Rectilinear Figures and Quadrilaterals
Introduction
Rectilinear figures are plane figures bounded by straight lines. Quadrilaterals are four-sided rectilinear figures. Understanding their properties is fundamental for ICSE Class 9 geometry.
Angle Sum of Polygons
Triangle (3 sides)
Sum of interior angles = (3 - 2) × 180° = 180°
Quadrilateral (4 sides)
Sum of interior angles = (4 - 2) × 180° = 360°
n-sided Polygon
Sum of interior angles = (n - 2) × 180°
Each Interior Angle of a Regular n-sided Polygon
Each interior angle = (n - 2) × 180° / n
<ICSEExample title="Find Angle Sum"> Find the sum of interior angles of a pentagon (5 sides). <Solution> Sum = (5 - 2) × 180° = 3 × 180° = 540° </Solution> </ICSEExample>Types of Quadrilaterals
| Quadrilateral | Key Properties | Diagonals |
|---|---|---|
| Parallelogram | Opposite sides parallel and equal | Bisect each other |
| Rectangle | All angles 90°, opposite sides equal | Equal and bisect each other |
| Rhombus | All sides equal, opposite sides parallel | Perpendicular bisectors |
| Square | All sides equal, all angles 90° | Equal, perpendicular bisectors |
| Trapezium | One pair of opposite sides parallel | Not necessarily equal |
Parallelogram
Properties:
- Opposite sides are equal: AB = CD, AD = BC
- Opposite angles are equal: A = C, B = D
- Diagonals bisect each other
- Adjacent angles are supplementary: A + B = 180°
Special Parallelograms
Rectangle
- All properties of parallelogram apply
- All angles are 90°
- Diagonals are equal
Rhombus
- All properties of parallelogram apply
- All sides are equal
- Diagonals bisect each other at 90°
- Diagonals bisect the interior angles
Square
- All properties of rectangle and rhombus apply
- All sides equal and all angles 90°
- Diagonals are equal and perpendicular bisectors
Trapezium
A trapezium has one pair of opposite sides parallel.
Isosceles Trapezium: Non-parallel sides are equal. Base angles are equal.
Proofs
Proof: Diagonals of a Parallelogram Bisect Each Other
Given: Parallelogram ABCD with diagonals AC and BD intersecting at O.
To prove: AO = OC and BO = OD
Proof:
- AB is parallel to CD (definition of parallelogram)
- AB = CD (opposite sides of parallelogram)
- In triangles AOB and COD:
- AB = CD (proved)
- Angle ABO = Angle CDO (alternate interior angles, AB parallel CD)
- Angle BAO = Angle DCO (alternate interior angles)
- Therefore, triangle AOB is congruent to triangle COD (ASA)
- By CPCT: AO = OC and BO = OD
Hence proved.
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Assuming all quadrilaterals are parallelograms | Only specific quadrilaterals have parallel opposite sides |
| Confusing rhombus and square | A square has 90° angles, a rhombus does not necessarily |
| Thinking diagonals of rectangle are perpendicular | Rectangle diagonals are equal but not perpendicular |
| Forgetting that a square is both a rectangle and a rhombus | A square inherits all properties of both |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Properties of quadrilaterals | 3-4 marks | Very common |
| Angle sum of polygons | 2-3 marks | Very common |
| Geometric proofs involving quadrilaterals | 4-5 marks | Common |
| Identification of quadrilateral types | 3 marks | Common |
Self-Test
Q1: The sum of interior angles of a polygon is 1080°. How many sides does it have?
Q2: In a parallelogram, one angle is 120°. Find all other angles.
Q3: Prove that the diagonals of a rectangle are equal.
Q4: A rhombus has a side of 10 cm and one diagonal of 12 cm. Find the other diagonal.
Q5: In parallelogram ABCD, diagonals intersect at O. If AO = 3x + 1 and OC = 5x - 7, find x.
