Practical Geometry – Study Notes (NMMS | SAT)
Purpose – Build an intuitive, picture‑driven understanding of geometry that can be applied to everyday problems (e.g. measuring room tiles, designing furniture, cutting out shapes from paper, etc.).
Scope – 2‑D plane figures, 3‑D solids, coordinate methods, and some simple trigonometry.
Audience – Students preparing for the NMMS Maths Test (SAT‑style) who need practical solutions, not just formulas.
1. Simple Explanation
Practical geometry means using geometry to solve real‑world tasks:
| Real‑world task | Geometry concepts used | Quick Idea |
|---|---|---|
| Cutting a circular pizza into equal slices | Angle‑bisectors, circle sectors | Divide (360^\circ) by the number of slices |
| Laying square tiles in a rectangular room | Area, perimeter | Count tiles = room area ÷ tile area |
| Building a pyramid from cardboard | Pyramid volume, surface area | Use (V=\frac13Bh) where (B)=base area |
| Finding the shortest path across a lake | Distance formula | Use the straight‑line segment (chord) |
Key to success: visualise the problem, draw a diagram, and pick the right formula from the table below.
2. Key Points (Cheat‑Sheet)
| Category | Concept | Formula (with notation) | Notes |
|---|---|---|---|
| Basic Elements | Triangle | ( \text{Area}= \frac12 \times \text{base} \times \text{height}) | Height is perpendicular to base |
| Rectangle | ( \text{Area}=l \times w) | (l) = length, (w) = width | |
| Square | ( \text{Area}=s^2) | (s) = side | |
| Circle | ( \text{Area}= \pi r^2) ( \text{Circumference}=2\pi r) |
(r)=radius | |
| Perimeter / Circumference | Polygon | Sum of side lengths | |
| Circle | (2\pi r) | ||
| Coordinate Geometry | Distance between ((x_1,y_1)) and ((x_2,y_2)) | (d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}) | |
| Midpoint | (M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)) | ||
| Angles & Circles | Central angle (\theta) (in degrees) | Sector area ( = \frac{\theta}{360} \pi r^2) | |
| Inscribed angle (\alpha) | ( \alpha = \frac12 \theta) | ||
| Triangles | Pythagorean Theorem | (a^2+b^2=c^2) | For right triangles |
| Similar triangles | ( \frac{a}{x}=\frac{b}{y}=\frac{c}{z}) | Ratios of corresponding sides | |
| Law of Cosines (optional) | (c^2=a^2+b^2-2ab\cos C) | ||
| 3‑D Solids | Cube | Volume = (s^3) Surface area = (6s^2) |
|
| Rectangular prism | Volume = (lwh) Surface area = (2(lw+lh+wh)) |
||
| Cylinder | Volume = (\pi r^2h) Surface area = (2\pi r(h+r)) |
||
| Pyramid | Volume = (\frac13Bh) | (B)=base area, (h)=height | |
| Applications | Finding missing side with similar triangles | Use similarity ratio | e.g., a ladder leaning against a wall |
| “Cut‑and‑paste” area problems | Subtract or add areas | e.g., net of a box |
Tip: Keep a “formula‑card” handy. When you see a shape, quickly decide if you need area, perimeter, volume, or a distance formula.
3. Worked Example (Real‑World)
Problem
A rectangular garden measures 12 m by 8 m.
You want to lay 30 cm × 30 cm square tiles.
Questions:
- How many tiles are needed?
- If each tile costs ₹45, what is the total cost?
- How many tiles must be cut (i.e., incomplete tiles) to fill the garden exactly?
Step‑by‑step Solution
-
Convert units
Tile size: 30 cm = 0.30 m.
Garden area: (12 \times 8 = 96 \text{ m}^2).
Tile area: (0.30 \times 0.30 = 0.09 \text{ m}^2). -
Total tiles (ideal, without cutting)
[ \frac{96}{0.09} = 1066.\overline{6} ] Since we can’t have a fraction of a tile, we round up to 1067 tiles. -
Count of whole tiles
Along length: ( \lceil 12 / 0.30 \rceil = \lceil 40 \rceil = 40) tiles.
Along width: ( \lceil 8 / 0.30 \rceil = \lceil 26.\overline{6} \rceil = 27) tiles.
Total whole tiles: (40 \times 27 = 1080).
But 1080 tiles over‑cover the garden area:
[ 1080 \times 0.09 = 97.2 \text{ m}^2 > 96 \text{ m}^2 ] So we have 3 m² of excess.