Lesson Notes By Weeks and Term v5 - Grade 10
Euclidean geometry – Week 8 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Euclidean geometry – Week 8 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 10
Term: 2nd Term
Week: 8
Theme: General lesson support
This page supports the lesson note with a companion video and a short classroom-ready summary.
For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.
Euclidean geometry, named after the ancient Greek mathematician Euclid, is the study of shapes, sizes, relative positions of figures, and the properties of space. In Grade 10, we build upon the geometric foundations laid in earlier grades, delving deeper into the relationships between angles and lines, particularly focusing on the properties of triangles. Understanding these principles isn't just about passing exams. It's about developing crucial problem-solving skills applicable in many areas of life. Think about architects designing buildings, engineers planning bridges, or even farmers dividing their land fairly.
2. 1. Sum of Interior Angles of a Triangle Theorem: The sum of the interior angles of any triangle is always 180°.
Explanation: Consider any triangle ABC. The angles at vertices A, B, and C are referred to as angle A, angle B, and angle C, respectively (or sometimes ∠A, ∠B, ∠C). The theorem states that ∠A + ∠B + ∠C = 180°. Why? Imagine tearing off the three angles of any triangle and placing them next to each other at a point. You'll find that they perfectly form a straight line, and a straight line contains 180 degrees.
Example 1: In triangle PQR, ∠P = 60° and ∠Q = 80°. Find ∠
R. Solution: ∠P + ∠Q + ∠R = 180° (Sum of interior angles of a triangle) 60° + 80° + ∠R = 180° 140° + ∠R = 180° ∠R = 180° - 140° ∠R = 40° Example 2: In triangle XYZ, ∠X = 2x, ∠Y = 3x, and ∠Z = x. Find the value of x and the measure of each angle.
Solution: ∠X + ∠Y + ∠Z = 180° (Sum of interior angles of a triangle) 2x + 3x + x = 180° 6x = 180° x = 180° / 6 x = 30° Therefore: ∠X = 2 * 30° = 60° ∠Y = 3 * 30° = 90° ∠Z = 30° 2.
2. Exterior Angle of a Triangle Theorem: The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Explanation: An exterior angle is formed when one side of a triangle is extended beyond the vertex. For example, extend side BC of triangle ABC to a point D. Then angle ACD is an exterior angle. The two opposite interior angles are angles A and B. The theorem states that ∠ACD = ∠A + ∠B. Why? The exterior angle and its adjacent interior angle are supplementary (add up to 180°), i.e., ∠ACD + ∠ACB = 180°. We also know that ∠A + ∠B + ∠ACB = 180°. Thus, ∠ACD = 180° - ∠ACB = ∠A + ∠
B. Example 3: In triangle DEF, side EF is extended to
G. If ∠D = 70° and ∠E = 50°, find ∠DF
G. Solution: ∠DFG = ∠D + ∠E (Exterior angle of a triangle) ∠DFG = 70° + 50° ∠DFG = 120° Example 4: In triangle RST, side ST is extended to
U. If ∠RSU = 130° and ∠R = 40°, find ∠
T. Solution: ∠RSU = ∠R + ∠T (Exterior angle of a triangle) 130° = 40° + ∠T ∠T = 130° - 40° ∠T = 90° 2.
3. Isosceles Triangles Theorem: Angles opposite equal sides in an isosceles triangle are equal. Conversely, if two angles in a triangle are equal, then the sides opposite those angles are equal (making it an isosceles triangle).
Explanation: An isosceles triangle has two sides of equal length. The angles opposite these equal sides are also equal. If a triangle has two equal angles, then the sides opposite these angles have equal length.
Example 5: In triangle ABC, AB = A
C. If ∠B = 65°, find ∠C and ∠
A. Solution: Since AB = AC, ∠C = ∠B (Angles opposite equal sides) ∠C = 65° ∠A + ∠B + ∠C = 180° (Sum of interior angles of a triangle) ∠A + 65° + 65° = 180° ∠A + 130° = 180° ∠A = 180° - 130° ∠A = 50° Example 6: In triangle PQR, ∠P = 70° and ∠Q = 70°. Find the length of side PR if side PQ is 10cm.
Solution: Since ∠P = ∠Q, PR = QR (Sides opposite equal angles) Therefore, triangle PQR is isosceles. We are given that PQ = 10cm.
However, we don't have direct information relating PQ to PR. The sides opposite the equal angles P and Q are QR and PR, respectively. Since ∠P=∠Q, we have PR = QR. Without further information about the triangle's shape, we cannot determine the length of P
R. There is insufficient information given in the prompt to give a valid answer.
Note: A common mistake is to assume PR = PQ. This would only be true if all three angles were equal (making it an equilateral triangle). Guided Practice (With Solutions)
Question 1: In triangle KLM, ∠K = 35° and ∠L = 75°. Calculate the size of ∠
M. Solution: ∠K + ∠L + ∠M = 180° (Sum of interior angles of a triangle) 35° + 75° + ∠M = 180° 110° + ∠M = 180° ∠M = 180° - 110° ∠M = 70°
Commentary: This is a straightforward application of the theorem about the sum of interior angles. We substitute the known values and solve for the unknown.
Question 2: In triangle ABC, side BC is extended to point
D. If ∠ABC = 48° and ∠BAC = 62°, find the measure of ∠AC
D. Solution: ∠ACD = ∠ABC + ∠BAC (Exterior angle of a triangle) ∠ACD = 48° + 62° ∠ACD = 110°
Commentary: This problem uses the exterior angle theorem. Ensure you understand which angles are the opposite interior angles.
Question 3: Triangle PQR is isosceles with PQ = P
R. If ∠P = 40°, find ∠Q and ∠
R. Solution: Since PQ = PR, ∠Q = ∠R (Angles opposite equal sides) ∠P + ∠Q + ∠R = 180° (Sum of interior angles of a triangle) 40° + ∠Q + ∠Q = 180° (Since ∠Q = ∠R) 40° + 2∠Q = 180° 2∠Q = 140° ∠Q = 70° Therefore, ∠R = 70°
Commentary: This problem combines the properties of isosceles triangles with the sum of interior angles theorem. Remember that equal sides imply equal opposite angles.
Question 4: In triangle XYZ, ∠X = x + 10, ∠Y = 2x - 20, and ∠Z = 3x. Find the value of x and the measure of each angle.