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Areas of Parallelograms and Triangles – CBSE NCERT Study Resources

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9th

9th - Mathematics

Areas of Parallelograms and Triangles

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Overview of the Chapter: Areas of Parallelograms and Triangles

This chapter introduces students to the fundamental concepts of calculating the areas of parallelograms and triangles. It builds upon prior knowledge of basic geometric shapes and their properties, focusing on the relationship between these two shapes in terms of area. The chapter emphasizes understanding through theorems, proofs, and practical applications.

Parallelogram: A quadrilateral with both pairs of opposite sides parallel.

Triangle: A three-sided polygon formed by three non-collinear points connected by line segments.

Key Concepts

  • Area of a parallelogram is the product of its base and corresponding height.
  • Area of a triangle is half the product of its base and corresponding height.
  • Parallelograms on the same base and between the same parallels have equal areas.
  • Triangles on the same base and between the same parallels have equal areas.

Theorems and Proofs

Theorem 1: Parallelograms on the Same Base and Between the Same Parallels

If two parallelograms lie on the same base and between the same parallels, their areas are equal.

Theorem 2: Triangles on the Same Base and Between the Same Parallels

If two triangles lie on the same base and between the same parallels, their areas are equal.

Formulas

Shape Area Formula
Parallelogram Base × Height
Triangle ½ × Base × Height

Applications

The concepts learned in this chapter are applied in various real-world scenarios, such as land measurement, architecture, and design. Understanding these principles helps in solving practical problems involving area calculations.

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 area of a parallelogram with base 8 cm and height 5 cm?
Answer:
40 cm²
Question 2:
If two parallelograms are on the same base and between the same parallels, their areas are _______.
Answer:
equal
Question 3:
Find the area of a triangle with base 10 cm and height 6 cm.
Answer:
30 cm²
Question 4:
A triangle and a parallelogram have the same base and area. What is the ratio of their heights?
Answer:
2:1
Question 5:
What is the area of a parallelogram if its base is 12 cm and height is half the base?
Answer:
72 cm²
Question 6:
If the area of a triangle is 24 cm² and its base is 8 cm, find its height.
Answer:
6 cm
Question 7:
Two triangles with the same base and between the same parallels have _______ areas.
Answer:
equal
Question 8:
A parallelogram has an area of 50 cm². If its height is 5 cm, find its base.
Answer:
10 cm
Question 9:
What is the area of a right-angled triangle with legs 3 cm and 4 cm?
Answer:
6 cm²
Question 10:
The median of a triangle divides it into two triangles of _______ area.
Answer:
equal
Question 11:
If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is _______ of the parallelogram.
Answer:
half
Question 12:
Find the height of a parallelogram with area 36 cm² and base 9 cm.
Answer:
4 cm
Question 13:
What is the formula for the area of a parallelogram?
Answer:

The area of a parallelogram is given by the formula: Area = base × height.
Here, the base is any one side, and the height is the perpendicular distance from the base to the opposite side.

Question 14:
If the base of a parallelogram is 8 cm and its height is 5 cm, what is its area?
Answer:

Using the formula for the area of a parallelogram:
Area = base × height
= 8 cm × 5 cm
= 40 cm².

Question 15:
How is the area of a triangle related to the area of a parallelogram with the same base and height?
Answer:

The area of a triangle is half the area of a parallelogram with the same base and height.
Mathematically, Area of Triangle = ½ × base × height.

Question 16:
A triangle and a parallelogram are on the same base and between the same parallels. If the area of the parallelogram is 24 cm², what is the area of the triangle?
Answer:

Since the triangle and parallelogram share the same base and height:
Area of Triangle = ½ × Area of Parallelogram
= ½ × 24 cm²
= 12 cm².

Question 17:
What is the area of a triangle with base 10 cm and height 6 cm?
Answer:

Using the formula for the area of a triangle:
Area = ½ × base × height
= ½ × 10 cm × 6 cm
= 30 cm².

Question 18:
If two parallelograms have the same base and lie between the same parallels, what can you say about their areas?
Answer:

Two parallelograms with the same base and lying between the same parallels have equal areas.
This is because their heights are the same (distance between the parallels).

Question 19:
What is the height of a parallelogram if its area is 56 cm² and its base is 7 cm?
Answer:

Using the formula for the area of a parallelogram:
Area = base × height
56 cm² = 7 cm × height
Height = 56 cm² ÷ 7 cm
= 8 cm.

