☰ خطة العام · الرياضيات (بالإنجليزية)

Term 1 review

الأسبوع 15 · مراجعة الفصل الدراسي الأول · 3 دروس
3

Review: geometry and statistics

Mathematics Prep 2 Term 1, Units 1–4, Review
GoalReview Units 3 and 4: congruence, Pythagoras, midsegments and medians, isosceles triangles and polygon angles, and the mean, median and mode of grouped data.
New words
congruence cases · SSS, SAS, ASA, AAS and RHS; AAA and SSA are not enoughmidsegment · parallel to the third side and half its lengthpolygon angles · interior sum (n − 2) × 180°; exterior sum 360°grouped data · mean from midpoints, median from the cumulative table, mode from the histogram
Explanation

Unit 3 facts. Triangles are congruent by SSS, SAS (the angle between the sides), ASA, AAS or RHS. Pythagoras: c² = a² + b²; the converse tests a triangle, and c² < a² + b² means acute, c² > a² + b² means obtuse. The midsegment joining two midpoints is parallel to the third side and half as long; the three medians meet at the centroid, which cuts each median 2 : 1 from the vertex; the median to the hypotenuse of a right-angled triangle is half the hypotenuse. In an isosceles triangle the base angles are equal; an equilateral triangle has three 60° angles; the 30°–60°–90° triangle has the short leg equal to half of the hypotenuse. A polygon has interior angle sum (n − 2) × 180°, exterior angle sum 360°, and a regular polygon has exterior angle 360° ÷ n. Unit 4 facts. Cumulative frequency is a running total plotted at the upper limits. Mean = total of f × m ÷ total of f; the median has rank total ÷ 2 and is found by proportion or from the curve; the mode comes from the crossing lines on the histogram.

TopicRule
Right-angled trianglec² = a² + b²
MidsegmentDE = ½ BC
CentroidAG = 2 GD
Polygon interior sum(n − 2) × 180°
Mean (grouped)Σ(f × m) ÷ Σf
Worked examples
3x + 2x + 3ABCDE
In Δ ABC, D and E are the midpoints of AB and AC, BC = 3x + 2 and DE = x + 3. Since BC = 2 DE, 3x + 2 = 2x + 6, so x = 4, BC = 14 and DE = 7. A right-angled triangle with legs 9 and 12 has hypotenuse 15 and the median from the right angle is 7.5.
Each interior angle of a regular polygon is 135°, so each exterior angle is 45° and n = 360 ÷ 45 = 8 sides. It has 8 × 5 ÷ 2 = 20 diagonals and an interior angle sum of 6 × 180° = 1080°. In an isosceles triangle with vertex angle 40° the base angles are (180° − 40°) ÷ 2 = 70°.
Marks of 20 pupils in the classes 0–, 10–, 20–, 30–40 have frequencies 4, 6, 8, 2. Mean: midpoints 5, 15, 25, 35 give 20 + 90 + 200 + 70 = 380, and 380 ÷ 20 = 19. Cumulative values: 0, 4, 10, 18, 20. Median: rank 10 is reached at 20, so the median is 20. Mode: modal class 20–, d₁ = 8 − 6 = 2 and d₂ = 8 − 2 = 6, so 20 + 10 × 2 ÷ 8 = 22.5. Less than 30: 18 ÷ 20 × 100 = 90%.
Class activity

Triangle detective. Draw a triangle on card with sides 6 cm, 8 cm and 10 cm, check that it is right-angled with the converse of Pythagoras, mark its three midpoints and cut out the midsegment triangle. Measure the sides of the small triangle and compare them with the halves of the sides of the big one.

Practice
1
A triangle has sides 7, 24 and 25. Is it right-angled? What is the median to the longest side?
2
The medians of a triangle meet at G and AD = 18 cm. Find AG and GD.
3
Find the interior angle sum of a decagon and each exterior angle of a regular decagon.
4
Cumulative values are 0, 6, 18, 32, 40 at 0, 10, 20, 30, 40. How many values are at least 20, and what percent is that?
5
Two triangles have sides 5, 7 and an angle of 40° that is NOT between the sides. Are they congruent?