Solved Examples

NCERT Solutions for Class 8 Maths Chapter 3: Understanding Quadrilaterals — Complete Guide

Complete solutions for all exercises — polygons, angle sum property, parallelograms, and special quadrilaterals with 30+ solved problems.

CBSEClass 8
The SparkEd Authors (IITian & Googler)15 March 202655 min read
NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals — SparkEd

Why Understanding Quadrilaterals Is Essential for Geometry

Chapter 3 of NCERT Class 8 Maths is one of the most visually rich and property-heavy chapters in the entire textbook. It takes you from general polygons to the specific world of quadrilaterals — four-sided figures that include some of the most important shapes in geometry: parallelograms, rectangles, rhombuses, squares, trapeziums, and kites.

What makes this chapter special is the beautiful hierarchy of quadrilaterals. A square is a special rectangle, which is a special parallelogram, which is a special quadrilateral. Understanding this hierarchy means you can apply properties efficiently — anything true for a parallelogram is automatically true for a rectangle, a rhombus, and a square.

This chapter typically carries 6-8 marks in CBSE exams and forms the foundation for the Quadrilaterals chapter in Class 9 (where you prove many of these properties rigorously) and the Coordinate Geometry chapter in Class 10 (where you verify properties using algebra).

The key ideas you will master in this chapter include the angle sum property of polygons, the exterior angle sum property, and the defining properties of each type of quadrilateral. Let us walk through every concept and exercise with detailed solutions!

Polygons: Classification and Basics

Before studying quadrilaterals specifically, the chapter begins with a broader look at polygons — closed figures made up of straight line segments.

A polygon with nn sides is called an **nn-gon**. Common names:
- 33 sides: Triangle
- 44 sides: Quadrilateral
- 55 sides: Pentagon
- 66 sides: Hexagon
- 77 sides: Heptagon
- 88 sides: Octagon
- 99 sides: Nonagon
- 1010 sides: Decagon

Polygons are classified as convex (all interior angles less than 180°180°, all diagonals lie inside) or concave (at least one interior angle greater than 180°180°, at least one diagonal lies partly outside).

A regular polygon has all sides equal and all angles equal. A regular triangle is an equilateral triangle; a regular quadrilateral is a square; a regular hexagon has all sides equal and all angles =120°= 120°.

A diagonal of a polygon is a line segment connecting two non-adjacent vertices. The number of diagonals of an nn-sided polygon is n(n3)2\dfrac{n(n-3)}{2}.

Solved Example 1: Counting Diagonals

Problem: How many diagonals does a decagon (1010-sided polygon) have?

Solution:

Using the formula: Number of diagonals =n(n3)2= \dfrac{n(n-3)}{2}

=10(103)2=10×72=702=35= \dfrac{10(10-3)}{2} = \dfrac{10 \times 7}{2} = \dfrac{70}{2} = 35

Answer: A decagon has 3535 diagonals.

Angle Sum Property of Polygons

The angle sum property tells us the total of all interior angles of a polygon. This is one of the most important formulas in geometry.

Theorem: The sum of interior angles of a polygon with nn sides is:

(n2)×180°(n - 2) \times 180°

Why does this work? Any polygon with nn sides can be divided into (n2)(n - 2) triangles by drawing diagonals from one vertex. Since each triangle has an angle sum of 180°180°, the total is (n2)×180°(n - 2) \times 180°.

Examples:
- Triangle (n=3n = 3): (32)×180°=180°(3-2) \times 180° = 180°
- Quadrilateral (n=4n = 4): (42)×180°=360°(4-2) \times 180° = 360°
- Pentagon (n=5n = 5): (52)×180°=540°(5-2) \times 180° = 540°
- Hexagon (n=6n = 6): (62)×180°=720°(6-2) \times 180° = 720°
- Octagon (n=8n = 8): (82)×180°=1080°(8-2) \times 180° = 1080°

For a regular polygon, since all angles are equal, each interior angle =(n2)×180°n= \dfrac{(n-2) \times 180°}{n}.