Question 20:
A triangle has an area of 18 cm² and a base of 6 cm. What is its height?
Answer:

Using the formula for the area of a triangle:
Area = ½ × base × height
18 cm² = ½ × 6 cm × height
Height = (18 cm² × 2) ÷ 6 cm
= 6 cm.

Question 21:
Why do all triangles on the same base and between the same parallels have equal areas?
Answer:

All triangles on the same base and between the same parallels have equal areas because:

  • They share the same base.
  • Their heights are equal (distance between the parallels).

Since Area = ½ × base × height, their areas must be equal.

Question 22:
If the area of a triangle is 20 cm² and its height is 5 cm, what is the length of its base?
Answer:

Using the formula for the area of a triangle:
Area = ½ × base × height
20 cm² = ½ × base × 5 cm
Base = (20 cm² × 2) ÷ 5 cm
= 8 cm.

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:
State the formula for the area of a parallelogram and explain the terms involved.
Answer:

The area of a parallelogram is given by the formula: Area = base × height.
Here, base refers to the length of any one side of the parallelogram.
Height is the perpendicular distance between the base and its opposite side.

Question 2:
If the base of a parallelogram is 12 cm and its height is 8 cm, calculate its area.
Answer:

Using the formula for the area of a parallelogram: Area = base × height.
Given: base = 12 cm, height = 8 cm.
Area = 12 cm × 8 cm = 96 cm².

Question 3:
A triangle and a parallelogram have the same base and area. If the height of the parallelogram is 6 cm, find the height of the triangle.
Answer:

Given that the areas are equal and the base is the same:
Area of parallelogram = Area of triangle
base × heightparallelogram = (base × heighttriangle) / 2.
Simplifying: heighttriangle = 2 × heightparallelogram.
Thus, heighttriangle = 2 × 6 cm = 12 cm.

Question 4:
Explain why the area of a parallelogram is the same as that of a rectangle with the same base and height.
Answer:

A parallelogram can be rearranged into a rectangle by cutting and pasting a triangular piece from one side to the other.
Since the base and height remain unchanged, the area (base × height) is the same for both shapes.

Question 5:
Calculate the area of a triangle with base 10 cm and height 5 cm.
Answer:

Using the formula for the area of a triangle: Area = (base × height) / 2.
Given: base = 10 cm, height = 5 cm.
Area = (10 cm × 5 cm) / 2 = 25 cm².

Question 6:
A triangle has an area of 24 cm² and a base of 6 cm. Find its height.
Answer:

Using the formula for the area of a triangle: Area = (base × height) / 2.
Given: Area = 24 cm², base = 6 cm.
Rearranging: height = (2 × Area) / base = (2 × 24 cm²) / 6 cm = 8 cm.

Question 7:
What is the area of a parallelogram if its base is 15 cm and height is one-third of the base?
Answer:

Given: base = 15 cm, height = (1/3) × base = (1/3) × 15 cm = 5 cm.
Using the formula for the area of a parallelogram: Area = base × height.
Area = 15 cm × 5 cm = 75 cm².

Question 8:
Prove that the area of a triangle formed by joining the midpoints of the sides of another triangle is one-fourth of the original triangle's area.
Answer:

When the midpoints of a triangle's sides are joined, the resulting smaller triangle is similar to the original and has sides half as long.
Since area scales with the square of the side ratio, the smaller triangle's area is (1/2)² = 1/4 of the original triangle's area.

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:
Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Answer:

Consider a parallelogram ABCD with diagonal AC.
In triangles ABC and ADC:
1. AB = CD (Opposite sides of a parallelogram are equal)
2. AD = BC (Opposite sides of a parallelogram are equal)
3. AC = AC (Common side)
By SSS congruency, △ABC ≅ △ADC.
Thus, the diagonal divides the parallelogram into two congruent triangles.

Question 2:
If the base of a triangle is 12 cm and its corresponding height is 8 cm, find its area. Also, state the formula used.
Answer:

The area of a triangle is given by:
Area = (1/2) × base × height
Given: Base = 12 cm, Height = 8 cm.
Substituting the values:
Area = (1/2) × 12 × 8
Area = 6 × 8
Area = 48 cm².
Thus, the area of the triangle is 48 cm².

