Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Apply the 180° sum of angles in a triangle.
• Use the exterior angle theorem for triangles.
• Identify properties of parallelograms, rectangles, squares, rhombuses, and kites.
• Solve problems using these geometric theorems.

Lesson summary

This week, we delve deeper into the fascinating world of geometry, focusing on the properties and theorems related to triangles and quadrilaterals. Understanding these theorems is crucial not only for succeeding in Mathematics but also for developing spatial reasoning skills applicable in various real-life situations. For example, architects use these principles to design buildings, engineers rely on them to construct bridges, and even artists utilize them to create balanced and visually appealing compositions.

Lesson notes

2.1 Triangles: Angle Sum and Exterior Angle Theorem 2.1.1 Angle Sum Theorem: The sum of the interior angles of any triangle is always equal to 180°. Why is this true? Imagine tearing off the three corners of a triangle and placing them together. They will always form a straight line, which represents 180°. Let's denote the three angles of a triangle as A, B, and

C. Then: A + B + C = 180° Example 1: In triangle ABC, angle A = 60° and angle B = 80°. Find angle

C. Solution: A + B + C = 180° 60° + 80° + C = 180° 140° + C = 180° C = 180° - 140° C = 40° 2.1.2 Exterior Angle Theorem: An exterior angle of a triangle is equal to the sum of the two opposite interior angles. An exterior angle is formed when one side of a triangle is extended. It's supplementary to the adjacent interior angle. Let's say we extend side BC of triangle ABC to point

D. Then, angle ACD is an exterior angle.

The Exterior Angle Theorem states that: Angle ACD = Angle A + Angle B Why is this true? Let's use the Angle Sum Theorem. We know A + B + C = 180°. We also know that Angle ACD + Angle C = 180° (because they form a straight line).

Therefore, Angle ACD + Angle C = A + B +

C. Subtracting Angle C from both sides gives us Angle ACD = A +

B. Example 2: In triangle PQR, angle P = 70° and angle Q = 50°. Side QR is extended to point

S. Find the measure of angle PR

S. Solution: Angle PRS = Angle P + Angle Q (Exterior Angle Theorem) Angle PRS = 70° + 50° Angle PRS = 120° 2.2 Quadrilaterals: Properties A quadrilateral is a closed shape with four sides. Specific quadrilaterals have unique properties. 2.2.1 Parallelogram: A quadrilateral with both pairs of opposite sides parallel. Opposite sides are equal in length. Opposite angles are equal. Diagonals bisect each other (they cut each other in half). 2.2.2 Rectangle: A parallelogram with all angles equal to 90°. All properties of a parallelogram apply. Diagonals are equal in length. 2.2.3 Square: A rectangle with all sides equal in length. All properties of a parallelogram and rectangle apply. Diagonals are equal and bisect each other at 90°. 2.2.4 Rhombus: A parallelogram with all sides equal in length. All properties of a parallelogram apply. Diagonals bisect each other at 90°. Diagonals bisect the angles of the rhombus. 2.2.5 Kite: A quadrilateral with two pairs of adjacent sides equal in length. One diagonal bisects the other diagonal. One diagonal bisects a pair of opposite angles. The diagonals are perpendicular.

Example 3: ABCD is a parallelogram. Angle A = 110°. Find angle C and angle

B. Solution: Since ABCD is a parallelogram, opposite angles are equal.

Therefore, angle C = angle A = 110°. Also, adjacent angles in a parallelogram are supplementary (add up to 180°).

Therefore, angle B = 180° - angle A = 180° - 110° = 70°.

Example 4: PQRS is a rhombus. If angle PQR = 60°, find angle PSR and angle QP

R. Solution: Since PQRS is a rhombus, opposite angles are equal.

Therefore, angle PSR = angle PQR = 60°. The diagonals of a rhombus bisect the angles.

Therefore, angle QPR = 1/2 angle PQR = 1/2 60° = 30°. Guided Practice (With Solutions)

Question 1: Triangle DEF has angle D = x, angle E = 2x, and angle F = 3x. Find the value of x and the measure of each angle.

Solution: We know that the sum of the angles in a triangle is 180°. x + 2x + 3x = 180° 6x = 180° x = 30° Therefore: Angle D = x = 30° Angle E = 2x = 2 * 30° = 60° Angle F = 3x = 3 * 30° = 90° Question 2: ABCD is a rectangle. If diagonal AC = 10 cm, find the length of diagonal B

D. Solution: In a rectangle, the diagonals are equal in length.

Therefore, BD = AC = 10 cm.

Question 3: PQRS is a parallelogram. Angle P = 80°. Find angle Q, angle R, and angle

S. Solution: In a parallelogram, opposite angles are equal and adjacent angles are supplementary. Angle R = Angle P = 80° Angle Q = 180° - Angle P = 180° - 80° = 100° Angle S = Angle Q = 100° Question 4: KITE is a kite with KE = K

I. Angle EKT = 40 degrees. Find angle IK

T. Solution: In a kite, one diagonal bisects the angle. Because KE = KI, then KT bisects angle EKI.

Therefore, angle IKT = angle EKT = 40 degrees. Independent Practice (Questions Only) In triangle XYZ, angle X = 45° and angle Y = 75°. Find angle Z. In triangle ABC, angle A = x + 10, angle B = 2x - 5, and angle C = 3x +

1

5. Find the value of x and the measure of each angle. Side AB of triangle ABC is extended to D. If angle BAC = 70° and angle ABC = 60°, find angle CBD. ABCD is a parallelogram. If angle A = 65°, find angle B, angle C, and angle D. PQRS is a rhombus. If diagonal PR bisects angle PQR, and angle PQR = 50°, find angle QPR and angle QRS. ABCD is a rectangle. If AC = 12 cm, find the length of BD. ABCD is a square. If AB = 5 cm, find the length of AC. KITE is a kite where KI = KE. Angle IKE = 80 degrees. Calculate angle ITE. ABCD is a parallelogram. If angle DAC = 40 degrees and angle BAC = 30 degrees, find the measures of all the interior angles of the parallelogram. Triangle PQR is isosceles with PQ = PR.