Solved Example 2: Sum of Interior Angles

Problem: Find the sum of interior angles of a polygon with 1212 sides.

Solution:

Sum=(n2)×180°=(122)×180°=10×180°=1800°\text{Sum} = (n-2) \times 180° = (12-2) \times 180° = 10 \times 180° = 1800°

Answer: The sum of interior angles of a 1212-sided polygon is 1800°1800°.

Solved Example 3: Each Angle of a Regular Polygon

Problem: Find each interior angle of a regular hexagon.

Solution:

Sum of interior angles =(62)×180°=720°= (6-2) \times 180° = 720°.

Since all angles are equal in a regular polygon:

Each angle=720°6=120°\text{Each angle} = \dfrac{720°}{6} = 120°

Answer: Each interior angle of a regular hexagon is 120°120°.

Solved Example 4: Finding Number of Sides from Interior Angle

Problem: Each interior angle of a regular polygon is 140°140°. Find the number of sides.

Solution:

Each exterior angle =180°140°=40°= 180° - 140° = 40°.

Sum of exterior angles of any convex polygon =360°= 360°.

n=360°40°=9n = \dfrac{360°}{40°} = 9

Answer: The polygon has 99 sides (nonagon).

Solved Example 5: Finding Missing Angle

Problem: Four angles of a pentagon are 80°80°, 120°120°, 100°100°, and 130°130°. Find the fifth angle.

Solution:

Sum of interior angles of a pentagon =(52)×180°=540°= (5-2) \times 180° = 540°.

Let the fifth angle =x= x.

80+120+100+130+x=54080 + 120 + 100 + 130 + x = 540

430+x=540430 + x = 540

x=110°x = 110°

Answer: The fifth angle is 110°110°.

Practice this topic on SparkEd — free visual solutions and AI coaching

Try Free

Exterior Angles of a Polygon

An exterior angle of a polygon is formed by one side of the polygon and the extension of an adjacent side. At each vertex, the interior angle and the exterior angle are supplementary (they add up to 180°180°).

Key Theorem: The sum of exterior angles of any convex polygon is always 360°360°, regardless of the number of sides.

This is a remarkably powerful result. Whether you have a triangle, a quadrilateral, or a 100100-sided polygon, the sum of all exterior angles (one at each vertex) is always 360°360°.

For a regular polygon with nn sides, each exterior angle =360°n= \dfrac{360°}{n}.

This gives us a quick way to find the number of sides: n=360°each exterior anglen = \dfrac{360°}{\text{each exterior angle}}.

Solved Example 6: Exterior Angles

Problem: The exterior angles of a quadrilateral are x°, 2x°2x°, 3x°3x°, and 4x°4x°. Find all the angles.

Solution:

Sum of exterior angles =360°= 360°:

x+2x+3x+4x=360x + 2x + 3x + 4x = 360

10x=36010x = 360

x=36°x = 36°

Exterior angles: 36°,72°,108°,144°36°, 72°, 108°, 144°.

Corresponding interior angles: 180°36°=144°180° - 36° = 144°, 180°72°=108°180° - 72° = 108°, 180°108°=72°180° - 108° = 72°, 180°144°=36°180° - 144° = 36°.

Verification: Sum of interior angles =144+108+72+36=360°= 144 + 108 + 72 + 36 = 360°

Solved Example 7: Is a Regular Polygon Possible?

Problem: Can a regular polygon have each interior angle equal to 100°100°?

Solution:

Each exterior angle =180°100°=80°= 180° - 100° = 80°.

Number of sides =360°80°=4.5= \dfrac{360°}{80°} = 4.5.

Since the number of sides must be a whole number, no regular polygon can have each interior angle equal to 100°100°.

Exercise 3.1 — Complete Solutions (Polygons and Angle Sums)

Exercise 3.1 covers the angle sum property of polygons and exterior angle calculations. Here are complete solutions for all problem types.

Solved Example 8: Angles of a Quadrilateral in Ratio

Problem: The angles of a quadrilateral are in the ratio 3:5:7:93:5:7:9. Find all the angles.