Question 3:
Explain why the area of a parallelogram is equal to the product of its base and height.
Answer:

A parallelogram can be divided into a rectangle and two congruent triangles by drawing a perpendicular from one vertex to the base.
1. The area of the rectangle is base × height.
2. The two triangles can be rearranged to form another rectangle of the same area.
Thus, the total area of the parallelogram is the sum of the areas of the rectangle and the rearranged triangles, which simplifies to base × height.
This proves that the area of a parallelogram is equal to the product of its base and height.

Question 4:
A triangle and a parallelogram have the same base and lie between the same parallels. Compare their areas.
Answer:

When a triangle and a parallelogram share the same base and lie between the same parallels:
1. The height of both shapes is equal because they lie between the same parallels.
2. The area of the parallelogram is base × height.
3. The area of the triangle is (1/2) × base × height.
Thus, the area of the parallelogram is twice the area of the triangle under these conditions.

Question 5:
Find the height of a parallelogram whose area is 54 cm² and base is 9 cm.
Answer:

The area of a parallelogram is given by:
Area = base × height
Given: Area = 54 cm², Base = 9 cm.
Substituting the values:
54 = 9 × height
height = 54 / 9
height = 6 cm.
Thus, the height of the parallelogram is 6 cm.

Question 6:
Explain why the area of a parallelogram is equal to the product of its base and corresponding height.
Answer:

A parallelogram can be divided into a triangle and a trapezium by drawing a diagonal.
However, a simpler proof involves rearranging it into a rectangle:
1. Cut a right triangle from one side of the parallelogram.
2. Move it to the opposite side to form a rectangle.
The base and height of the parallelogram become the length and width of the rectangle.
Since the area of the rectangle is base × height, the parallelogram's area is also base × height.

Question 7:
Two triangles have the same base and equal areas. Will their heights also be equal? Justify your answer.
Answer:

Yes, their heights must be equal.
The area of a triangle is given by:
Area = (1/2) × base × height.
If two triangles have the same base and equal areas, then:
(1/2) × base × height₁ = (1/2) × base × height₂.
Canceling the common terms (1/2 and base):
height₁ = height₂.
Thus, their heights must be equal.

Question 8:
A farmer has a triangular field with sides 50 m, 65 m, and 65 m. Find its area using Heron's formula.
Answer:

First, calculate the semi-perimeter (s):
s = (a + b + c)/2 = (50 + 65 + 65)/2 = 180/2 = 90 m.
Now, apply Heron's formula:
Area = √[s(s - a)(s - b)(s - c)]
Area = √[90(90 - 50)(90 - 65)(90 - 65)]
Area = √[90 × 40 × 25 × 25]
Area = √(2,250,000) = 1500 m².
Thus, the area of the field is 1500 m².

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 a diagonal of a parallelogram divides it into two congruent triangles. Use the properties of parallelograms and triangle congruence criteria.
Answer:
Introduction

We studied that a parallelogram has two pairs of parallel sides and its diagonals bisect each other.


Argument 1

Consider parallelogram ABCD with diagonal AC. In triangles ABC and ADC, AB = CD (opposite sides of parallelogram are equal).


Argument 2

AD = BC (opposite sides) and AC is common. By SSS congruency, ΔABC ≅ ΔADC. Thus, the diagonal divides the parallelogram into two congruent triangles.


Conclusion

This proves that diagonals of parallelograms create congruent triangles, as shown in NCERT Example 9.1.

Question 2:
A farmer has a triangular field with base 24m and height 15m. Find its area. If he divides it into two equal parts by a fence parallel to the base, find the height where the fence is built.
Answer:
Introduction

We know the area of a triangle is (1/2)×base×height. Our textbook shows how to divide triangles into smaller similar triangles.


Argument 1

Total area = (1/2)×24×15 = 180m². When divided equally, each part has area 90m².


Argument 2

The fence creates a smaller similar triangle. Area ratio = 1:2, so height ratio = 1:√2. Thus, height of fence = 15/√2 ≈ 10.61m.


Conclusion

The fence is built at approximately 10.61m height, demonstrating real-life application of area division.

