Quadrilaterals | Grade 9th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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9th

9th - Mathematics

Quadrilaterals

Overview

This chapter introduces the concept of quadrilaterals, their properties, types, and theorems related to them. Students will learn about different types of quadrilaterals such as parallelograms, rectangles, rhombuses, squares, trapeziums, and kites, along with their defining characteristics and proofs of related theorems.

Quadrilateral: A quadrilateral is a closed figure with four sides, four vertices, and four angles. The sum of the interior angles of a quadrilateral is 360°.

Types of Quadrilaterals

Parallelogram: A quadrilateral with both pairs of opposite sides parallel.
Rectangle: A parallelogram with all angles equal to 90°.
Rhombus: A parallelogram with all sides equal in length.
Square: A rectangle with all sides equal in length.
Trapezium: A quadrilateral with at least one pair of parallel sides.
Kite: A quadrilateral with two distinct pairs of adjacent sides equal in length.

Properties of Quadrilaterals

Properties of a Parallelogram:

Opposite sides are equal and parallel.
Opposite angles are equal.
Diagonals bisect each other.

Properties of a Rectangle:

All properties of a parallelogram apply.
All angles are right angles.
Diagonals are equal in length.

Properties of a Rhombus:

All properties of a parallelogram apply.
All sides are equal.
Diagonals bisect each other at right angles.

Properties of a Square:

All properties of a rectangle and rhombus apply.
Diagonals are equal and bisect each other at right angles.

Theorems Related to Quadrilaterals

The diagonal of a parallelogram divides it into two congruent triangles.
In a parallelogram, opposite sides are equal.
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
The diagonals of a parallelogram bisect each other.
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Mid-Point Theorem

Mid-Point Theorem: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Summary

This chapter covers the fundamental concepts of quadrilaterals, their types, properties, and related theorems. Understanding these concepts helps in solving geometric problems and proofs effectively.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

What is the sum of the angles of a quadrilateral?

Answer:

Numeric answer:
360°

Question 2:

Name the quadrilateral with only one pair of parallel sides.

Answer:

Trapezium

Question 3:

If three angles of a quadrilateral are 70°, 80°, and 110°, find the fourth angle.

Answer:

Numeric answer:
100°

Question 4:

What is the mid-point theorem in quadrilaterals?

Answer:

Line joining mid-points of two sides is parallel to third side.

Question 5:

A parallelogram has one angle 60°. Find its opposite angle.

Answer:

Numeric answer:
60°

Question 6:

What is the diagonal property of a rectangle?

Answer:

Diagonals are equal and bisect each other.

Question 7:

If diagonals of a quadrilateral bisect each other at 90°, it is a _____.

Answer:

Rhombus

Question 8:

Find the perimeter of a square with side 5 cm.

Answer:

Numeric answer:
20 cm

Question 9:

What is the area of a rectangle with length 8 cm and breadth 3 cm?

Answer:

Numeric answer:
24 cm²

Question 10:

A rhombus has diagonals 6 cm and 8 cm. Find its area.

Answer:

Numeric answer:
24 cm²

Question 11:

In a kite, one diagonal is 10 cm. If area is 30 cm², find the other diagonal.

Answer:

Numeric answer:
6 cm

Question 12:

If a quadrilateral is both rectangle and rhombus, what is it called?

Answer:

Square

Question 13:

Define a quadrilateral.

Answer:

A quadrilateral is a polygon with four sides, four vertices, and four angles. The sum of its interior angles is always 360°.

Question 14:

What is the sum of the interior angles of a quadrilateral?

Answer:

The sum of the interior angles of any quadrilateral is 360°.

Question 15:

Name the quadrilateral whose diagonals are equal and bisect each other at right angles.

Answer:

The quadrilateral is a square or a rhombus.
In a square, diagonals are equal and bisect at 90°.
In a rhombus, diagonals bisect at 90° but may not be equal.

Question 16:

State the property of a parallelogram related to its opposite sides.

Answer:

In a parallelogram, opposite sides are equal in length and parallel.

Question 17:

If one angle of a parallelogram is 90°, what type of quadrilateral is it?

Answer:

If one angle of a parallelogram is 90°, it is a rectangle because all angles become 90° due to the properties of a parallelogram.

Question 18:

What is the difference between a trapezium and a parallelogram?

Answer:

A trapezium has only one pair of parallel sides.
A parallelogram has both pairs of opposite sides parallel.

Question 19:

In a rhombus, if one diagonal is 10 cm and the other is 24 cm, find its area.

Answer:

Area of a rhombus = (d₁ × d₂) / 2
Given: d₁ = 10 cm, d₂ = 24 cm
Area = (10 × 24) / 2
Area = 120 cm².

Question 20:

What is the condition for a quadrilateral to be a rectangle?

Answer:

A quadrilateral is a rectangle if:

All angles are 90°.
Opposite sides are equal and parallel.
Diagonals are equal and bisect each other.

Question 21:

If the diagonals of a quadrilateral bisect each other at 90°, what type of quadrilateral is it?

Answer:

It is a rhombus or a square because their diagonals bisect each other at 90°.

Question 22:

Prove that the diagonals of a rectangle are equal in length.

Answer:

In a rectangle ABCD:
Consider triangles ABC and ABD.
AB = AB (common side)
AD = BC (opposite sides of rectangle)
∠DAB = ∠CBA = 90°
By SAS congruency, ΔABC ≅ ΔABD.
Thus, AC = BD (CPCT).
Hence, diagonals are equal.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

Define a quadrilateral and list its basic properties.

Answer:

A quadrilateral is a polygon with four sides, four vertices, and four angles. Its basic properties include:

The sum of all interior angles is 360°.
It can be convex or concave.
It has two diagonals.

Question 2:

What is the sum of the interior angles of a quadrilateral? Justify your answer.

Answer:

The sum of the interior angles of a quadrilateral is 360°.
Justification: A quadrilateral can be divided into two triangles by drawing one diagonal.
Since the sum of angles in a triangle is 180°, for two triangles, it becomes 180° × 2 = 360°.