Solution:

Let the angles be 3x,5x,7x,9x3x, 5x, 7x, 9x.

Sum of angles of a quadrilateral =360°= 360°:

3x+5x+7x+9x=360°3x + 5x + 7x + 9x = 360°

24x=360°24x = 360°

x=15°x = 15°

The angles are: 3×15°=45°3 \times 15° = 45°, 5×15°=75°5 \times 15° = 75°, 7×15°=105°7 \times 15° = 105°, 9×15°=135°9 \times 15° = 135°.

Verification: 45+75+105+135=360°45 + 75 + 105 + 135 = 360°

Solved Example 9: Sum of Interior Angles of Octagon

Problem: Find the sum of interior angles of a polygon with 88 sides (octagon).

Solution:

Sum=(n2)×180°=(82)×180°=6×180°=1080°\text{Sum} = (n-2) \times 180° = (8-2) \times 180° = 6 \times 180° = 1080°

Answer: The sum of interior angles of an octagon is 1080°1080°.

Solved Example 10: Finding Missing Angles in a Quadrilateral

Problem: Three angles of a quadrilateral are 90°90°, 85°85°, and 95°95°. Find the fourth angle.

Solution:

Let the fourth angle =x= x.

90+85+95+x=36090 + 85 + 95 + x = 360

270+x=360270 + x = 360

x=90°x = 90°

Answer: The fourth angle is 90°90°.

Solved Example 11: Exterior Angle of Regular Polygon

Problem: Find each exterior angle of a regular polygon with 1515 sides.

Solution:

Each exterior angle=360°n=360°15=24°\text{Each exterior angle} = \dfrac{360°}{n} = \dfrac{360°}{15} = 24°

Answer: Each exterior angle is 24°24°.

Properties of Parallelograms

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. This simple definition leads to several powerful properties:

Property 1: Opposite sides are equal.
In parallelogram ABCDABCD: AB=CDAB = CD and AD=BCAD = BC.

Property 2: Opposite angles are equal.
A=C\angle A = \angle C and B=D\angle B = \angle D.

Property 3: Adjacent angles are supplementary.
A+B=180°\angle A + \angle B = 180°, B+C=180°\angle B + \angle C = 180°, etc.

Property 4: Diagonals bisect each other.
If diagonals ACAC and BDBD intersect at OO, then AO=OCAO = OC and BO=ODBO = OD.

Converse Properties (used to prove a quadrilateral is a parallelogram):
- If both pairs of opposite sides are equal, it is a parallelogram.
- If both pairs of opposite angles are equal, it is a parallelogram.
- If diagonals bisect each other, it is a parallelogram.
- If one pair of opposite sides is both equal and parallel, it is a parallelogram.

Solved Example 12: Finding Angles of a Parallelogram

Problem: In a parallelogram ABCDABCD, A=65°\angle A = 65°. Find B\angle B, C\angle C, and D\angle D.

Solution:

Adjacent angles are supplementary:

B=180°A=180°65°=115°\angle B = 180° - \angle A = 180° - 65° = 115°

Opposite angles are equal:

C=A=65°\angle C = \angle A = 65°

D=B=115°\angle D = \angle B = 115°

Verification: 65+115+65+115=360°65 + 115 + 65 + 115 = 360°

Solved Example 13: Using Opposite Sides Property

Problem: In a parallelogram PQRSPQRS, PQ=3x+5PQ = 3x + 5 and RS=5x7RS = 5x - 7. Find PQPQ.

Solution:

Opposite sides of a parallelogram are equal:

PQ=RSPQ = RS

3x+5=5x73x + 5 = 5x - 7

5+7=5x3x5 + 7 = 5x - 3x

12=2x12 = 2x

x=6x = 6

PQ=3(6)+5=23PQ = 3(6) + 5 = 23

Answer: PQ=23PQ = 23 units.

Solved Example 14: Diagonals Bisect Each Other

Problem: The diagonals of a parallelogram ABCDABCD intersect at OO. If AO=3x+1AO = 3x + 1 and OC=5x3OC = 5x - 3, find ACAC.

