Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Prove and apply theorems on similar triangles.
• Apply Pythagoras' theorem in complex figures.
• Solve problems involving geometric means.
• Use properties of similar triangles to solve for scale and ratio.
• Construct geometric figures applying these theorems.

Lesson summary

Euclidean geometry provides the foundations for understanding shapes, sizes, and spatial relationships in the world around us. In Grade 12, we delve deeper into the concepts of similarity and Pythagoras' theorem, building upon what you've learned in previous grades. This week, we will be focusing on applying these concepts to solve more complex problems and prove geometrical theorems. Mastery of these concepts is crucial, not only for success in Mathematics but also for applications in fields like architecture, engineering, surveying, and even art and design.

Lesson notes

2.1 Similarity Two geometric figures are similar if they have the same shape but potentially different sizes. This means their corresponding angles are equal (congruent) and their corresponding sides are in proportion. We denote similarity using the symbol ~.

Criteria for Triangle Similarity: There are three main criteria for proving that two triangles are similar: AAA (Angle-Angle-Angle)

Similarity: If all three angles of one triangle are congruent to the corresponding three angles of another triangle, then the triangles are similar. Crucially, proving two pairs of angles are congruent is sufficient because the third angle is then automatically congruent (angles in a triangle sum to 180 degrees).

SSS (Side-Side-Side)

Similarity: If all three pairs of corresponding sides of two triangles are proportional, then the triangles are similar. That is, if a/d = b/e = c/f, where a, b, c are the sides of one triangle and d, e, f are the sides of the other triangle, then the triangles are similar.

SAS (Side-Angle-Side)

Similarity: If two pairs of corresponding sides of two triangles are proportional, and the included angles are congruent, then the triangles are similar.

Theorems related to Similarity: Proportionality Theorem (also known as the Basic Proportionality Theorem or Thales' Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally. In other words, if DE || BC in triangle ABC, then AD/DB = AE/E

C. Converse of the Proportionality Theorem: If a line divides two sides of a triangle proportionally, then the line is parallel to the third side. If AD/DB = AE/EC in triangle ABC, then DE || B

C. Midpoint Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Example 1: Proving Similarity and Finding Lengths In triangle ABC, D and E are points on AB and AC respectively such that DE || B

C. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find E

C. Solution: Since DE || BC, by the Proportionality Theorem, AD/DB = AE/E

C. Substitute the given values: 4/6 = 5/EC Cross-multiply: 4 EC = 6 5 Simplify: 4 * EC = 30 Divide by 4: EC = 30/4 = 7.5 cm Example 2: Applying the AAA Similarity Criterion Given triangle ABC and triangle DEF, if angle A = 50 degrees, angle B = 70 degrees, angle D = 50 degrees, and angle E = 70 degrees, prove that triangle ABC is similar to triangle DE

F. Solution: Angle A = Angle D = 50 degrees (given) Angle B = Angle E = 70 degrees (given) Therefore, by the AAA similarity criterion, triangle ABC ~ triangle DEF. 2.2 Pythagoras' Theorem Pythagoras' Theorem 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 c is the hypotenuse and a and b are the other two sides.

Converse of Pythagoras' Theorem: If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right-angled triangle.

Example 3: Finding the Hypotenuse In a right-angled triangle, the two shorter sides are 3 cm and 4 cm. Find the length of the hypotenuse.

Solution: Let a = 3 cm, b = 4 cm, and c be the hypotenuse.

By Pythagoras' Theorem: a² + b² = c² Substitute the values: 3² + 4² = c² Simplify: 9 + 16 = c² Therefore, c² = 25 Take the square root of both sides: c = √25 = 5 cm Example 4: Checking for a Right Angle A triangle has sides of length 5 cm, 12 cm, and 13 cm. Is it a right-angled triangle?

Solution: The longest side is 13 cm, so if it's a right-angled triangle, 13 cm must be the hypotenuse. Check if 5² + 12² = 13² Calculate: 25 + 144 = 169 Since 169 = 169, the triangle is a right-angled triangle by the converse of Pythagoras' Theorem. 2.3 Geometric Mean The geometric mean of two positive numbers 'a' and 'b' is the positive number 'x' such that a/x = x/b.

Therefore, x² = ab, and x = √(ab). In the context of similar triangles formed by the altitude to the hypotenuse of a right triangle, the altitude is the geometric mean between the two segments it creates on the hypotenuse.

Example 5: Altitude to Hypotenuse In right triangle ABC, angle B is the right angle. BD is the altitude to the hypotenuse A

C. If AD = 4 and DC = 9, find the length of B

D. Solution: BD is the geometric mean between AD and D

C. Therefore, BD = √(AD * DC)

Substitute the values: BD = √(4 * 9)

Simplify: BD = √36 = 6 Guided Practice (With Solutions)

Question 1: In triangle PQR, ST is parallel to QR, where S is on PQ and T is on P

R. If PS = 3 cm, SQ = 6 cm, and PT = 2 cm, find T

R. Solution: Since ST || QR, by the Proportionality Theorem, PS/SQ = PT/TR Substitute the values: 3/6 = 2/TR Cross-multiply: 3 TR = 6 2 Simplify: 3 * TR = 12 Divide by 3: TR = 12/3 = 4 cm

Commentary: This problem directly applies the Proportionality Theorem.