Question 3:

State the condition for a quadrilateral to be a parallelogram.

Answer:

A quadrilateral is a parallelogram if:

Both pairs of opposite sides are parallel and equal.
Opposite angles are equal.
Diagonals bisect each other.

Question 4:

If one angle of a parallelogram is 75°, find the measure of its adjacent angle.

Answer:

In a parallelogram, adjacent angles are supplementary (sum to 180°).
Given one angle = 75°,
Adjacent angle = 180° - 75° = 105°.

Question 5:

Differentiate between a rhombus and a rectangle based on their diagonals.

Answer:

Rhombus: Diagonals are perpendicular and bisect each other at 90°.
Rectangle: Diagonals are equal in length and bisect each other but not necessarily at 90°.

Question 6:

Prove that the diagonals of a square are equal and bisect each other at right angles.

Answer:

In a square:
1. All sides are equal, and angles are 90°.
2. Diagonals are equal (by congruency of triangles).
3. Diagonals bisect each other (property of parallelogram).
4. Diagonals intersect at 90° (angles formed are equal and sum to 180°).

Question 7:

What is a kite in quadrilaterals? State its properties.

Answer:

A kite is a quadrilateral with two distinct pairs of adjacent sides that are equal.
Properties:

One pair of opposite angles is equal.
Diagonals are perpendicular.
One diagonal bisects the other.

Question 8:

In a trapezium ABCD, AB || CD. If ∠A = 60°, find ∠D.

Answer:

In a trapezium with AB || CD, ∠A and ∠D are co-interior angles.
Sum of co-interior angles = 180°.
Given ∠A = 60°,
∠D = 180° - 60° = 120°.

Question 9:

If the diagonals of a quadrilateral bisect each other at right angles, what type of quadrilateral is it?

Answer:

It is a rhombus or a square.
Both have diagonals that bisect each other at 90°.
If all angles are 90°, it is a square; otherwise, it is a rhombus.

Question 10:

A quadrilateral has three angles measuring 80°, 95°, and 120°. Find the fourth angle.

Answer:

Sum of all interior angles of a quadrilateral = 360°.
Given angles: 80° + 95° + 120° = 295°.
Fourth angle = 360° - 295° = 65°.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Define a parallelogram and list any two of its properties.

Answer:

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

Properties:

Opposite sides are equal and parallel.
Opposite angles are equal.

Question 2:

State the conditions under which a rhombus becomes a square.

Answer:

A rhombus becomes a square when:

All its angles are equal to 90° (each angle becomes a right angle).
Its diagonals are of equal length (since squares have equal diagonals).

Question 3:

In a trapezium ABCD, AB || CD. If ∠A = 60°, find ∠D.

Answer:

In a trapezium with AB || CD, the consecutive angles on the same side of the leg are supplementary.

Thus, ∠A + ∠D = 180°.
Given ∠A = 60°,
∠D = 180° - 60° = 120°.

Question 4:

Explain why a kite is not a parallelogram.

Answer:

A kite is not a parallelogram because:

Only one pair of opposite sides is parallel in a kite, whereas in a parallelogram, both pairs are parallel.
A kite has two distinct pairs of adjacent sides equal, while a parallelogram has opposite sides equal.

Question 5:

Prove that the diagonals of a parallelogram bisect each other.

Answer:

In a parallelogram ABCD, the diagonals AC and BD intersect at point O.
Using the properties of a parallelogram, we know that opposite sides are equal and parallel.
Triangles AOB and COD are congruent by the ASA (Angle-Side-Angle) rule because:

∠OAB = ∠OCD (alternate angles)
AB = CD (opposite sides of a parallelogram)
∠OBA = ∠ODC (alternate angles)

Thus, AO = OC and BO = OD, proving that the diagonals bisect each other.

Question 6:

If one angle of a rhombus is 60°, find the measures of its other angles.

Answer:

In a rhombus, opposite angles are equal, and adjacent angles are supplementary (sum to 180°).
Given one angle is 60°, its opposite angle is also 60°.
The adjacent angles will be:
180° - 60° = 120°.
Thus, the four angles of the rhombus are 60°, 120°, 60°, and 120°.

Question 7:

Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

Answer:

Let ABCD be a quadrilateral with midpoints P, Q, R, and S of sides AB, BC, CD, and DA, respectively.
Joining P to R and Q to S, we form two line segments intersecting at O.
Using the Midpoint Theorem, PQRS forms a parallelogram because opposite sides are equal and parallel.
In a parallelogram, the diagonals bisect each other, so PR and QS bisect at O.

Question 8:

In a trapezium ABCD with AB || CD, if ∠A = 70° and ∠D = 110°, find the measures of ∠B and ∠C.

Answer:

In a trapezium ABCD with AB || CD, the consecutive angles between the non-parallel sides are supplementary.
Given ∠A = 70° and ∠D = 110°, we can find the other angles:
∠A + ∠D = 70° + 110° = 180° (which confirms they are supplementary).
Similarly, ∠B + ∠C = 180°.
Since ∠A and ∠B are consecutive angles, ∠B = 180° - ∠A = 110°.
Similarly, ∠C = 180° - ∠D = 70°.

Question 9:

Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic.

Answer:

Let ABCD be a quadrilateral with internal angle bisectors forming another quadrilateral PQRS.
We know that the sum of the angles of any quadrilateral is 360°.
For quadrilateral PQRS, the angles are formed by the bisectors of the original angles.
Thus, the sum of opposite angles of PQRS will be:
∠P + ∠R = 180° and ∠Q + ∠S = 180° (since they are half the sum of the original quadrilateral's angles).
This satisfies the condition for a quadrilateral to be cyclic (opposite angles are supplementary).

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Prove that the diagonals of a rhombus bisect each other at right angles. Use a diagram and step-wise reasoning.