Solution:

Diagonals of a parallelogram bisect each other, so AO=OCAO = OC:

3x+1=5x33x + 1 = 5x - 3

1+3=5x3x1 + 3 = 5x - 3x

4=2x4 = 2x

x=2x = 2

AO=3(2)+1=7AO = 3(2) + 1 = 7

AC=2×AO=2×7=14AC = 2 \times AO = 2 \times 7 = 14

Answer: AC=14AC = 14 units.

Solved Example 15: Angle Bisector in a Parallelogram

Problem: In a parallelogram ABCDABCD, the bisector of A\angle A meets BCBC at EE. If A=70°\angle A = 70°, find AEB\angle AEB.

Solution:

Since the bisector of A\angle A divides it into two equal parts:

BAE=70°2=35°\angle BAE = \dfrac{70°}{2} = 35°

Since ABDCAB \parallel DC and ADAD is a transversal:
A+B=180°\angle A + \angle B = 180° (co-interior angles), so B=110°\angle B = 110°.

In ABE\triangle ABE:

BAE+ABE+AEB=180°\angle BAE + \angle ABE + \angle AEB = 180°

35°+110°+AEB=180°35° + 110° + \angle AEB = 180°

AEB=180°145°=35°\angle AEB = 180° - 145° = 35°

Answer: AEB=35°\angle AEB = 35°.

Exercise 3.2 — Complete Solutions (Parallelogram Properties)

Exercise 3.2 focuses on applying the properties of parallelograms to find unknown angles, sides, and diagonal lengths.

Solved Example 16: Angles in Terms of x

Problem: In a parallelogram ABCDABCD, A=(2x+25)°\angle A = (2x + 25)° and B=(3x5)°\angle B = (3x - 5)°. Find all angles.

Solution:

Adjacent angles of a parallelogram are supplementary:

A+B=180°\angle A + \angle B = 180°

(2x+25)+(3x5)=180(2x + 25) + (3x - 5) = 180

5x+20=1805x + 20 = 180

5x=1605x = 160

x=32x = 32

A=2(32)+25=89°\angle A = 2(32) + 25 = 89°
B=3(32)5=91°\angle B = 3(32) - 5 = 91°
C=A=89°\angle C = \angle A = 89°
D=B=91°\angle D = \angle B = 91°

Verification: 89+91+89+91=360°89 + 91 + 89 + 91 = 360°

Solved Example 17: Perimeter of Parallelogram

Problem: Two adjacent sides of a parallelogram are 1212 cm and 88 cm. Find its perimeter.

Solution:

In a parallelogram, opposite sides are equal. So the four sides are 12,8,12,812, 8, 12, 8 cm.

Perimeter=2(12+8)=2×20=40 cm\text{Perimeter} = 2(12 + 8) = 2 \times 20 = 40 \text{ cm}

Answer: The perimeter is 4040 cm.

Solved Example 18: Using Diagonal Properties

Problem: The diagonals of a parallelogram ABCDABCD intersect at OO. If BO=5BO = 5 cm and AC=16AC = 16 cm, find ODOD and OAOA.

Solution:

Diagonals bisect each other:

OD=BO=5 cmOD = BO = 5 \text{ cm}

OA=AC2=162=8 cmOA = \dfrac{AC}{2} = \dfrac{16}{2} = 8 \text{ cm}

Answer: OD=5OD = 5 cm, OA=8OA = 8 cm.

Special Quadrilaterals: Rectangle, Rhombus, and Square

The most important part of this chapter is understanding how rectangle, rhombus, and square are special types of parallelograms with additional properties.

Rectangle: A parallelogram with all angles =90°= 90°.
- All parallelogram properties apply.
- Additional: Diagonals are equal (AC=BDAC = BD).

Rhombus: A parallelogram with all sides equal.
- All parallelogram properties apply.
- Additional: Diagonals bisect each other at right angles (ACBDAC \perp BD).
- Additional: Diagonals bisect the vertex angles.