Question 3:
Show that the area of a triangle is half the area of a parallelogram on the same base and between the same parallels. Use a diagram and step-by-step reasoning.
Answer:
Introduction

We studied that parallelograms and triangles on same base between parallels have related areas, as in NCERT Theorem 9.1.


Argument 1

[Diagram: Triangle ABC and parallelogram ABCD sharing base BC, with same height h]. Area of parallelogram = base × height = BC × h.


Argument 2

Area of triangle ABC = (1/2)×base×height = (1/2)×BC×h. Thus, triangle's area is half of parallelogram's area.


Conclusion

This relationship is fundamental in geometry and used in many construction calculations.

Question 4:
A farmer has a triangular field with base 24 m and height 15 m. Find its area. If he divides it into two equal parts by a fence parallel to the base, what is the area of each smaller triangle?
Answer:
Introduction

Our textbook shows that the area of a triangle is given by ½ × base × height.


Argument 1
  • Area of the field = ½ × 24 m × 15 m = 180 m².
  • When divided equally, each part has area = 180 m² ÷ 2 = 90 m².

Argument 2
  • The fence divides the height proportionally, so the smaller triangle has height = 15 m ÷ √2 ≈ 10.6 m.
  • But area remains 90 m² as per equal division.

Conclusion

Each smaller triangle has an area of 90 m².

Question 5:
Explain why the area of a triangle is half the area of a parallelogram on the same base and between the same parallels. Derive this using a diagram.
Answer:
Introduction

We know that both shapes share the same base and height when between parallels.


Argument 1
  • Draw parallelogram ABCD and triangle ABE on base AB.
  • Height for both is the perpendicular distance between parallels.

Argument 2
  • Area of parallelogram = base × height.
  • Area of triangle = ½ × base × height, hence half.

Conclusion

Thus, the triangle’s area is half the parallelogram’s area under the given conditions.

Question 6:
A farmer has a triangular field with base 24 m and height 10 m. He divides it into two parts by drawing a line parallel to the base. If the area of the smaller triangle is 30 m², find the height at which the line is drawn.
Answer:
Introduction

Our textbook explains how areas of triangles with the same height are proportional to their bases. Here, we apply this concept to solve a real-life problem.


Argument 1
  • Total area of the field = (1/2) × 24 × 10 = 120 m².
  • Area of the smaller triangle = 30 m², so the remaining area = 90 m².

Argument 2
  • Since the triangles are similar, (height of smaller triangle / total height)² = (30/120).
  • Solving, height of smaller triangle = 5 m.

Conclusion

The line is drawn at a height of 5 m from the base, dividing the field as required.

Question 7:
Derive the formula for the area of a parallelogram using the area of a triangle. Explain each step with a diagram.
Answer:
Introduction

We know that a parallelogram can be split into two triangles by its diagonal. Our textbook uses this to derive its area formula.


Argument 1
  • Let ABCD be a parallelogram with diagonal AC.
  • Area of ΔABC = (1/2) × base × height = (1/2) × AB × h.

Argument 2
  • Since ΔABC ≅ ΔADC, total area of parallelogram = 2 × (1/2) × AB × h = AB × h.
  • [Diagram: Parallelogram ABCD with height h marked perpendicular to AB.]

Conclusion

Thus, the area of a parallelogram is base × height, derived using triangles.

Question 8:
A farmer has a triangular field with base 24 m and height 15 m. Adjacent to it is a parallelogram-shaped field with the same base and height. Compare their areas and justify your answer using the formula.
Answer:
Introduction

We know the area of a triangle is half the product of base and height, while a parallelogram's area is base × height. Our textbook shows examples of such comparisons.


Argument 1
  • Area of triangle = (1/2) × 24 m × 15 m = 180 m².
  • Area of parallelogram = 24 m × 15 m = 360 m².

Argument 2
  • The parallelogram's area is double the triangle's area.
  • This matches the formula derivation we learned.

Conclusion

Hence, the parallelogram field has twice the area of the triangular field.

Question 9:
Explain why two parallelograms on the same base and between the same parallels have equal areas. Support your answer with a diagram and logical steps.
Answer:
Introduction

We studied that parallelograms between the same parallels have equal heights. Our textbook shows this using the area formula.


Argument 1
  • Let ABCD and ABEF be two parallelograms on base AB and between parallels AB and DE.
  • Both have equal heights as they lie between the same parallels.

