Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply the Theorem of Pythagoras to solve problems.
• Prove triangles are similar using AA, SSS, and SAS.
• Apply similarity to solve problems with proportional sides.
• Solve riders combining Pythagoras and Similarity.

Lesson summary

Euclidean geometry is a fundamental branch of mathematics that deals with shapes, sizes, and relative positions of figures in space. It's the geometry we experience in our everyday world. This week, we focus on two crucial concepts: similarity and the Theorem of Pythagoras, and how they interact. Understanding these principles is essential not only for success in Mathematics but also for a deeper appreciation of the world around us. Think about architects designing buildings, engineers constructing bridges, or even artists creating visually appealing compositions - they all rely on these geometric principles.

Lesson notes

2.1 The Theorem of Pythagoras The Theorem of Pythagoras is one of the most fundamental theorems in Euclidean geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (legs).

Mathematically: a² + b² = c² where: `a` and `b` are the lengths of the legs of the right triangle. `c` is the length of the hypotenuse. Important

Note: This theorem only applies to right-angled triangles. Proof Outline (for understanding, not required for application in this section): Visual proofs are often used to demonstrate the validity of the theorem. One common proof involves constructing squares on each side of the right triangle and showing that the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

Example 1: In a right-angled triangle, one leg is 3 cm and the other leg is 4 cm. Find the length of the hypotenuse.

Solution: a = 3 cm b = 4 cm c = ?

Applying the Theorem of Pythagoras: 3² + 4² = c² 9 + 16 = c² 25 = c² c = √25 c = 5 cm Therefore, the length of the hypotenuse is 5 cm.

Example 2: A ladder 5 meters long leans against a wall. The base of the ladder is 1.5 meters away from the wall. How high up the wall does the ladder reach?

Solution: c (hypotenuse) = 5 m (length of the ladder) b = 1.5 m (distance from the wall) a = ? (height up the wall)

Applying the Theorem of Pythagoras: a² + 1.5² = 5² a² + 2.25 = 25 a² = 25 - 2.25 a² = 22.75 a = √22.75 a ≈ 4.77 m Therefore, the ladder reaches approximately 4.77 meters up the wall. 2.2 Similarity of Triangles Two triangles are said to be similar if they have the same shape but may differ in size. Formally, two triangles are similar if: AA (Angle-Angle): Two angles of one triangle are congruent (equal in measure) to two angles of the other triangle. If two angles are the same, the third must also be the same because the angles in a triangle add to 180 degrees.

SSS (Side-Side-Side): All three sides of one triangle are proportional to the corresponding three sides of the other triangle. That is, the ratios of corresponding sides are equal.

SAS (Side-Angle-Side): Two sides of one triangle are proportional to the corresponding two sides of the other triangle, and the included angles (the angles between those sides) are congruent.

Properties of Similar Triangles: Corresponding angles are congruent (equal in measure). Corresponding sides are proportional (the ratios of their lengths are equal). Important

Note: The order of the vertices in the similarity statement (e.g., ∆ABC ~ ∆DEF) indicates which vertices correspond to each other. So, ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F, and AB/DE = BC/EF = AC/D

F. Example 3: ∆ABC and ∆PQR are given. ∠A = 50°, ∠B = 70°, ∠P = 50°, and ∠Q = 70°. Are the triangles similar?

Solution: Since ∠A = ∠P = 50° and ∠B = ∠Q = 70°, by the AA similarity theorem, ∆ABC ~ ∆PQ

R. Example 4: ∆ABC has sides AB = 4 cm, BC = 6 cm, and AC = 8 cm. ∆DEF has sides DE = 6 cm, EF = 9 cm, and DF = 12 cm. Are the triangles similar?

Solution: Calculate the ratios of corresponding sides: AB/DE = 4/6 = 2/3 BC/EF = 6/9 = 2/3 AC/DF = 8/12 = 2/3 Since the ratios of all three corresponding sides are equal, by the SSS similarity theorem, ∆ABC ~ ∆DE

F. Example 5: In ∆ABC, AB = 3 cm, AC = 5 cm, and ∠A = 60°. In ∆PQR, PQ = 6 cm, PR = 10 cm, and ∠P = 60°. Are the triangles similar?

Solution: AB/PQ = 3/6 = 1/2 AC/PR = 5/10 = 1/2 ∠A = ∠P = 60° Since two sides are proportional and the included angle is congruent, by the SAS similarity theorem, ∆ABC ~ ∆PQR. 2.3 Combining Pythagoras and Similarity Often, problems require the combined application of the Pythagorean Theorem and the concepts of similar triangles. These problems test your ability to recognize geometric relationships and apply the appropriate theorems.

Example 6: In the diagram, ∆ABC is right-angled at B. BD is perpendicular to A

C. Prove that ∆ABC ~ ∆ADB ~ ∆BDC and find the length of BD if AB = 4 cm and BC = 3 cm.

Solution: Proof of Similarity: In ∆ABC and ∆ADB: ∠ABC = ∠ADB = 90° and ∠A is common.

Therefore, ∆ABC ~ ∆ADB (AA similarity). In ∆ABC and ∆BDC: ∠ABC = ∠BDC = 90° and ∠C is common.

Therefore, ∆ABC ~ ∆BDC (AA similarity). Since ∆ABC ~ ∆ADB and ∆ABC ~ ∆BDC, it follows that ∆ADB ~ ∆BD

C. Finding BD: First, find AC using the Theorem of Pythagoras in ∆ABC: AC² = AB² + BC² = 4² + 3² = 16 + 9 =

2

5. Therefore, AC = 5 cm. Since ∆ABC ~ ∆ADB, we have AB/AC = AD/AB, or 4/5 = AD/

4. Therefore, AD = (44)/5 = 3.2 cm. Now, using the Theorem of Pythagoras in ∆ADB: AB² = AD² + BD², or 4² = 3.2² + BD².

Therefore, BD² = 16 - 10.24 = 5.76, and BD = √5.76 = 2.4 cm. Guided Practice (With Solutions)

Question 1: Triangle PQR is a right-angled triangle with ∠Q = 90°. If PQ = 8 cm and PR = 17 cm, calculate the length of Q

R. Solution: Identify the sides: PR is the hypotenuse (c), PQ is one leg (a), and QR is the other leg (b).