Square: A parallelogram with all sides equal AND all angles =90°= 90°.
- It is simultaneously a rectangle AND a rhombus.
- Diagonals are equal, bisect each other at right angles, and bisect the vertex angles.

The Hierarchy:

SquareRectangleParallelogramQuadrilateral\text{Square} \subset \text{Rectangle} \subset \text{Parallelogram} \subset \text{Quadrilateral}

SquareRhombusParallelogramQuadrilateral\text{Square} \subset \text{Rhombus} \subset \text{Parallelogram} \subset \text{Quadrilateral}

This means every property of a parallelogram is automatically true for rectangles, rhombuses, and squares.

Solved Example 19: Diagonals of a Rhombus

Problem: The diagonals of a rhombus are 1616 cm and 1212 cm. Find the side of the rhombus.

Solution:

Diagonals of a rhombus bisect each other at right angles. Half-diagonals: 162=8\dfrac{16}{2} = 8 cm and 122=6\dfrac{12}{2} = 6 cm.

Using Pythagoras theorem in the right triangle formed:

Side=82+62=64+36=100=10 cm\text{Side} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}

Answer: Each side of the rhombus is 1010 cm.

Solved Example 20: Rectangle Diagonal

Problem: ABCDABCD is a rectangle with AB=12AB = 12 cm and BC=5BC = 5 cm. Find the length of diagonal ACAC.

Solution:

In a rectangle, each angle is 90°90°. In ABC\triangle ABC:

AC=AB2+BC2=144+25=169=13 cmAC = \sqrt{AB^2 + BC^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}

Since diagonals of a rectangle are equal: BD=13BD = 13 cm also.

Answer: AC=BD=13AC = BD = 13 cm.

Solved Example 21: Properties of a Square

Problem: ABCDABCD is a square with side 77 cm. Find: (a) the length of each diagonal, (b) the angle each diagonal makes with a side.

Solution:

(a) Using Pythagoras theorem:

Diagonal=72+72=49+49=98=72 cm\text{Diagonal} = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \text{ cm}

(b) Since all sides are equal, the diagonal divides the 90°90° corner angle into two 45°45° angles.

Answer: (a) Diagonal =72= 7\sqrt{2} cm. (b) The diagonal makes 45°45° with each side.

Solved Example 22: Area of Rhombus from Diagonals

Problem: The diagonals of a rhombus are 2424 cm and 1010 cm. Find the area and the perimeter.

Solution:

Area of a rhombus =12×d1×d2=12×24×10=120= \dfrac{1}{2} \times d_1 \times d_2 = \dfrac{1}{2} \times 24 \times 10 = 120 cm2^2.

Perimeter: First find the side. Half-diagonals: 1212 cm and 55 cm.

Side=122+52=144+25=169=13 cm\text{Side} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}

Perimeter=4×13=52 cm\text{Perimeter} = 4 \times 13 = 52 \text{ cm}

Answer: Area =120= 120 cm2^2, Perimeter =52= 52 cm.

Trapezium and Kite

These two quadrilaterals are NOT parallelograms, but they have their own special properties.

Trapezium (Trapezoid): A quadrilateral with exactly one pair of parallel sides.
- The parallel sides are called bases and the non-parallel sides are called legs.
- An isosceles trapezium has equal legs, equal base angles, and equal diagonals.

Kite: A quadrilateral with two pairs of consecutive equal sides.
- Properties: One pair of opposite angles are equal (the angles between the unequal sides).
- Diagonals are perpendicular to each other.
- One diagonal bisects the other (but not vice versa).
- One diagonal is the axis of symmetry.

Solved Example 23: Trapezium Angles

Problem: In a trapezium ABCDABCD, ABDCAB \parallel DC. If A=70°\angle A = 70° and B=80°\angle B = 80°, find C\angle C and D\angle D.

Solution:

Since ABDCAB \parallel DC:
- A+D=180°\angle A + \angle D = 180° (co-interior angles), so D=180°70°=110°\angle D = 180° - 70° = 110°.
- B+C=180°\angle B + \angle C = 180° (co-interior angles), so C=180°80°=100°\angle C = 180° - 80° = 100°.