Answer:

Introduction

We studied that a rhombus is a special type of parallelogram with all sides equal. Our textbook shows that its diagonals have unique properties.

Argument 1

Let ABCD be a rhombus with diagonals AC and BD intersecting at O.
In ΔAOB and ΔAOD, AB = AD (sides of rhombus), AO is common, and OB = OD (diagonals bisect).
By SSS congruency, ΔAOB ≅ ΔAOD, so ∠AOB = ∠AOD = 90°.

Conclusion

Thus, diagonals bisect each other at right angles. [Diagram: Rhombus with diagonals intersecting at 90°]

Question 2:

A playground is in the shape of a rectangle with length 40m and breadth 25m. Find the cost of fencing it at ₹20 per meter, including a 2m wide gate.

Answer:

Introduction

Our textbook explains that perimeter calculations are essential for real-life problems like fencing.

Argument 1

Perimeter of rectangle = 2 × (length + breadth) = 2 × (40m + 25m) = 130m.
Subtract gate width: Fencing length = 130m − 2m = 128m.

Argument 2

Cost = Fencing length × Rate = 128m × ₹20/m = ₹2560.

Conclusion

The total cost is ₹2560, demonstrating practical applications of quadrilaterals.

Question 3:

Show that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

Answer:

Introduction

We learned about the Mid-Point Theorem in quadrilaterals, which connects triangles and parallelograms.

Argument 1

Let ABC be a triangle with D and E as mid-points of AB and AC.
Extend DE to F such that DE = EF and join CF.
By SAS congruency, ΔADE ≅ ΔCFE, so AD ∥ CF.

Argument 2

Since D is mid-point, BD = AD = CF, making BDFC a parallelogram.
Thus, DF ∥ BC and DE = ½ BC.

Conclusion

This proves the theorem, as shown in NCERT Example 8.7.

Question 4:

Prove that the diagonals of a rhombus bisect each other at right angles. Use properties of congruent triangles.

Answer:

Introduction

We studied that a rhombus is a special type of parallelogram with all sides equal. Our textbook shows that its diagonals have unique properties.

Argument 1

Let ABCD be a rhombus with diagonals AC and BD intersecting at O.
In ΔAOB and ΔAOD, AB = AD (sides of rhombus), AO is common, and OB = OD (diagonals bisect).
By SSS congruency, ΔAOB ≅ ΔAOD, so ∠AOB = ∠AOD = 90°.

Conclusion

Thus, diagonals bisect each other at right angles, a key property of rhombuses.

Question 5:

A quadrilateral has angles in the ratio 3:5:9:13. Find all its angles. Show steps.

Answer:

Introduction

We know the sum of angles in a quadrilateral is 360°. Our textbook uses ratio problems to find unknown angles.

Argument 1

Let angles be 3x, 5x, 9x, and 13x.
Sum: 3x + 5x + 9x + 13x = 30x = 360°.
Solving, x = 12°.

Argument 2

Angles: 3x = 36°, 5x = 60°, 9x = 108°, 13x = 156°.
Verification: 36° + 60° + 108° + 156° = 360°.

Conclusion

The angles are 36°, 60°, 108°, and 156°, satisfying the quadrilateral angle sum property.

Question 6:

Explain why a rectangle is a parallelogram but a parallelogram may not be a rectangle. Give an example.

Answer:

Introduction

We studied that rectangles and parallelograms are quadrilaterals with specific properties. Our textbook compares their features.

Argument 1

A rectangle has all angles as 90° and opposite sides equal and parallel, fulfilling parallelogram conditions.
Example: A book’s shape is both a rectangle and parallelogram.

Argument 2

A parallelogram may have angles ≠ 90° (e.g., rhombus), so it’s not always a rectangle.
Example: A tilted door frame is a parallelogram but not a rectangle.

Conclusion

Thus, all rectangles are parallelograms, but the converse isn’t true due to angle differences.

Question 7:

Prove that the diagonals of a parallelogram bisect each other. Use a diagram and step-by-step reasoning.

Answer:

Introduction

We studied that a parallelogram has both pairs of opposite sides parallel. Let us prove its diagonals bisect each other.

Argument 1

Consider parallelogram ABCD with diagonals AC and BD intersecting at O. By ASA congruency, ∆AOB ≅ ∆COD (AB = CD, ∠OAB = ∠OCD, ∠OBA = ∠ODC).

Argument 2

From congruency, AO = OC and BO = OD. Thus, diagonals bisect each other.

Conclusion

This proves our textbook theorem that diagonals of a parallelogram bisect each other.

Question 8:

A playground is in the shape of a rectangle with length 40m and width 30m. Find the length of its diagonal. Show the calculation.

Answer:

Introduction

Our textbook shows that rectangles have right angles, allowing diagonal calculations using Pythagoras theorem.

Argument 1

Given: Length (l) = 40m, Width (w) = 30m. Diagonal (d) = √(l² + w²).

Argument 2

Substituting values: d = √(40² + 30²) = √(1600 + 900) = √2500 = 50m.

Conclusion

Thus, the playground's diagonal is 50 meters, a real-life application of quadrilateral properties.

Question 9:

Explain why a rhombus is considered a special type of parallelogram. Give two properties unique to rhombuses.

Answer:

Introduction

We studied that all rhombuses are parallelograms but have additional features.

Argument 1

Like parallelograms, rhombuses have opposite sides parallel and equal. However, all sides of a rhombus are equal (AB = BC = CD = DA).

Argument 2

Another unique property is that their diagonals bisect each other at right angles, unlike general parallelograms.

Conclusion

Thus, rhombuses are special parallelograms with equal sides and perpendicular diagonals.

Question 10:

Prove that the diagonals of a rhombus bisect each other at right angles. Support your answer with a diagram and properties studied in NCERT.

Answer:

Introduction

We studied that a rhombus is a special type of parallelogram with all sides equal. Our textbook shows its diagonals intersect at 90°.

