Lesson Notes By Weeks and Term v5 - Grade 9
Geometry: theorems about triangles and quadrilaterals – Week 2 focus
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Geometry: theorems about triangles and quadrilaterals – Week 2 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 9
Term: 3rd Term
Week: 2
Theme: General lesson support
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Geometry is all around us! From the shape of your house to the soccer ball you kick, understanding shapes and their properties is essential. This week, we delve deeper into the fascinating world of triangles and quadrilaterals, building on what you learned last week. A strong grasp of these concepts is vital for further studies in mathematics, physics, engineering, architecture, and even art! Imagine designing a safe and efficient building – geometry is your foundation. Moreover, knowing how shapes interact helps with everyday problem-solving, like calculating areas for tiling your room or understanding angles for a construction project.
2.1 The Sum of Interior Angles of a Triangle Theorem: The sum of the interior angles of any triangle is always 180°. Why? Think of tearing off the three corners of a triangle and placing them next to each other. They will always form a straight line, which is 180°.
Example 1: In triangle ABC, angle A = 60° and angle B = 80°. Find angle
C. Solution: Angle A + Angle B + Angle C = 180° (Sum of interior angles of a triangle) 60° + 80° + Angle C = 180° 140° + Angle C = 180° Angle C = 180° - 140° Angle C = 40° Example 2: In a right-angled triangle, one of the acute angles is 30°. Find the other acute angle.
Solution: One angle is 90° (right angle). Let the other acute angle be x. 90° + 30° + x = 180° 120° + x = 180° x = 180° - 120° x = 60° 2.2 Exterior Angle of a Triangle Theorem: The exterior angle of a triangle is equal to the sum of the two interior opposite angles. Why? An exterior angle and its adjacent interior angle are supplementary (add up to 180°). The sum of all three interior angles is also 180°.
Therefore, the exterior angle must equal the sum of the other two interior angles.
Example 1: In triangle PQR, side QR is extended to
S. Angle P = 70° and angle Q = 50°. Find angle PRS (the exterior angle).
Solution: Angle PRS = Angle P + Angle Q (Exterior angle of a triangle) Angle PRS = 70° + 50° Angle PRS = 120° Example 2: In triangle XYZ, side YZ is extended to
W. Angle XYZ = 65° and angle XZW = 110°. Find angle YX
Z. Solution: Angle XZW = Angle XYZ + Angle YXZ (Exterior angle of a triangle) 110° = 65° + Angle YXZ Angle YXZ = 110° - 65° Angle YXZ = 45° 2.3 Properties of Parallelograms A parallelogram is a quadrilateral with two pairs of parallel sides.
Properties: Opposite sides are parallel and equal in length. Opposite angles are equal. Diagonals bisect each other (they cut each other in half).
Example 1: ABCD is a parallelogram. AB = 8cm, BC = 5cm, and angle ABC = 70°. Find the length of CD, the length of AD, and the measure of angle AD
C. Solution: CD = AB = 8cm (Opposite sides of a parallelogram are equal) AD = BC = 5cm (Opposite sides of a parallelogram are equal) Angle ADC = Angle ABC = 70° (Opposite angles of a parallelogram are equal)
Example 2: PQRS is a parallelogram. The diagonals PR and QS intersect at point
O. PO = 6cm. Find the length of P
R. Solution: PR = 2 PO (Diagonals of a parallelogram bisect each other) PR = 2 6cm PR = 12cm 2.4 Properties of Special Quadrilaterals Rhombus: A parallelogram with all four sides equal.
Properties: All sides equal, opposite angles equal, diagonals bisect each other at right angles, diagonals bisect the angles.
Rectangle: A parallelogram with all four angles equal to 90°.
Properties: All angles are right angles, opposite sides equal, diagonals are equal and bisect each other.
Square: A parallelogram with all four sides equal and all four angles equal to 90°.
Properties: All sides equal, all angles are right angles, diagonals are equal and bisect each other at right angles, diagonals bisect the angles.
Kite: A quadrilateral with two pairs of adjacent sides equal.
Properties: One diagonal bisects the other diagonal, one pair of opposite angles are equal, diagonals intersect at right angles.
Trapezium (Trapezoid): A quadrilateral with one pair of parallel sides.
Properties: Only one pair of sides is parallel.
Example 1: Rhombus ABCD is a rhombus. Angle ABC = 120°. Find angle BA
C. Solution: The diagonals of a rhombus bisect the angles. Thus, angle ABD = 120/2 = 60 degrees. Since AB = BC (rhombus), triangle ABC is isosceles, so angle BAC = angle BCA. Angle BAC + Angle BCA + Angle ABC = 180 degrees. 2 angle BAC + 120 = 180 2 angle BAC = 60 Angle BAC = 30 degrees.
Example 2: Rectangle PQRS is a rectangle. PR = 10cm. Find Q
S. Solution: QS = PR (Diagonals of a rectangle are equal) QS = 10cm Guided Practice (With Solutions)
Question 1: In triangle KLM, angle K = 35° and angle L = 75°. Calculate the size of angle
M. Solution: Angle K + Angle L + Angle M = 180° (Sum of interior angles of a triangle) 35° + 75° + Angle M = 180° 110° + Angle M = 180° Angle M = 180° - 110° Angle M = 70°
Commentary: This is a direct application of the theorem. It tests the understanding of the sum of interior angles.
Question 2: In triangle ABC, side BC is extended to
D. Angle ABC = 60° and angle ACD = 130°. Calculate the size of angle BA
C. Solution: Angle ACD = Angle ABC + Angle BAC (Exterior angle of a triangle) 130° = 60° + Angle BAC Angle BAC = 130° - 60° Angle BAC = 70°
Commentary: This problem applies the exterior angle theorem. Students need to identify the exterior angle correctly.
Question 3: PQRS is a parallelogram. Angle PQR = 110°. Calculate the size of angle PS
R. Solution: Angle PSR = Angle PQR (Opposite angles of a parallelogram are equal) Angle PSR = 110°
Commentary: This problem tests the understanding of the properties of a parallelogram, specifically the equality of opposite angles.
Question 4: ABCD is a rhombus. If angle BAC = 32°, find angle BCA.