Verification: 70+80+100+110=360°70 + 80 + 100 + 110 = 360°

Solved Example 24: Kite Properties

Problem: In a kite ABCDABCD, AB=AD=5AB = AD = 5 cm and BC=CD=8BC = CD = 8 cm. If B=110°\angle B = 110°, find D\angle D.

Solution:

In a kite, the angles between the pairs of unequal sides are equal:

B=D=110°\angle B = \angle D = 110°

Wait — this needs careful consideration. In a kite ABCDABCD where AB=ADAB = AD and BC=CDBC = CD, the equal angles are B\angle B and D\angle D (the angles between unequal sides).

Actually, in this configuration, BB and DD are the vertices where unequal sides meet, so B=D\angle B = \angle D.

B=D=110°\angle B = \angle D = 110°

Sum of all angles =360°= 360°:

A+B+C+D=360°\angle A + \angle B + \angle C + \angle D = 360°

A+110°+C+110°=360°\angle A + 110° + \angle C + 110° = 360°

A+C=140°\angle A + \angle C = 140°

The diagonal ACAC (connecting the vertices where equal sides meet) is the axis of symmetry, and it bisects angles AA and CC. Without additional information, we know A+C=140°\angle A + \angle C = 140°.

Answer: D=110°\angle D = 110°.

Exercise 3.3 — Complete Solutions (Special Quadrilaterals)

Exercise 3.3 focuses on applying the properties of special quadrilaterals — rectangles, rhombuses, squares, trapeziums, and kites.

Solved Example 25: Identifying the Quadrilateral

Problem: A quadrilateral has all four sides equal and all four angles equal. What type is it?

Solution:

All sides equal \Rightarrow it could be a rhombus or a square.
All angles equal \Rightarrow each angle =360°4=90°= \dfrac{360°}{4} = 90°, so it is a rectangle.

Since it satisfies BOTH conditions (all sides equal AND all angles 90°90°), it is a square.

A square is simultaneously a rectangle and a rhombus.

Solved Example 26: Equal Diagonals Identification

Problem: ABCDABCD is a parallelogram. If AC=BDAC = BD, what can you conclude about ABCDABCD?

Solution:

In a parallelogram, diagonals bisect each other. If they are also equal (AC=BDAC = BD), then all four segments from the center to the vertices are equal (AO=BO=CO=DOAO = BO = CO = DO).

A parallelogram with equal diagonals is a rectangle.

Answer: ABCDABCD is a rectangle.

Solved Example 27: Perpendicular Diagonals Identification

Problem: ABCDABCD is a parallelogram with ACBDAC \perp BD. What type of quadrilateral is it?

Solution:

A parallelogram whose diagonals are perpendicular to each other is a rhombus.

If additionally all angles were 90°90°, it would be a square. But with only the perpendicularity condition, we can only conclude it is a rhombus.

Answer: ABCDABCD is a rhombus.

Solved Example 28: Square Diagonal Properties

Problem: The diagonal of a square is 10210\sqrt{2} cm. Find the side and the perimeter.

Solution:

Let the side =a= a. Diagonal of a square =a2= a\sqrt{2}.

a2=102a\sqrt{2} = 10\sqrt{2}

a=10 cma = 10 \text{ cm}

Perimeter=4a=40 cm\text{Perimeter} = 4a = 40 \text{ cm}

Answer: Side =10= 10 cm, Perimeter =40= 40 cm.

Solved Example 29: Combined Properties

Problem: ABCDABCD is a rhombus where A=60°\angle A = 60°. The side of the rhombus is 66 cm. Find the lengths of the diagonals.

Solution:

Since ABCDABCD is a rhombus, opposite angles are equal: A=C=60°\angle A = \angle C = 60°.
Adjacent angles are supplementary: B=D=180°60°=120°\angle B = \angle D = 180° - 60° = 120°.

The diagonals bisect the vertex angles:
- Diagonal ACAC makes 30°30° with each side at vertices AA and CC.
- Diagonal BDBD makes 60°60° with each side at vertices BB and DD.

