Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Prove triangles are similar using AA, SSS, and SAS criteria.
• Apply similarity to solve for unknown sides and angles.
• Apply the Pythagorean Theorem in 2D and 3D problems.
• Solve problems combining similarity and Pythagoras.
• Use proportionality theorems to solve problems.

Lesson summary

Euclidean geometry, named after the Greek mathematician Euclid, is a cornerstone of mathematics. This week, we will deepen our understanding of similarity and the Pythagorean Theorem, essential tools for solving problems involving shapes, sizes, and spatial relationships. Mastering these concepts equips you with vital problem-solving skills applicable in various fields, from architecture and engineering to surveying and even art. Think about how surveyors use geometric principles to map land boundaries or how architects design stable and aesthetically pleasing structures.

Lesson notes

2.1 Similarity of Triangles: Two triangles are said to be similar if their corresponding angles are equal (equiangular) and their corresponding sides are in proportion. The order of vertices is important when stating similarity (e.g., ΔABC ~ ΔDEF means angle A = angle D, angle B = angle E, angle C = angle F, and AB/DE = BC/EF = AC/DF).

Similarity Criteria: AA (Angle-Angle)

Similarity: If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. (Remember: if two angles are equal, the third must also be equal since the sum of angles in a triangle is always 180°).

SSS (Side-Side-Side)

Similarity: If the corresponding sides of two triangles are in proportion, then the triangles are similar. That is, if AB/DE = BC/EF = CA/FD, then ΔABC ~ ΔDE

F. SAS (Side-Angle-Side)

Similarity: If two sides of one triangle are proportional to two sides of another triangle, and the included angles are equal, then the triangles are similar. That is, if AB/DE = AC/DF and angle A = angle D, then ΔABC ~ ΔDE

F. Example 1 (AA Similarity): In ΔABC, angle A = 60° and angle B = 80°. In ΔPQR, angle P = 60° and angle Q = 80°. Prove that ΔABC ~ ΔPQ

R. Solution: Angle A = Angle P = 60° (Given) Angle B = Angle Q = 80° (Given) Therefore, ΔABC ~ ΔPQR (AA Similarity)

Example 2 (SSS Similarity): In ΔABC, AB = 3, BC = 4, and AC =

5. In ΔDEF, DE = 6, EF = 8, and DF =

1

0. Prove that ΔABC ~ ΔDE

F. Solution: AB/DE = 3/6 = 1/2 BC/EF = 4/8 = 1/2 AC/DF = 5/10 = 1/2 Since AB/DE = BC/EF = AC/DF, ΔABC ~ ΔDEF (SSS Similarity)

Example 3 (SAS Similarity): In ΔABC, AB = 4, AC = 6, and angle A = 50°. In ΔDEF, DE = 8, DF = 12, and angle D = 50°. Prove that ΔABC ~ ΔDE

F. Solution: AB/DE = 4/8 = 1/2 AC/DF = 6/12 = 1/2 Angle A = Angle D = 50° (Given) Since AB/DE = AC/DF and angle A = angle D, ΔABC ~ ΔDEF (SAS Similarity) 2.2 Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. If c is the hypotenuse and a and b are the other two sides, then: a² + b² = c² Application in 2D and 3D: The Pythagorean Theorem is not limited to 2D problems. It can be extended to find distances in 3D space.

Example 4 (2D): A ladder leans against a wall. The ladder is 5 meters long, and the base of the ladder is 3 meters from the wall. How high up the wall does the ladder reach?

Solution: Let h be the height the ladder reaches on the wall.

Using the Pythagorean Theorem: 3² + h² = 5² 9 + h² = 25 h² = 16 h = 4 meters.

Example 5 (3D): A rectangular prism has dimensions length = 4 cm, width = 3 cm, and height = 2 cm. Find the length of the diagonal from one corner to the opposite corner.

Solution: First, find the diagonal of the base (d): d² = 4² + 3² = 25, so d = 5 cm. Now, consider the right-angled triangle formed by the height (2 cm), the base diagonal (5 cm), and the space diagonal (D). D² = 5² + 2² = 29 D = √29 cm ≈ 5.39 cm. 2.3 Proportionality Theorems The Intercept Theorem (also known as Thales' Theorem) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.

Theorem: If DE || BC, then AD/DB = AE/EC Converse: If AD/DB = AE/EC, then DE || BC Example 6 In triangle ABC, DE is parallel to B

C. AD = 4 cm, DB = 6 cm, and AE = 5 cm. Find the length of E

C. Solution: Since DE || BC, by the Intercept Theorem, AD/DB = AE/EC Therefore, 4/6 = 5/EC Cross-multiplying, 4 * EC = 30 EC = 30/4 = 7.5 cm Guided Practice (With Solutions)

Question 1: In ΔABC, AB = 8 cm, AC = 6 cm, and BC = 10 cm. In ΔPQR, PQ = 4 cm, PR = 3 cm, and QR = 5 cm. Prove that ΔABC ~ ΔPQ

R. Solution: AB/PQ = 8/4 = 2 AC/PR = 6/3 = 2 BC/QR = 10/5 = 2 Since AB/PQ = AC/PR = BC/QR, ΔABC ~ ΔPQR (SSS Similarity).

Commentary: This problem directly applies the SSS similarity criterion. Notice how calculating the ratios of the corresponding sides is the key to proving similarity.

Question 2: Triangle KLM is right-angled at

L. If KL = 5 cm and KM = 13 cm, find the length of L

M. Solution: Using the Pythagorean Theorem: KL² + LM² = KM² 5² + LM² = 13² 25 + LM² = 169 LM² = 144 LM = 12 cm.

Commentary: This is a straightforward application of the Pythagorean Theorem in a right-angled triangle. Recognizing the hypotenuse is crucial.

Question 3: In ΔABC, DE is parallel to BC, where D lies on AB and E lies on A

C. If AD = 2, DB = 3, and AE = 4, find E

C. Solution: Since DE || BC, by the Intercept Theorem, AD/DB = AE/EC 2/3 = 4/EC 2*EC = 12 EC = 6 Question 4: Two right-angled triangles ABC and PQR are given where Angle B= Angle Q = 90°. If AB = 3, BC=4 and PQ=

6. What must the length of QR be to make these two triangles similar?

Solution: If triangle ABC is similar to triangle PQR, the ratios of the corresponding sides must be equal.

Therefore AB/PQ = BC/QR 3/6 = 4/QR 3 * QR = 24 QR = 8 Independent Practice (Questions Only) In ΔXYZ, angle X = 40° and angle Y = 70°. In ΔLMN, angle L = 40° and angle M = 70°.