Pythagoras Applications
Apply Pythagoras' theorem to 3D problems, find diagonals of rectangles and solids, and use the distance formula on the Cartesian plane.
Review of Pythagoras' Theorem
Pythagoras' theorem states that in a right-angled triangle, the square of the hypotenuse (the longest side) equals the sum of the squares of the other two sides.
Pythagoras' Theorem
c² = a² + b²
where c is the hypotenuse and a, b are the other two sides.
Applying Pythagoras in 3D
Pythagoras' theorem can be applied in three dimensions by breaking 3D problems into two separate right-triangle steps. A common example is finding the diagonal of a rectangular box (cuboid).
Diagonal of a Rectangular Box
For a box with length l, width w, and height h:
d = √(l² + w² + h²)
Method: First find the diagonal of the base: dbase = √(l² + w²). Then use Pythagoras again with the height: d = √(dbase² + h²).
The Distance Formula
The distance formula is derived from Pythagoras' theorem. It gives the straight-line distance between two points (x1, y1) and (x2, y2) on the Cartesian plane.
Distance Formula
d = √[(x2 − x1)² + (y2 − y1)²]
d = √(4² + 4²) = √32 = 4√2 ≈ 5.66 units.
Key Vocabulary
| Term | Definition |
|---|---|
| Hypotenuse | The longest side of a right-angled triangle; the side opposite the right angle. |
| Pythagoras' theorem | In a right-angled triangle: c² = a² + b², where c is the hypotenuse. |
| Distance formula | d = √[(x2−x1)² + (y2−y1)²]. Finds distance between two Cartesian points. |
| Space diagonal | The longest diagonal inside a 3D solid, passing through the interior of the shape. |
Worked Examples
Diagonal of a Rectangle
A rectangle is 8 cm wide and 6 cm tall. Find the length of the diagonal.
Step 1: Identify the right-angled triangle: legs are 8 and 6, diagonal is hypotenuse.
Step 2: c² = 8² + 6² = 64 + 36 = 100
Step 3: c = √100 = 10 cm
Space Diagonal of a Box
A box is 3 m long, 4 m wide and 12 m tall. Find the space diagonal.
Step 1: Base diagonal. dbase = √(3² + 4²) = √25 = 5 m
Step 2: Space diagonal. d = √(5² + 12²) = √(25 + 144) = √169 = 13 m
Distance Between Two Points
Find the distance between A(2, 1) and B(7, 13).
Step 1: Δx = 7 − 2 = 5, Δy = 13 − 1 = 12
Step 2: d = √(5² + 12²) = √(25 + 144) = √169 = 13 units
Knowledge Check
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Key Concepts Summary
- ●Pythagoras' theorem: c² = a² + b². Applies to any right-angled triangle.
- ●For 3D problems, apply Pythagoras twice: once for the base diagonal, then again with the height.
- ●Space diagonal of a cuboid: d = √(l² + w² + h²).
- ●Distance formula: d = √[(x2−x1)² + (y2−y1)²].
- ●Common Pythagorean triples to remember: 3-4-5, 5-12-13, 8-15-17.