Argument 1

Let ABCD be a rhombus with diagonals AC and BD intersecting at O.
In ΔAOB and ΔAOD, AB = AD (sides of rhombus), AO is common, and OB = OD (diagonals bisect).

Argument 2

By SSS congruency, ΔAOB ≅ ΔAOD, so ∠AOB = ∠AOD = 90°.
Thus, diagonals bisect at right angles.

Conclusion

This proves the diagonals of a rhombus bisect each other perpendicularly, a key property in NCERT.

[Diagram: Rhombus ABCD with diagonals intersecting at O]

Question 11:

A playground is in the shape of a rectangle with length 40m and breadth 25m. Find the cost of fencing it at ₹20 per meter. Justify your steps using quadrilateral properties.

Answer:

Introduction

We know a rectangle has equal opposite sides. Our textbook uses perimeter for fencing problems.

Argument 1

Given: Length (l) = 40m, Breadth (b) = 25m.
Perimeter = 2(l + b) = 2(40 + 25) = 130m (using rectangle property).

Argument 2

Cost per meter = ₹20.
Total cost = 130 × 20 = ₹2600.

Conclusion

Thus, fencing cost is ₹2600, applying rectangle perimeter formula from NCERT.

Question 12:

Show that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it. Use a diagram and NCERT Theorem 8.9.

Answer:

Introduction

Our textbook (Theorem 8.9) states the mid-segment of a triangle is parallel to the third side.

Argument 1

Let ABC be a triangle with D and E as mid-points of AB and AC.
Join DE. By midpoint theorem, DE || BC.

Argument 2

Construct CF || BA meeting DE at F.
Using congruency (ASA), ΔADE ≅ ΔCFE, so DE = EF = ½ BC.

Conclusion

Hence, DE is parallel and half of BC, as per NCERT Theorem 8.9.

[Diagram: Triangle ABC with mid-points D, E and line DE]

Question 13:

Prove that the diagonals of a rectangle are equal in length using the properties of quadrilaterals. Support your answer with a diagram.

Answer:

A rectangle is a type of parallelogram where all angles are right angles (90°). To prove that its diagonals are equal, we can use the properties of rectangles and the concept of congruent triangles.

Step 1: Construction
Draw rectangle ABCD with diagonals AC and BD intersecting at point O.

Step 2: Properties Used
In a rectangle:

Opposite sides are equal (AB = CD and AD = BC).
All angles are 90°.

Step 3: Proving Triangles Congruent
Consider triangles ABD and BAC:

AB is common to both triangles.
AD = BC (opposite sides of rectangle).
Angle A = Angle B = 90°.

By the SAS (Side-Angle-Side) congruence rule, ΔABD ≅ ΔBAC.

Step 4: Conclusion
Since the triangles are congruent, their corresponding sides BD and AC must be equal.
Hence, the diagonals of a rectangle are equal in length.

Note: A labeled diagram should show rectangle ABCD with diagonals intersecting at O, highlighting the congruent triangles.

Question 14:

Explain why a rhombus is considered a special type of parallelogram. List the additional properties that make it unique.

Answer:

A rhombus is a special type of parallelogram because it inherits all the properties of a parallelogram but has additional unique features.

Properties of a Parallelogram present in a Rhombus:

Opposite sides are equal and parallel.
Opposite angles are equal.
Diagonals bisect each other at 90°.

Additional Unique Properties of a Rhombus:

All four sides are of equal length.
Diagonals bisect the angles of the rhombus.
Diagonals are perpendicular to each other (intersect at 90°).

Explanation: Since a rhombus satisfies all the conditions of a parallelogram (opposite sides parallel and equal, diagonals bisect each other) and also has all sides equal and diagonals perpendicular, it is a special case of a parallelogram.

Example: If a quadrilateral has sides AB = BC = CD = DA and diagonals AC ⊥ BD, it is a rhombus and also a parallelogram.

Question 15:

Explain why a rhombus is considered a special type of parallelogram. List its unique properties and justify your answer with a diagram.

Answer:

A rhombus is a special type of parallelogram because it inherits all properties of a parallelogram while having additional unique features.

Properties of a Rhombus:

All sides are equal: Unlike a general parallelogram, a rhombus has all four sides of equal length.
Diagonals bisect each other at 90°: The diagonals are perpendicular and bisect each other.
Diagonals bisect the angles: Each diagonal divides the angles of the rhombus into two equal parts.

Why is it a Special Parallelogram?

A rhombus satisfies all properties of a parallelogram:
- Opposite sides are parallel and equal.
- Opposite angles are equal.
- Diagonals bisect each other.
However, the additional properties (equal sides and perpendicular diagonals) make it unique.

Justification with Diagram:
A labeled diagram of rhombus ABCD should show:

All sides marked as equal (AB = BC = CD = DA).
Diagonals AC and BD intersecting at 90°.
Angles bisected by the diagonals (e.g., Angle ABC split into two equal angles by diagonal BD).

Note: The uniqueness of a rhombus lies in its symmetry and equal side lengths, distinguishing it from a general parallelogram.

Question 16:

Prove that the diagonals of a rectangle are equal in length. Support your answer with a diagram and step-by-step reasoning.

Answer:

To prove that the diagonals of a rectangle are equal, let us consider a rectangle ABCD with diagonals AC and BD.

Step 1: Draw rectangle ABCD with sides AB = CD and AD = BC, and all angles as right angles (90°).

Step 2: Consider triangles ABD and BAC:

Both triangles share side AB.
AD = BC (opposite sides of a rectangle are equal).
Angle BAD = Angle ABC (both are 90°).

Step 3: By the SAS (Side-Angle-Side) congruence rule, triangles ABD and BAC are congruent.

Step 4: Since the triangles are congruent, their corresponding sides BD and AC must be equal.

Thus, the diagonals of a rectangle are equal in length.

Diagram: (Draw rectangle ABCD with diagonals intersecting at O, labeling all sides and angles appropriately.)

