️ Answer/Explanation

\( c^2 = a^2 + b^2 \), where c = 10 m (ladder), a = 6 m (distance from wall).

\( 10^2 = 6^2 + h^2 \Rightarrow 100 = 36 + h^2 \Rightarrow h^2 = 64 \Rightarrow h = 8 \)

Answer: Ladder reaches \(\boxed{8 \text{ m}}\).

Example : 

Find the diagonal of a cuboid with sides 4 cm, 3 cm, and 12 cm.

▶️ Answer/Explanation

Diagonal \( d = \sqrt{x^2 + y^2 + z^2} = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \text{ cm}\).

Answer: \( \boxed{13 \text{ cm}} \).

Example:

The shape $ABCDEFGH$ is a cuboid.

$AB = 6\,\text{cm},\quad BG = 3\,\text{cm},\quad FG = 2\,\text{cm}$ 
$AF = 7\,\text{cm}$

Calculate the angle between $AF$ and the plane $ABCD$.

▶️ Answer/Explanation

Construct right triangle $\triangle AFC$:

$AF = 7\,\text{cm}$ (hypotenuse)
$FC = 3\,\text{cm}$ (perpendicular to base)
Let angle between $AF$ and base $ABCD$ be $x$

Using:

$
\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{FC}{AF} = \frac{3}{7}
$

$
\sin x \approx 0.428571
$

$
x = \sin^{-1}(0.428571) \approx \boxed{25.4^\circ}
\quad \text{(3 significant figures)}
$

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