Loci & 3D Solids
Interactive lessons, worked examples, and problem-solving techniques to help Kenyan students master KJSEA, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Construct and describe loci in 2D and 3D, calculate 3D distance between coordinates, and evaluate surface areas and volumes of 3D solids.
A locus (plural: loci) is the path or region formed by all points satisfying a given geometric condition.
(a) Common Geometric Loci
- Fixed Distance from a Point: Circle in 2D; Sphere in 3D (\( r = \text{radius} \)).
- Fixed Distance from a Line: Pair of parallel lines in 2D; Cylindrical surface in 3D.
- Equidistant from Two Points: Perpendicular bisector in 2D; Bisector plane in 3D.
(b) 3D Distance Formula
Extending Pythagoras to three dimensions gives the straight-line distance between \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \):
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]Module 1: Loci Gallery
Choose a condition and watch the locus form. Adjust the distance slider.
Module 2: Shape Unfolder
Select a solid and unfold it into its net. Remember: a net is NOT a cross-section.
Key Formulas
Worked Examples
- Apply formula: \( d = \sqrt{(4-1)^2 + (6-2)^2 + (3-3)^2} \).
- Simplify: \( d = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16 + 0} = \sqrt{25} \).
- Calculate root: \( d = 5 \).
Answer: 5
- Apply surface area formula: \( A = 2(lw + wh + hl) \).
- Substitute values: \( A = 2(8 \times 5 + 5 \times 3 + 3 \times 8) = 2(40 + 15 + 24) \).
- Sum terms: \( A = 2(79) = 158\text{ cm}^2 \).
Answer: 158
- Use formula \( V = \pi r^2 h \implies 2000 = \pi \cdot r^2 \cdot 20 \).
- Divide by \( 20\pi \): \( r^2 = 2000 / (20\pi) = 100 / \pi \approx 31.831 \).
- Take square root: \( r = \sqrt{31.831} \approx 5.64\text{ cm} \).
Answer: 5.64
Common Mistakes
Real World
Practice