Geometry and Measurement
Geometry is the branch of mathematics that deals with shapes, sizes, positions, and dimensions. For the Post-UTME examination, you must master the properties of 2D shapes, the Pythagorean theorem, perimeters, areas, volumes, and the use of conversion factors. Geometry word problems are common and often require combining geometric knowledge with algebra.
Basic Geometric Figures
Triangle
A triangle has three sides, three angles, and the sum of its interior angles is always 180°.
Types of Triangles:
| Type | By Sides | By Angles |
|---|---|---|
| Equilateral | All sides equal; all angles = 60° | All angles < 90° |
| Isosceles | Two sides equal | Two angles equal |
| Scalene | No sides equal | One angle > 90° (obtuse) |
Right-angled triangle: One angle = 90°. The side opposite is the hypotenuse (longest side).
The Pythagorean Theorem
For a right-angled triangle with legs a and b and hypotenuse c: a² + b² = c²
Pythagorean Triples (whole number solutions):
- (3, 4, 5) — most common
- (5, 12, 13); (8, 15, 17); (7, 24, 25); (6, 8, 10)
Example: A ladder 13 m long leans against a wall, foot 5 m from wall. Height? c = 13, a = 5; b² = 13² − 5² = 169 − 25 = 144; b = 12 m
Perimeter and Area Formulas
| Shape | Perimeter | Area |
|---|---|---|
| Square | P = 4s | A = s² |
| Rectangle | P = 2(l+w) | A = lw |
| Triangle | P = a+b+c | A = ½bh |
| Circle | C = 2πr | A = πr² |
| Parallelogram | P = 2(a+b) | A = bh |
| Trapezium | P = sum of all sides | A = ½(a+b)h |
| Rhombus | P = 4s | A = ½(d₁×d₂) |
Example: Triangle with base 8 cm and height 5 cm. A = ½ × 8 × 5 = 20 cm²
Example: Circle with radius 7 cm (π = 22/7). Circumference = 2 × (22/7) × 7 = 44 cm; Area = (22/7) × 49 = 154 cm²
Circles: Key Properties
- Radius (r): Center to any point on circle
- Diameter (d): 2r
- Circumference (C): 2πr or πd
- Area: πr²
- Angle in a Semicircle: Always 90° (right angle)
- Tangent: Perpendicular to the radius at the point of contact
Solid Figures: Surface Area and Volume
| Solid | Volume | Surface Area |
|---|---|---|
| Cube | V = s³ | SA = 6s² |
| Cuboid | V = lwh | SA = 2(lw+lh+wh) |
| Cylinder | V = πr²h | CSA = 2πrh; TSA = 2πr(r+h) |
| Sphere | V = (4/3)πr³ | SA = 4πr² |
| Cone | V = (1/3)πr²h | CSA = πrl |
Example: Cylindrical tank: radius 3 m, height 7 m. Volume (π = 22/7)? V = (22/7) × 3² × 7 = 22 × 9 = 198 m³
Angle Properties
- Angles on a straight line: Sum to 180°
- Angles around a point: Sum to 360°
- Vertically opposite angles: Equal
- Angles in a triangle: Sum to 180°
- Angles in a quadrilateral: Sum to 360°
Congruent and Similar Triangles
Congruent triangles: Same shape AND same size (all corresponding sides and angles equal). Tests: SSS, SAS, ASA, AAS, RHS (for right-angled triangles).
Similar triangles: Same shape but different sizes (corresponding angles equal, sides in proportion).
If two triangles are similar: (corresponding side of triangle 1)/(corresponding side of triangle 2) = constant ratio.
Key Post-UTME Exam Facts
- Pythagorean theorem: a² + b² = c² — only for right-angled triangles
- Triangle angles: Always sum to 180°
- Circle: C = 2πr; A = πr²
- Volume of cylinder: V = πr²h; Volume of cone = (1/3)πr²h
- Similar triangles: Corresponding angles equal; sides in proportion
- ⚡ Exam tip: Pythagorean triples (3,4,5) and their multiples appear frequently — memorize them!
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
- Pythagoras: a² + b² = c² for right-angled triangles only
- Area formulas: Triangle = ½bh; Rectangle = lw; Circle = πr²
- Perimeter: Square = 4s; Rectangle = 2(l+w); Circle = 2πr
- Volume: Cylinder = πr²h; Cone = (1/3)πr²h; Sphere = (4/3)πr³
- Angle facts: Straight line = 180°; Triangle = 180°; Quadrilateral = 360°; Around point = 360°
- ⚡ Exam tip: For “semicircle angle” problems, the angle at the circumference subtended by the diameter is always 90°
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Practical Geometry Word Problems
Example: A rectangular garden is 20 m by 15 m. A path 2 m wide is built around it. Find the area of the path.
- Outer rectangle: (20+4) × (15+4) = 24 × 19 = 456 m²
- Inner rectangle: 20 × 15 = 300 m²
- Path area: 456 − 300 = 156 m²
Example: A cone has base radius 3 cm and slant height 5 cm. Find its total surface area. TSA = πr(r + l) = (22/7) × 3 × (3+5) = (22/7) × 24 = 528/7 ≈ 75.43 cm²
Geometry of Circles
Arc Length and Sector Area
- Arc length: (θ/360) × 2πr (where θ is the angle in degrees)
- Sector area: (θ/360) × πr²
Example: Find the area of a sector of radius 6 cm with angle 60°. = (60/360) × π × 36 = (1/6) × 36π = 6π cm²
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Scale Drawing and Map Problems
Example: A map has scale 1:50,000. The distance between two towns is 8 cm on the map. What is the actual distance in km?
- 1 cm = 50,000 cm = 500 m = 0.5 km
- Actual = 8 × 0.5 km = 4 km
More Solid Geometry
Pyramid
- Volume: V = (1/3) × Base Area × Height
Example: A square-based pyramid has base side 6 cm and height 4 cm. Base area = 36 cm²; V = (1/3) × 36 × 4 = 48 cm³
Locus Problems
The locus of points equidistant from a fixed point is a circle. The locus of points equidistant from two intersecting lines is the bisectors (angle bisector). The locus of points equidistant from two fixed points is the perpendicular bisector of the line joining them.
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