Question 17:

Explain why a rhombus is considered a special type of parallelogram. Highlight the unique properties of a rhombus with examples and comparisons.

Answer:

A rhombus is a special type of parallelogram because it inherits all the properties of a parallelogram while having additional unique features.

Properties of a Parallelogram:

Opposite sides are equal and parallel.
Opposite angles are equal.
Diagonals bisect each other.

Unique Properties of a Rhombus:

All sides are equal (unlike a general parallelogram where only opposite sides are equal).
Diagonals bisect each other at right angles (90°).
Diagonals bisect the vertex angles, dividing the rhombus into four congruent right-angled triangles.

Example: Consider a rhombus PQRS with sides PQ = QR = RS = SP. Its diagonals PR and QS intersect at O at 90° and divide the rhombus into four equal triangles.

Comparison: While a parallelogram only requires opposite sides to be equal, a rhombus enforces equality on all sides, making it a more symmetric and specialized quadrilateral.

Question 18:

Prove that the diagonals of a rectangle are equal in length using the properties of quadrilaterals. Support your answer with a labeled diagram.

Answer:

To prove that the diagonals of a rectangle are equal, let us consider a rectangle ABCD with diagonals AC and BD.

Step 1: Draw rectangle ABCD with sides AB, BC, CD, and DA, where all angles are right angles (90°).
Step 2: Draw diagonals AC and BD, intersecting at point O.
Step 3: In triangles ABC and BAD:
- AB is common to both triangles.
- BC = AD (opposite sides of a rectangle are equal).
- ∠ABC = ∠BAD = 90° (all angles in a rectangle are right angles).
Step 4: By the SAS (Side-Angle-Side) congruence rule, △ABC ≅ △BAD.
Step 5: Since corresponding parts of congruent triangles are equal, AC = BD.

Thus, the diagonals of a rectangle are equal in length.

Diagram: (Draw rectangle ABCD with diagonals AC and BD intersecting at O, labeling all sides and angles correctly.)

Additional Insight: This property is unique to rectangles among parallelograms, as other parallelograms (like rhombuses) do not necessarily have equal diagonals.

Question 19:

Prove that the diagonals of a rectangle are equal in length. Using this property, show that if the diagonals of a parallelogram are equal, then it must be a rectangle.

Answer:

To prove that the diagonals of a rectangle are equal, consider rectangle ABCD with diagonals AC and BD.

Step 1: In triangles ABC and BAD,
- AB is common.
- ∠ABC = ∠BAD = 90° (since it is a rectangle).
- BC = AD (opposite sides of a rectangle are equal).
Thus, by SAS congruence rule, ΔABC ≅ ΔBAD.

Step 2: Therefore, AC = BD (by CPCT).

Now, to prove the second part:
Let ABCD be a parallelogram with equal diagonals AC = BD.
Step 1: In triangles ABC and BAD,
- AB is common.
- AD = BC (opposite sides of a parallelogram).
- AC = BD (given).
Thus, by SSS congruence rule, ΔABC ≅ ΔBAD.

Step 2: Therefore, ∠ABC = ∠BAD (by CPCT).
But ∠ABC + ∠BAD = 180° (co-interior angles).
Hence, ∠ABC = ∠BAD = 90°.
Since one angle of the parallelogram is 90°, ABCD must be a rectangle.

Question 20:

In a quadrilateral ABCD, the angles are in the ratio 3:5:9:13. Find all the angles of the quadrilateral. Also, identify the type of quadrilateral formed and justify your answer.

Answer:

Let the angles of quadrilateral ABCD be 3x, 5x, 9x, and 13x.

Step 1: Sum of angles in a quadrilateral is 360°.
So, 3x + 5x + 9x + 13x = 360°.
30x = 360° ⇒ x = 12°.

Step 2: Calculate each angle:
- ∠A = 3x = 36°
- ∠B = 5x = 60°
- ∠C = 9x = 108°
- ∠D = 13x = 156°

Step 3: Identify the type of quadrilateral:
Since no sides or angles are equal, and opposite angles are not supplementary, ABCD is an irregular quadrilateral.

Note: The angles do not satisfy any special properties of parallelograms, rectangles, or trapeziums, confirming it is irregular.

Question 21:

Prove that the diagonals of a rectangle are equal in length using the properties of quadrilaterals. Also, state the converse of this theorem.

Answer:

To prove that the diagonals of a rectangle are equal, let us consider a rectangle ABCD with diagonals AC and BD.

Step 1: In triangles ABC and BAD,
AB = BA (Common side)
∠ABC = ∠BAD = 90° (Angles of a rectangle)
BC = AD (Opposite sides of a rectangle are equal)

Step 2: By the SAS congruence rule, ΔABC ≅ ΔBAD.

Step 3: Therefore, AC = BD (Corresponding parts of congruent triangles).

Thus, the diagonals of a rectangle are equal in length.

Converse of the theorem: If the diagonals of a parallelogram are equal, then it is a rectangle.

Question 22:

In a parallelogram ABCD, the bisectors of ∠A and ∠D meet at point O. Prove that ∠AOD = 90°.

Answer:

Given: Parallelogram ABCD where the bisectors of ∠A and ∠D meet at O.

Step 1: In a parallelogram, consecutive angles are supplementary, so ∠A + ∠D = 180°.

Step 2: The bisectors divide ∠A and ∠D into two equal angles each.
Let ∠OAD = ∠OAB = x (since AO is the bisector of ∠A)
Let ∠ADO = ∠ODC = y (since DO is the bisector of ∠D)

Step 3: From Step 1, 2x + 2y = 180° ⇒ x + y = 90°.

Step 4: In ΔAOD, the sum of angles is 180°.
∠OAD + ∠ADO + ∠AOD = 180°
x + y + ∠AOD = 180°
90° + ∠AOD = 180° (from Step 3)

Step 5: Therefore, ∠AOD = 180° - 90° = 90°.

