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For Class 8 Mathematics, this chapter in ICSE Class 8 Maths Chapter 23 Quadrilaterals and Polygons provides a detailed overview of important concepts. We highly recommend using this text alongside the ICSE Solutions for Class 8 Mathematics to learn the exercise questions provided at the end of the chapter.
Chapter 23 Quadrilaterals and Polygons ICSE Book Class Class 8 PDF (2026-27)
Chapter 23 - Quadrilaterals And Polygons
Quadrilaterals
A closed plane figure bounded by four line segments is called a quadrilateral. In the adjoining diagram, ABCD is a quadrilateral. It has:
four sides - AB, BC, CD and DA
four (interior) angles - ∠A, ∠B, ∠C and ∠D
four vertices - A, B, C and D
two diagonals - AC and BD.
Sum of (interior) angles of a quadrilateral is 360°
In the adjoining figure, ABCD is any quadrilateral. Diagonal AC divides it into two triangles. We know that the sum of angles of a triangle is 180°,
in △ABC, ∠1 + ∠B + ∠2 = 180° ...(i)
in △ACD, ∠4 + ∠D + ∠3 = 180° ...(ii)
On adding (i) and (ii), we get
∠1 + ∠4 + ∠B + ∠D + ∠2 + ∠3 = 360°
⇒ ∠A + ∠B + ∠D + ∠C = 360° (from figure)
Hence, the sum of (interior) angles of a quadrilateral is 360°.
Example 1
From the adjoining diagram, calculate the value of x.
Solution. As the sum of (interior) angles of a quadrilateral is 360°,
90° + 110° + 83° + ∠ABC = 360°
⇒ ∠ABC = 360° - 90° - 110° - 83° = 77°.
As ABE is a straight line,
x + ∠ABC = 180°
⇒ x = 180° - 77°
⇒ x = 103°.
Example 2
If the angles of a quadrilateral are in the ratio 5 : 8 : 11 : 12, find the angles.
Solution. Since the angles of the quadrilateral are in the ratio 5 : 8 : 11 : 12, let these angles be 5x, 8x, 11x and 12x.
As the sum of angles of a quadrilateral is 360°,
5x + 8x + 11x + 12x = 360°
⇒ 36x = 360°
⇒ x = 10°
∴ The angles of the quadrilateral are 5 × 10°, 8 × 10°, 11 × 10° and 12 × 10° i.e. 50°, 80°, 110° and 120°.
Teacher's Note
Understanding quadrilateral angles helps in architecture and construction, where builders must ensure that corners and walls meet at correct angles to create stable structures.
Exercise 23.1
1. If three angles of a quadrilateral are 70°, 83° and 112°, find the fourth angle.
2. From the adjoining diagram, find the value of x.
3. If two angles of a quadrilateral are 76° and 138° and the other two angles are equal, find the measure of equal angles.
4. A quadrilateral has three interior angles each equal to 95°. Find the size of the fourth interior angle.
5. If one of the angles of a quadrilateral is 210° and the remaining three angles are equal, find the measure of the equal angles. Note. It is a re-entrant quadrilateral.
6. If the angles of a quadrilateral are x°, (x - 20)°, (x - 30)° and (x + 10)°, find (i) x (ii) the angles of the quadrilateral.
7. If the angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6, find the angles.
8. Three angles of a quadrilateral are in the ratio 3 : 5 : 6. If the fourth angle is 80°, find the other angles of the quadrilateral.
9. Two angles of a quadrilateral are 78° and 87°. If the other two angles are in the ratio 5 : 8, find the size of each of them.
10. In a quadrilateral ABCD, AB ∥ DC. If ∠A : ∠D = 2 : 3 and ∠B : ∠C = 7 : 8, find the measure of each angle.
11. From the adjoining figure, find (i) x (ii) ∠DAB (iii) ∠ADB
Parallelogram
A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. In the adjoining quadrilateral, AB ∥ DC and AD ∥ BC, so ABCD is a parallelogram.
Teacher's Note
Parallelograms appear everywhere in design and nature - from the shape of building facades to the structure of certain crystals and the way some plants arrange their leaves.
Theorem 1
(i) The opposite sides of a parallelogram are equal.
(ii) The opposite angles of a parallelogram are equal.
(iii) Each diagonal bisects the parallelogram.
Given. A parallelogram ABCD.
To prove. (i) AB = DC and AD = BC
(ii) ∠B = ∠D and ∠A = ∠C
(iii) Area of △ ABC = area of △ ACD and area of △ ABD = area of △ BCD
Construction. Join AC and BD.
Proof.
| Statements | Reasons |
|---|---|
| In △ABC and △CDA | |
| 1. ∠1 = ∠4 | 1. Alt. ∠s, as AB ∥ DC and AC cuts them |
| 2. ∠3 = ∠2 | 2. Alt. ∠s, as AD ∥ BC and AC cuts them |
| 3. AC = AC | 3. Common |
| 4. △ABC ≅ △ CDA | 4. A.S.A. axiom of congruency 'c.p.c.t.' |
| (i) AB = DC and AD = BC | 'c.p.c.t.' |
| (ii) ∠B = ∠D and ∠A = ∠C | Adding 1 and 2, ∠1 + ∠2 = ∠3 + ∠4 △ABC ≅ △ CDA and congruent triangles have equal area. |
| (iii) area of △ABC = area of △ CDA = \(\frac{1}{2}\)(area of parallelogram ABCD) | |
| ⇒ AC bisects parallelogram ABCD Similarly, △ABD ≅ △ CDB ⇒ area of △ABD = area of △ CDB ⇒ BD bisects parallelogram ABCD Q.E.D. |
Theorem 2
The diagonals of a parallelogram bisect each other.
Given. A parallelogram ABCD whose diagonals AC and BD intersect at O.
To prove. OA = OC and OB = OD.
Proof.
| Statements | Reasons |
|---|---|
| In △OAB and △OCD | |
| 1. ∠1 = ∠2 | 1. Alt. ∠s, as AB ∥ DC and AC cuts them |
| 2. ∠3 = ∠4 | 2. Alt. ∠s, as AB ∥ DC and BD cuts them |
| 3. AB = DC | 3. Opp. sides of a ∥ gm are equal, Theorem 1 |
| 4. △ OAB ≅ △ OCD | 4. A.S.A. axiom of congruency 'c.p.c.t.' |
| ∴ OA = OC and OB = OD Q.E.D. |
Teacher's Note
The property that diagonals of a parallelogram bisect each other is used in engineering and robotics to create balanced mechanisms and stable structures.
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ICSE Book Class 8 Mathematics Chapter 23 Quadrilaterals and Polygons
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