In AOB\triangle AOB (where OO is the intersection of diagonals):
OAB=30°\angle OAB = 30°, ABO=60°\angle ABO = 60°, AOB=90°\angle AOB = 90° (diagonals perpendicular).

AO=ABcos30°=6×32=33AO = AB \cos 30° = 6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3} cm.
BO=ABcos60°=6×12=3BO = AB \cos 60° = 6 \times \dfrac{1}{2} = 3 cm.

AC=2×AO=63 cm10.39 cmAC = 2 \times AO = 6\sqrt{3} \text{ cm} \approx 10.39 \text{ cm}

BD=2×BO=6 cmBD = 2 \times BO = 6 \text{ cm}

Answer: AC=63AC = 6\sqrt{3} cm, BD=6BD = 6 cm.

Solved Example 30: Prove It Is a Parallelogram

Problem: In quadrilateral ABCDABCD, AB=CD=7AB = CD = 7 cm and ABCDAB \parallel CD. Prove that ABCDABCD is a parallelogram.

Solution:

In quadrilateral ABCDABCD:
- AB=CDAB = CD (given: one pair of opposite sides are equal)
- ABCDAB \parallel CD (given: the same pair is parallel)

A quadrilateral in which one pair of opposite sides is both equal and parallel is a parallelogram.

Therefore, ABCDABCD is a parallelogram. \square

The Quadrilateral Hierarchy — Summary Table

This table summarizes all properties at a glance. Use it for quick revision.

PropertyParallelogramRectangleRhombusSquareTrapeziumKite
Opp. sides parallelBoth pairsBoth pairsBoth pairsBoth pairsOne pairNo
Opp. sides equalYesYesYes (all equal)Yes (all equal)NoNo
All sides equalNoNoYesYesNoTwo pairs
Opp. angles equalYesYes (all 90°)YesYes (all 90°)NoOne pair
All angles 90°NoYesNoYesNoNo
Diag. bisect each otherYesYesYesYesNoOne bisects other
Diag. equalNoYesNoYesNo*No
Diag. perpendicularNoNoYesYesNoYes
Diag. bisect vertex anglesNoNoYesYesNoOne diagonal

*Isosceles trapezium has equal diagonals.

Note: Every property that a parallelogram has is automatically inherited by rectangles, rhombuses, and squares. The special quadrilaterals simply add MORE properties on top.

Common Mistakes Students Make in Quadrilaterals

Here are the most frequent errors — learn these and you will avoid losing marks:

1. Confusing the hierarchy:
* Mistake: Saying "a rectangle is not a parallelogram" or "a square is not a rhombus".
* Fix: Remember: Square \subset Rectangle \subset Parallelogram and Square \subset Rhombus \subset Parallelogram. Every square IS a rectangle AND a rhombus.

2. Using wrong angle sum:
* Mistake: Using 180°180° for the angle sum of a quadrilateral (that is for a triangle).
* Fix: Angle sum of a quadrilateral =360°= 360°. General formula: (n2)×180°(n-2) \times 180°.

3. Confusing interior and exterior angles:
* Mistake: Adding interior and exterior angles incorrectly, or forgetting they are supplementary.
* Fix: Interior angle ++ Exterior angle =180°= 180° at each vertex. Sum of all exterior angles =360°= 360°.

4. Not stating properties used:
* Mistake: Finding the answer but not mentioning which property was applied.
* Fix: Always write: "Since opposite angles of a parallelogram are equal..." or "By the angle sum property of a quadrilateral..."

5. Assuming perpendicular diagonals for all parallelograms:
* Mistake: Claiming diagonals of a parallelogram are perpendicular (that is only true for rhombuses).
* Fix: Perpendicular diagonals \Rightarrow Rhombus. Equal diagonals \Rightarrow Rectangle. Both \Rightarrow Square.

6. Incorrect Pythagoras in rhombus problems:
* Mistake: Using full diagonal lengths instead of half-diagonals when applying Pythagoras theorem.
* Fix: Diagonals of a rhombus bisect each other. Use HALF of each diagonal as the two legs of the right triangle.