Hence, it is proved that ∠AOD = 90°.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A farmer has a rhombus-shaped field with diagonals 16 m and 12 m. He wants to fence it.
Problem Interpretation: Calculate the perimeter of the field using properties of quadrilaterals.
Mathematical Modeling: Recall that diagonals of a rhombus bisect each other at 90°.

Answer:

Problem Interpretation: We need to find the side length of the rhombus using its diagonals and then calculate the perimeter.
Mathematical Modeling: The diagonals divide the rhombus into 4 right triangles. Each triangle has legs of 8 m and 6 m (half of diagonals).
Solution: Using Pythagoras theorem, side = √(8² + 6²) = 10 m. Perimeter = 4 × 10 = 40 m.

Question 2:

In a parallelogram ABCD, ∠A = (3x + 10)° and ∠D = (5x - 30)°.
Problem Interpretation: Find the measures of all angles using quadrilateral properties.
Mathematical Modeling: Recall that consecutive angles in a parallelogram are supplementary.

Answer:

Problem Interpretation: We know ∠A + ∠D = 180° as they are consecutive angles.
Mathematical Modeling: So, (3x + 10) + (5x - 30) = 180. Solving gives 8x - 20 = 180 → x = 25.
Solution: ∠A = 85°, ∠D = 95°. Opposite angles are equal, so ∠B = 95° and ∠C = 85°.

Question 3:

A farmer has a rectangular field with length 15 m and breadth 10 m. He wants to fence it diagonally to divide it into two triangular plots.
Problem Interpretation: Find the length of the diagonal fencing required.
Mathematical Modeling: Use the properties of rectangles and the Pythagoras theorem.

Answer:

Problem Interpretation: We need to find the diagonal of a rectangle.
Mathematical Modeling: In a rectangle, diagonals are equal and bisect each other. Using Pythagoras theorem:
Solution: Diagonal = √(length² + breadth²) = √(15² + 10²) = √(225 + 100) = √325 ≈ 18.03 m. Thus, the farmer needs ≈ 18.03 m of fencing.

Question 4:

In quadrilateral ABCD, ∠A = 80°, ∠B = 100°, ∠C = 70°.
Problem Interpretation: Find ∠D using the angle sum property of quadrilaterals.
Mathematical Modeling: Our textbook shows that the sum of all angles in a quadrilateral is 360°.

Answer:

Problem Interpretation: We need to find the missing angle in quadrilateral ABCD.
Mathematical Modeling: Sum of angles = ∠A + ∠B + ∠C + ∠D = 360°.
Solution: 80° + 100° + 70° + ∠D = 360° ⇒ 250° + ∠D = 360° ⇒ ∠D = 110°. Thus, ∠D = 110°.

Question 5:

A farmer has a rhombus-shaped field with diagonals 16 m and 12 m. He wants to fence it.
Problem Interpretation: Calculate the perimeter of the field using properties of a rhombus.
Mathematical Modeling: Recall that diagonals of a rhombus bisect each other at 90°.

Answer:

Problem Interpretation: We need to find the side length of the rhombus using its diagonals and then calculate the perimeter.
Mathematical Modeling: The diagonals (16 m and 12 m) divide the rhombus into 4 right triangles with legs 8 m and 6 m.
Solution: Using Pythagoras theorem, side = √(8² + 6²) = 10 m. Perimeter = 4 × 10 = 40 m.

Question 6:

In a parallelogram ABCD, ∠A = (3x + 10)° and ∠D = (5x - 30)°.
Problem Interpretation: Find the measures of all angles.
Mathematical Modeling: Recall that consecutive angles in a parallelogram are supplementary.

Answer:

Problem Interpretation: We need to find x using the property of consecutive angles and then determine all angles.
Mathematical Modeling: ∠A + ∠D = 180° ⇒ (3x + 10) + (5x - 30) = 180.
Solution: Solving, 8x - 20 = 180 ⇒ x = 25. Thus, ∠A = 85°, ∠D = 95°. Opposite angles are equal, so ∠B = 95° and ∠C = 85°.

Question 7:

A farmer has a rhombus-shaped field with diagonals 16 m and 12 m. He wants to fence it with barbed wire. Calculate the length of wire needed if he fences the field 3 times. (Use NCERT concepts of quadrilaterals).

Answer:

Problem Interpretation

We need to find the perimeter of the rhombus and multiply it by 3 for fencing.

Mathematical Modeling

Diagonals (d₁, d₂) = 16 m, 12 m
Side (s) = √((d₁/2)² + (d₂/2)²)

Solution

Side = √(8² + 6²) = 10 m. Perimeter = 4 × 10 = 40 m. Total wire = 3 × 40 = 120 m.

Question 8:

In a parallelogram ABCD, ∠A = (3x + 10)° and ∠D = (5x - 30)°. Find all angles using the properties studied in NCERT.

Answer:

Problem Interpretation

We know adjacent angles of a parallelogram are supplementary.

Mathematical Modeling

∠A + ∠D = 180°
3x + 10 + 5x - 30 = 180

Solution

Solving: 8x - 20 = 180 → x = 25. ∠A = 85°, ∠D = 95°. Opposite angles are equal, so ∠B = 95°, ∠C = 85°.

Question 9:

A farmer has a rhombus-shaped field with diagonals 16 m and 12 m. He wants to fence it with barbed wire.
Problem Interpretation: Calculate the perimeter of the field.
Mathematical Modeling: Use properties of rhombus.

Answer:

Problem Interpretation: We need to find the perimeter of the rhombus using its diagonals.
Mathematical Modeling: In a rhombus, diagonals bisect each other at 90°. Using Pythagoras theorem, side = √((8)² + (6)²) = 10 m.
Solution: Perimeter = 4 × side = 4 × 10 = 40 m. Our textbook shows similar problems in Chapter 8.

Question 10:

In a parallelogram ABCD, ∠A = (3x + 10)° and ∠D = (5x - 30)°.
Problem Interpretation: Find all angles.
Mathematical Modeling: Use properties of parallelogram.

