Solved Exercises on Types of Quadrilaterals and Their Properties in Class 9

Master quadrilaterals: parallelograms, rectangles, squares, rhombuses, trapeziums, and kites through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Parallelogram Properties
Exercise 1
In parallelogram ABCD, AB = 8 cm, BC = 6 cm, and angle A = 60°. Find the length of CD, AD, and measure of angle C.
Definition:

Parallelogram: A quadrilateral with both pairs of opposite sides parallel and equal

Properties used:
  1. Opposite sides are equal: AB = CD and BC = AD
  2. Opposite angles are equal: ∠A = ∠C and ∠B = ∠D
  3. Adjacent angles are supplementary: ∠A + ∠B = 180°
Given
AB = 8cm, BC = 6cm, ∠A = 60°
Find
CD, AD, ∠C
Step 1: Apply opposite sides property

In parallelogram ABCD: AB = CD and BC = AD

Therefore, CD = AB = 8 cm

And AD = BC = 6 cm

Step 2: Apply opposite angles property

In parallelogram ABCD: ∠A = ∠C

Therefore, ∠C = ∠A = 60°

Step 3: Verify adjacent angles property

∠A + ∠B = 180° (adjacent angles in parallelogram)

60° + ∠B = 180°, so ∠B = 120°

And ∠D = ∠B = 120° (opposite angles)

CD = 8 cm, AD = 6 cm, ∠C = 60°
Final answer:

CD = 8 cm, AD = 6 cm, ∠C = 60°

Applied rules:

• Opposite sides equal: AB = CD and BC = AD

• Opposite angles equal: ∠A = ∠C and ∠B = ∠D

• Adjacent angles supplementary: ∠A + ∠B = 180°

2 Rectangle Properties
Exercise 2
In rectangle PQRS, diagonal PR = 10 cm. Find the length of diagonal QS and prove that diagonals bisect each other.
Definition:

Rectangle: A parallelogram with all angles equal to 90°

Given
Rectangle PQRS, PR = 10 cm
Find
QS and proof of bisecting
Step 1: Apply rectangle diagonal property

In rectangle, diagonals are equal: PR = QS

Therefore, QS = PR = 10 cm

Step 2: Prove diagonals bisect each other

Let diagonals PR and QS intersect at point O

Since rectangle is also a parallelogram, its diagonals bisect each other

Therefore, PO = OR = PR/2 = 10/2 = 5 cm

And QO = OS = QS/2 = 10/2 = 5 cm

Step 3: Additional properties verification

Diagonals of rectangle are equal and bisect each other

Also, they are congruent and create 4 triangles of equal area

Diagonal QS = 10 cm, diagonals bisect at midpoint
Final answer:

QS = 10 cm, diagonals bisect each other at their midpoints

Applied rules:

• Equal diagonals: Diagonals of rectangle are equal in length

• Bisecting property: Diagonals of parallelogram bisect each other

• All angles 90°: Each interior angle measures 90°

3 Rhombus Properties
Exercise 3
In rhombus ABCD, side AB = 5 cm and one diagonal AC = 8 cm. Find the length of the other diagonal BD and the area of the rhombus.
Definition:

Rhombus: A parallelogram with all sides equal

Given
Rhombus ABCD, AB = 5cm, AC = 8cm
Find
BD and Area
Step 1: Apply rhombus properties

In rhombus: all sides are equal, so AB = BC = CD = DA = 5 cm

Diagonals of rhombus bisect each other at right angles

Step 2: Use Pythagorean theorem

Let diagonals AC and BD intersect at O

Then AO = AC/2 = 8/2 = 4 cm

And BO = BD/2 (to find)

In triangle AOB: AB² = AO² + BO²

5² = 4² + BO²

25 = 16 + BO²

BO² = 9, so BO = 3 cm

Step 3: Calculate BD and area

Since BO = 3 cm, then BD = 2 × BO = 2 × 3 = 6 cm

Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × AC × BD

Area = (1/2) × 8 × 6 = 24 cm²

BD = 6 cm, Area = 24 cm²
Final answer:

BD = 6 cm, Area of rhombus = 24 cm²

Applied rules:

• All sides equal: AB = BC = CD = DA

• Diagonals perpendicular: Diagonals intersect at 90°

• Area formula: (1/2) × d₁ × d₂

Quadrilateral Types and Properties
Sum of interior angles = 360°
Angle Sum Property
Parallelogram
▰
  • Opposite sides parallel & equal
  • Opposite angles equal
  • Adjacent angles supplementary
  • Diagonals bisect each other
Rectangle
▭
  • All angles 90°
  • Opposite sides equal
  • Diagonals equal & bisect
  • Special parallelogram
Square
□
  • All sides equal
  • All angles 90°
  • Diagonals equal & perpendicular
  • Both rectangle & rhombus
Rhombus
◆
  • All sides equal
  • Opposite angles equal
  • Diagonals ⊥ & bisect angles
  • Special parallelogram
Angle Relationships
∠A + ∠B + ∠C + ∠D = 360°

Sum of all interior angles

Side Relationships
AB = CD, BC = AD

In parallelogram

Area = base × height
Parallelogram Area
Area = (1/2) × d₁ × d₂
Rhombus/Kite Area
Solution: Exercises 4 to 5
4 Trapezium Properties
Exercise 4
In trapezium ABCD, AB || DC, AB = 10 cm, DC = 6 cm, and height = 4 cm. Find the area of the trapezium and prove that the line joining midpoints of non-parallel sides is parallel to bases.
Definition:

Trapezium: A quadrilateral with exactly one pair of parallel sides

Given
AB || DC, AB = 10cm, DC = 6cm, h = 4cm
Find
Area and midpoint line
Step 1: Apply trapezium area formula

Area of trapezium = (1/2) × (sum of parallel sides) × height

Area = (1/2) × (AB + DC) × h

Area = (1/2) × (10 + 6) × 4 = (1/2) × 16 × 4 = 32 cm²

Step 2: Midpoint line property proof

Let E and F be midpoints of non-parallel sides AD and BC respectively

Line EF is called the median or mid-segment of trapezium

EF is parallel to both AB and DC

Length of EF = (1/2)(AB + DC) = (1/2)(10 + 6) = 8 cm

Step 3: Verify the property

This is a standard theorem: The line segment joining the midpoints of the non-parallel sides of a trapezium is parallel to the parallel sides and equal to half their sum

Area = 32 cm², EF || AB || DC, EF = 8 cm
Final answer:

Area = 32 cm², line joining midpoints is parallel to bases and measures 8 cm

Applied rules:

• Area formula: (1/2) × (sum of parallel sides) × height

• Midsegment property: Parallel to bases, length = average of bases

• One pair parallel: Exactly one set of parallel sides

5 Kite Properties
Exercise 5
In kite ABCD, AB = AD = 5 cm, BC = CD = 7 cm, and diagonal AC = 8 cm. Find the length of diagonal BD and the area of the kite.
Definition:

Kite: A quadrilateral with two distinct pairs of adjacent sides equal

Given
AB = AD = 5cm, BC = CD = 7cm, AC = 8cm
Find
BD and Area
Step 1: Apply kite properties

In kite: two pairs of adjacent sides are equal (AB = AD, BC = CD)

Diagonals of kite are perpendicular to each other

One diagonal bisects the other

Step 2: Use perpendicular diagonals

Let diagonals AC and BD intersect at O perpendicularly

AC bisects BD, so BO = OD

AO and CO are parts of AC where AO + CO = AC = 8 cm

Step 3: Apply Pythagorean theorem

In right triangle AOB: AB² = AO² + BO²

5² = AO² + BO² → 25 = AO² + BO²

In right triangle BOC: BC² = CO² + BO²

7² = CO² + BO² → 49 = CO² + BO²

Step 4: Solve the system of equations

From equation 1: BO² = 25 - AO²

From equation 2: BO² = 49 - CO²

So: 25 - AO² = 49 - CO²

CO² - AO² = 24

Since AO + CO = 8, let CO = 8 - AO

(8-AO)² - AO² = 24

64 - 16AO + AO² - AO² = 24

64 - 16AO = 24

16AO = 40, so AO = 2.5 cm

Therefore CO = 8 - 2.5 = 5.5 cm

Step 5: Calculate BO and BD

BO² = 25 - AO² = 25 - 6.25 = 18.75

BO = √18.75 ≈ 4.33 cm

Since BO = OD, BD = 2 × BO ≈ 2 × 4.33 = 8.66 cm

Step 6: Calculate area

Area of kite = (1/2) × d₁ × d₂ = (1/2) × AC × BD

Area = (1/2) × 8 × 8.66 ≈ 34.64 cm²

BD ≈ 8.66 cm, Area ≈ 34.64 cm²
Final answer:

BD ≈ 8.66 cm, Area of kite ≈ 34.64 cm²

Applied rules:

• Adjacent sides equal: Two pairs of consecutive equal sides

• Perpendicular diagonals: Diagonals intersect at 90°

• Area formula: (1/2) × d₁ × d₂

Quadrilateral Properties and Formulas Guide
Interior Angle Sum = (n-2)×180°
Polygon Formula (n=4)
Key definitions:

Quadrilateral: A polygon with four sides and four vertices

Parallelogram: Both pairs of opposite sides parallel

Rectangle: Parallelogram with all angles 90°

Square: Rectangle with all sides equal

Rhombus: Parallelogram with all sides equal

Trapezium: One pair of opposite sides parallel

Kite: Two distinct pairs of adjacent sides equal

Problem-solving methodology:
  1. Identify the quadrilateral type: Look for specific properties
  2. List known properties: Use relevant theorems and formulas
  3. Apply geometric relationships: Use angle sums, side relationships
  4. Use algebra when needed: Set up equations using given measurements
  5. Verify results: Check if answer satisfies all properties
Tip 1: Remember: Every square is a rectangle, rhombus, and parallelogram!
Tip 2: In any parallelogram, diagonals bisect each other.
Tip 3: For area of rhombus/kite: use (1/2) × d₁ × d₂ where d₁, d₂ are diagonals.
Tip 4: Always check if your answer makes sense geometrically.
Key formulas: Area of parallelogram = base × height, Area of trapezium = (1/2)(sum of parallel sides) × height
Memory trick: "Some People Have Curly Brown Hair Through Proper Brushing" - Square, Parallelogram, Rectangle, Rhombus, Trapezium, Kite
Important theorems to remember:

• Opposite sides of parallelogram are equal and parallel

• Opposite angles of parallelogram are equal

• Diagonals of parallelogram bisect each other

• Diagonals of rectangle are equal and bisect each other

• Diagonals of rhombus bisect each other at right angles

• Diagonals of square are equal and bisect at right angles

• In kite, one diagonal bisects the other at right angles

Questions & Answers

Question: I'm confused about when a parallelogram becomes a rectangle or rhombus. What are the additional properties that make this happen?

Answer: Great question! Starting from a parallelogram, we add specific properties to get special types:

  • Parallelogram (basic): Opposite sides parallel and equal, opposite angles equal, diagonals bisect each other
  • Rectangle (parallelogram +): All properties of parallelogram PLUS all angles are 90°
  • Rhombus (parallelogram +): All properties of parallelogram PLUS all sides equal
  • Square (rectangle + rhombus): All properties of rectangle AND rhombus (all sides equal, all angles 90°)

Think of it hierarchically: Square ⊂ Rectangle ⊂ Parallelogram and Square ⊂ Rhombus ⊂ Parallelogram

So every square is automatically a rectangle, rhombus, and parallelogram, but not vice versa.

Question: Why do the diagonals of a rhombus bisect each other at right angles? Can you explain the proof?

Answer: Let's prove that diagonals of a rhombus are perpendicular:

Consider rhombus ABCD with diagonals AC and BD intersecting at O.

Proof steps:

  1. AB = BC = CD = DA (all sides equal in rhombus)
  2. AO = OC and BO = OD (diagonals of parallelogram bisect each other)
  3. In triangles AOB and COB: AB = CB (sides of rhombus), AO = CO (from step 2), OB = OB (common)
  4. By SSS congruence, ΔAOB ≅ ΔCOB
  5. Therefore ∠AOB = ∠COB
  6. But ∠AOB + ∠COB = 180° (linear pair)
  7. So 2∠AOB = 180°, which means ∠AOB = 90°

This proves that diagonals intersect at right angles.

Question: How can I quickly identify which quadrilateral I'm dealing with in a problem? Are there quick recognition tips?

Answer: Yes, here are quick recognition tips:

  • Parallelogram: Look for 2 pairs of parallel sides OR opposite sides equal
  • Rectangle: Look for 4 right angles OR parallelogram with 1 right angle
  • Rhombus: Look for 4 equal sides OR parallelogram with 2 adjacent sides equal
  • Square: Look for 4 equal sides AND 4 right angles
  • Trapezium: Look for exactly 1 pair of parallel sides
  • Kite: Look for 2 pairs of adjacent equal sides

Always start by identifying what's given (sides, angles, diagonals) and match to the defining properties of each quadrilateral.

Drawing a quick sketch often helps visualize the properties.