Congruence, Similarity, Pythagoras
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Identify congruent and similar shapes, and apply Pythagoras' theorem to solve geometric problems.
Concrete Scenario: A matatu driver in Nairobi needs a ramp to load heavy cargo onto a lorry. The vertical height of the lorry bed is \(3\,\text{m}\) and the ground distance is \(4\,\text{m}\). What length of wooden timber must be cut for the ramp? Using Pythagoras' theorem (\(a^2 + b^2 = c^2\)), the driver calculates \(\sqrt{3^2 + 4^2} = 5\,\text{m}\).
Geometric Insight:
- Congruence (\(\cong\)): Identical in both shape and size. Corresponding angles and corresponding side lengths are exactly equal.
- Similarity (\(\sim\)): Same shape, but different sizes. Corresponding angles are equal, while corresponding sides are in the same constant ratio \(k\) (scale factor).
- Pythagoras' Theorem: In any right-angled triangle, the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides.
Visualizing Pythagoras (3-4-5 Triangle)
\(a^2 + b^2 = c^2 \implies 9 + 16 = 25 \implies c = 5\)
Key Formulas
Worked Examples
Problem: Triangle A has sides 5 cm, 7 cm, and 9 cm. Triangle B has sides 9 cm, 5 cm, and 7 cm. Are they congruent?
- Compare corresponding side lengths: both triangles have side lengths of 5 cm, 7 cm, and 9 cm.
- Apply the SSS (Side-Side-Side) rule: Since all three corresponding sides are equal, the two triangles are congruent.
Problem: Triangles ABC and DEF are similar (\(\triangle ABC \sim \triangle DEF\)). If \(AB = 8\,\text{cm}\), \(DE = 12\,\text{cm}\), and \(BC = 6\,\text{cm}\), find \(EF\).
- Calculate linear scale factor \(k = \frac{DE}{AB} = \frac{12}{8} = 1.5\).
- Multiply corresponding side \(BC\) by \(k\): \[EF = BC \times k = 6 \times 1.5 = 9\,\text{cm}\]
Problem: In a right-angled triangle with legs \(a = 6\,\text{cm}\) and \(b = 8\,\text{cm}\), calculate the length of the altitude \(h\) drawn from the right angle to the hypotenuse.
- Calculate the hypotenuse \(c\): \[c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\,\text{cm}\]
- Express the area of the triangle in two ways: \[\text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 6 \times 8 = 24\,\text{cm}^2\] \[\text{Area} = \frac{1}{2} \times c \times h = \frac{1}{2} \times 10 \times h = 5h\]
- Equate the areas and solve for \(h\): \[5h = 24 \implies h = 4.8\,\text{cm}\]
Common Mistakes
Real World
Practice