Answer:

Problem Interpretation: We studied that consecutive angles in a parallelogram are supplementary.
Mathematical Modeling: ∠A + ∠D = 180° ⇒ (3x + 10) + (5x - 30) = 180 ⇒ x = 25.
Solution: ∠A = 85°, ∠D = 95°. Opposite angles are equal, so ∠B = 95° and ∠C = 85°. [Diagram: Parallelogram ABCD]

Question 11:

A farmer has a piece of land in the shape of a quadrilateral ABCD. He wants to divide it into two parts of equal area by building a fence from vertex A to a point E on side CD. The coordinates of the vertices are A(0, 0), B(4, 0), C(4, 3), and D(0, 4).

Task: Find the coordinates of point E such that the fence divides the land into two regions of equal area. Justify your answer with proper steps.

Answer:

To divide quadrilateral ABCD into two equal areas, we first calculate its total area using the shoelace formula.

Step 1: Calculate area of ABCD
Vertices in order: A(0,0), B(4,0), C(4,3), D(0,4).
Using shoelace formula:
Area = ½ |(0×0 + 4×3 + 4×4 + 0×0) - (0×4 + 0×3 + 3×0 + 4×0)|
= ½ |(0 + 12 + 16 + 0) - (0 + 0 + 0 + 0)|
= ½ × 28 = 14 sq. units.

Step 2: Find area to be enclosed by fence AE
Half of total area = 7 sq. units.

Step 3: Locate point E on CD
Let E divide CD in ratio k:1. Coordinates of E:
E = ((k×0 + 1×4)/(k+1), (k×4 + 1×3)/(k+1)) = (4/(k+1), (4k+3)/(k+1)).

Area of triangle ADE must be 7 sq. units (since ΔABD + ΔADE = 7).
Vertices of ΔADE: A(0,0), D(0,4), E(4/(k+1), (4k+3)/(k+1)).
Area = ½ |(0×4 + 0×y_E + 4/(k+1)×0) - (0×0 + 4×4/(k+1) + y_E×0)|
= ½ |0 - 16/(k+1)| = 8/(k+1).
Set 8/(k+1) = 7 → k+1 = 8/7 → k = 1/7.

Final Coordinates of E:
E = (4/(1/7 + 1), (4×(1/7)+3)/(1/7 + 1)) = (4/(8/7), (4/7 + 21/7)/(8/7))
= (3.5, 25/8) = (3.5, 3.125).

Verification: Recalculate area with E(3.5, 3.125) to confirm 7 sq. units.

Question 12:

A kite-shaped quadrilateral PQRS has diagonals PR and QS intersecting at O. Given that PQ = QR = 5 cm, PS = RS = 6 cm, and PR = 8 cm.

Task: Prove that the diagonals of kite PQRS are perpendicular to each other and find the length of QS. Show all steps.

Answer:

Proof of Perpendicular Diagonals:
In a kite, one diagonal bisects the other at 90°. Here, diagonal QS bisects PR at O, making PO = OR = 4 cm.

Step 1: Verify perpendicularity using Pythagoras theorem
For ΔPOQ and ΔROQ:
PQ² = PO² + OQ² → 5² = 4² + OQ² → OQ = 3 cm.
QR² = RO² + OQ² → 5² = 4² + OQ² → OQ = 3 cm (consistent).

Similarly, for ΔPOS and ΔROS:
PS² = PO² + OS² → 6² = 4² + OS² → OS = √20 ≈ 4.47 cm.
RS² = RO² + OS² → 6² = 4² + OS² → OS ≈ 4.47 cm (consistent).

Since OQ and OS lie on QS, and ∠POQ = ∠ROQ = 90°, diagonals are perpendicular.

Step 2: Calculate length of QS
QS = OQ + OS = 3 + √20 ≈ 3 + 4.47 = 7.47 cm.

Note: The exact value of QS is 3 + 2√5 cm. Kite properties ensure symmetry and perpendicular diagonals.

Question 13:

Rahul drew a quadrilateral ABCD where AB = 5 cm, BC = 7 cm, CD = 6 cm, and AD = 4 cm. He also measured the diagonal AC = 8 cm.

Using the properties of quadrilaterals, determine whether ABCD is a parallelogram or not. Justify your answer with proper reasoning.

Answer:

To check if quadrilateral ABCD is a parallelogram, we need to verify if both pairs of opposite sides are equal.

Given:
AB = 5 cm, CD = 6 cm
BC = 7 cm, AD = 4 cm

Since AB ≠ CD and BC ≠ AD, the opposite sides are not equal.

Conclusion: ABCD is not a parallelogram because a parallelogram must have both pairs of opposite sides equal.

Additional Note: Even though diagonal AC is given, it does not affect the condition for a parallelogram unless the diagonals bisect each other, which is not applicable here.

Question 14:

In a quadrilateral PQRS, the angles are in the ratio 3:5:7:9.

(a) Find the measure of each angle.
(b) Identify the type of quadrilateral based on the angle measures.

Answer:

(a) Let the angles be 3x, 5x, 7x, and 9x.

Sum of angles in a quadrilateral = 360°
3x + 5x + 7x + 9x = 360°
24x = 360°
x = 15°

Thus, the angles are:
3x = 45°
5x = 75°
7x = 105°
9x = 135°

(b) Since all angles are different and none are 90°, the quadrilateral is a general quadrilateral (no special properties).

Additional Note: If any angle were 90°, it could indicate a rectangle or square, but here it is an irregular quadrilateral.

Question 15:

Rahul drew a quadrilateral ABCD where AB = 5 cm, BC = 7 cm, CD = 5 cm, and DA = 7 cm. He claims it is a parallelogram. Verify his claim using properties of quadrilaterals.

Answer:

To verify if quadrilateral ABCD is a parallelogram, we check the following properties:

1. Opposite sides are equal: Here, AB = CD = 5 cm and BC = DA = 7 cm.
2. Opposite sides are parallel: Since Rahul did not provide angle measures, we assume he measured them and confirmed parallelism.