7. Forgetting that a trapezium is NOT a parallelogram:
* Mistake: Applying parallelogram properties (like opposite sides equal) to a trapezium.
* Fix: A trapezium has only ONE pair of parallel sides. Only the co-interior angle property (A+D=180°\angle A + \angle D = 180° between parallel sides) applies.

Exam Strategy: How to Score Full Marks in Chapter 3

Chapter 3 is a highly scoring chapter because the questions are visual and property-based. Here is your strategy:

Weightage: This chapter typically carries 6-8 marks in CBSE exams across MCQs, short-answer, and diagram-based questions.

Typical Question Patterns:

* 1 Mark (MCQ/VSA): "How many diagonals does a hexagon have?" or "Each interior angle of a regular polygon is 120°120°. How many sides?" or "Which quadrilateral has perpendicular diagonals?"

* 2-3 Marks (SA): Finding missing angles of a quadrilateral; angles in ratio problems; finding sides/diagonals of special quadrilaterals using Pythagoras.

* 3-4 Marks (LA): Proving a quadrilateral is a parallelogram; multi-step problems involving diagonal properties; identifying types from given conditions.

High-Priority Topics:
1. Angle sum property of polygons — (n2)×180°(n-2) \times 180°
2. Exterior angle sum =360°= 360° and its applications
3. Finding angles of parallelograms (opposite equal, adjacent supplementary)
4. Rhombus diagonal problems using Pythagoras
5. Identifying quadrilateral types from properties

Time Allocation: Spend 33-44 minutes on a 22-mark question and 55-77 minutes on a 44-mark proof/multi-step problem. Always draw clean diagrams.

Pro Tips:
- Draw and label every diagram. Mark parallel sides with arrows, equal sides with tick marks, and right angles with small squares.
- State every property you use. "Opposite angles of a parallelogram are equal" earns marks.
- For "identify the quadrilateral" questions, check properties systematically: Are opposite sides parallel? Equal? Are diagonals equal? Perpendicular?

Practice on SparkEd's Quadrilaterals page for exam-ready confidence!

Connections to Other Chapters and Higher Classes

Understanding quadrilaterals connects to several other topics:

Within Class 8:
- Chapter 9 (Mensuration): Area formulas for trapeziums and general quadrilaterals build directly on this chapter.
- Chapter 5 (Squares and Square Roots): Pythagoras theorem used for finding diagonals.

In Class 9:
- Chapter 8 (Quadrilaterals): You will formally PROVE many of the properties stated in this chapter (like the midpoint theorem and properties of parallelograms).
- Chapter 9 (Areas of Parallelograms and Triangles): Area relationships between parallelograms and triangles sharing the same base and height.

In Class 10:
- Coordinate Geometry: Verifying that four given points form a parallelogram, rectangle, rhombus, or square using distance and slope formulas.
- Triangles: Similarity and congruence proofs involving quadrilateral diagonals.

A thorough understanding of quadrilateral properties now makes geometry much smoother in Classes 9 and 10!

Boost Your Preparation with SparkEd

You have just gone through the entire Understanding Quadrilaterals chapter — every property, every exercise, every quadrilateral type. But reading alone will not get you full marks; practice will.

Here is how SparkEd can help you ace this chapter:

* Practice by Difficulty: On our Quadrilaterals practice page, work through problems sorted by difficulty. Start with basic angle sums and build up to multi-step diagonal problems.

* AI Math Solver: Stuck on a quadrilateral property problem? Paste it into our AI Solver and get step-by-step solutions.

* AI Coach: Get personalized recommendations on which quadrilateral types need more practice based on your performance.

* Cross-Topic Connections: Quadrilaterals connect to Mensuration (Chapter 9) and Squares and Square Roots (Chapter 5). Explore all chapters on our programs page.

Head over to sparkedmaths.com and start practicing today!

Frequently Asked Questions

Try SparkEd Free

Visual step-by-step solutions, three difficulty levels of practice, and an AI-powered Spark coach to guide you when you are stuck. Pick your class and board to start.

Start Practicing Now