Since both conditions are satisfied, ABCD is indeed a parallelogram. Key takeaway: A quadrilateral with both pairs of opposite sides equal is a parallelogram.

Question 16:

In quadrilateral PQRS, the diagonals PR and QS intersect at O. If PO = OR and QO = OS, identify the type of quadrilateral and justify your answer.

Answer:

Given that the diagonals PR and QS bisect each other (PO = OR and QO = OS), we can conclude that PQRS is a parallelogram.

Reason: A quadrilateral whose diagonals bisect each other is a parallelogram (as per NCERT).

Additional insight: If the diagonals are also perpendicular, the quadrilateral becomes a rhombus. However, since no angle information is given, we stick to the basic property.

Question 17:

Rahul drew a quadrilateral ABCD where AB = 5 cm, BC = 6 cm, CD = 7 cm, and DA = 8 cm. He also measured the diagonal AC as 9 cm. Using the properties of quadrilaterals, determine whether ABCD is a parallelogram or not. Justify your answer with proper reasoning.

Answer:

To check if quadrilateral ABCD is a parallelogram, we use the property that in a parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of all four sides.

Given:
AB = 5 cm, BC = 6 cm, CD = 7 cm, DA = 8 cm, and AC = 9 cm.

First, let's assume BD is the other diagonal. According to the property:
AC² + BD² = AB² + BC² + CD² + DA²

Substitute the given values:
9² + BD² = 5² + 6² + 7² + 8²
81 + BD² = 25 + 36 + 49 + 64
81 + BD² = 174
BD² = 174 - 81 = 93
BD = √93 ≈ 9.64 cm

Now, for a parallelogram, the diagonals must bisect each other. However, without knowing the exact point of intersection or the other diagonal's properties, we cannot confirm this solely based on side lengths and one diagonal. Thus, ABCD does not necessarily satisfy all conditions of a parallelogram.

Conclusion: Since the diagonals do not necessarily bisect each other based on the given data, quadrilateral ABCD is not confirmed to be a parallelogram.

Question 18:

In a quadrilateral PQRS, the angles are in the ratio 3:5:7:9. Find the measure of each angle and identify the type of quadrilateral based on the angle measures. Justify your answer.

Answer:

Given the ratio of angles in quadrilateral PQRS as 3:5:7:9, let the angles be 3x, 5x, 7x, and 9x.

We know that the sum of the interior angles of a quadrilateral is 360°.
So, 3x + 5x + 7x + 9x = 360°
24x = 360°
x = 360° / 24 = 15°

Now, calculate each angle:
∠P = 3x = 3 × 15° = 45°
∠Q = 5x = 5 × 15° = 75°
∠R = 7x = 7 × 15° = 105°
∠S = 9x = 9 × 15° = 135°

To identify the type of quadrilateral:

Since all angles are different, it cannot be a rectangle or square (which require equal angles).
It does not satisfy the properties of a parallelogram or rhombus (opposite angles must be equal).
Thus, PQRS is a general quadrilateral with no specific special properties based on the given angle measures.

Conclusion: The angles are 45°, 75°, 105°, and 135°, and the quadrilateral is a general quadrilateral.

Question 19:

A farmer has a quadrilateral-shaped field ABCD with AB = 5 m, BC = 12 m, CD = 14 m, DA = 15 m, and diagonal AC = 13 m. He wants to divide the field into two triangular plots by fencing along the diagonal AC.

(i) Prove that ΔABC is a right-angled triangle.
(ii) Find the area of the entire field ABCD.

Answer:

(i) Proving ΔABC is right-angled:
Using the Pythagoras theorem, check if AB² + BC² = AC²:
AB² = 5² = 25 m²
BC² = 12² = 144 m²
AC² = 13² = 169 m²
Now, AB² + BC² = 25 + 144 = 169 = AC²
Since the condition is satisfied, ΔABC is right-angled at B.

(ii) Area of field ABCD:
First, find the area of ΔABC (right-angled at B):
Area = (1/2) × AB × BC = (1/2) × 5 × 12 = 30 m²
Next, find the area of ΔADC using Heron's formula:
Semi-perimeter (s) = (13 + 14 + 15)/2 = 21 m
Area = √[s(s-a)(s-b)(s-c)] = √[21 × (21-13) × (21-14) × (21-15)]
= √[21 × 8 × 7 × 6] = √7056 = 84 m²
Total area of ABCD = Area of ΔABC + Area of ΔADC = 30 + 84 = 114 m².

Question 20:

In a quadrilateral PQRS, the diagonals PR and QS intersect at O such that PO = RO and QO = SO.

(i) What special type of quadrilateral is PQRS? Justify your answer.
(ii) If ∠QPR = 40° and ∠PRQ = 70°, find ∠QSR.

Answer:

(i) Type of quadrilateral:
PQRS is a parallelogram because its diagonals bisect each other (PO = RO and QO = SO).
In a parallelogram, diagonals divide each other into two equal parts, which is the given condition.

(ii) Finding ∠QSR:
First, find ∠PQR in ΔPQR:
∠PQR = 180° - (∠QPR + ∠PRQ) = 180° - (40° + 70°) = 70°
Since PQRS is a parallelogram, opposite angles are equal. Thus, ∠PSR = ∠PQR = 70°.
Now, in ΔQSR, ∠QSR = 180° - (∠SQR + ∠SRQ).
But ∠SQR = ∠PRQ = 70° (alternate angles as PQ || SR) and ∠SRQ = ∠QPR = 40° (alternate angles as PS || QR).
Thus, ∠QSR = 180° - (70° + 40°) = 70°.

Quadrilaterals – CBSE NCERT Study Resources

Overview

Types of Quadrilaterals

Properties of Quadrilaterals

Theorems Related to Quadrilaterals

Mid-Point Theorem

Summary

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)