Argument 2
  • Area of ABCD = base AB × height.
  • Area of ABEF = base AB × same height.

Conclusion

Thus, their areas are equal. [Diagram: Two parallelograms sharing base AB between parallel lines.]

Question 10:
Prove that parallelograms on the same base and between the same parallels are equal in area. Use a diagram and step-wise reasoning.
Answer:
Introduction

We studied that two parallelograms sharing the same base and lying between the same parallels must have equal areas.


Argument 1

Consider parallelograms ABCD and ABEF on base AB between parallels AB and DE. [Diagram: Two parallelograms sharing AB as base]


Argument 2

Since heights are equal (same parallels) and base AB is common, area = base × height. Thus, ar(ABCD) = ar(ABEF). Our textbook shows this in Theorem 9.1.


Conclusion

Hence, parallelograms on same base and between same parallels are equal in area.

Question 11:
A farmer has a triangular field ABC with base BC = 24 m and height 10 m. He divides it into two parts by a line parallel to BC. If the area of the smaller triangle is 30 m², find the height of the smaller triangle.
Answer:
Introduction

We can solve this using the property that triangles on the same base and between same parallels have proportional areas.


Argument 1

Total area of ΔABC = ½ × 24 × 10 = 120 m². Smaller ΔADE has area 30 m².


Argument 2

Since ΔADE ~ ΔABC, (height of ΔADE / height of ΔABC)² = 30/120. Solving, height of ΔADE = 5 m.


Conclusion

Thus, the height of the smaller triangle is 5 m, as derived from area ratios.

Question 12:
Explain how the area of a triangle is half that of a parallelogram with the same base and height. Give a real-life example.
Answer:
Introduction

Our textbook shows that a triangle’s area is half of a parallelogram’s when both share the same base and height.


Argument 1

If we draw a diagonal in parallelogram ABCD, it divides into two congruent triangles (ΔABC and ΔADC). Thus, ar(ΔABC) = ½ ar(ABCD).


Argument 2

Real-life example: A sandwich cut diagonally forms two triangular pieces, each half the area of the original rectangular slice.


Conclusion

This proves the relationship between triangle and parallelogram areas.

Question 13:
Prove that a diagonal of a parallelogram divides it into two congruent triangles. Support your answer with a diagram and step-by-step reasoning.
Answer:

To prove that a diagonal of a parallelogram divides it into two congruent triangles, let us consider a parallelogram ABCD with diagonal AC.


Step 1: Draw the Diagram
Draw parallelogram ABCD with points A, B, C, D in order. Join diagonal AC.


Step 2: Identify Properties
In a parallelogram:

  • Opposite sides are equal (AB = CD and AD = BC).
  • Opposite sides are parallel (AB || CD and AD || BC).


Step 3: Prove Triangle Congruence
Consider triangles ABC and ADC:

  • AC is common to both triangles.
  • AB = CD (opposite sides of parallelogram).
  • AD = BC (opposite sides of parallelogram).

By the SSS (Side-Side-Side) congruence rule, △ABC ≅ △ADC.


Conclusion:
The diagonal AC divides the parallelogram into two congruent triangles. This property is useful in calculating areas and solving geometric problems.

Question 14:
Prove that a diagonal of a parallelogram divides it into two congruent triangles. Support your answer with a labeled diagram and step-by-step reasoning.
Answer:

To prove that a diagonal of a parallelogram divides it into two congruent triangles, let us consider a parallelogram ABCD with diagonal AC.


Step 1: Draw the Diagram
Sketch parallelogram ABCD with points A, B, C, D in order. Draw diagonal AC connecting vertices A and C.


Step 2: Identify Properties of a Parallelogram
In a parallelogram:

  • Opposite sides are equal and parallel (AB = CD and AD = BC).
  • Opposite angles are equal (∠A = ∠C and ∠B = ∠D).


Step 3: Prove Triangle Congruence
Consider triangles ABC and ADC:

  • AB = CD (Opposite sides of a parallelogram are equal).
  • AD = BC (Opposite sides of a parallelogram are equal).
  • AC = AC (Common side).

By the SSS (Side-Side-Side) congruence rule, △ABC ≅ △ADC.


Conclusion: Since all corresponding sides and angles are equal, the diagonal AC divides the parallelogram into two congruent triangles.


Value-Added Insight: This property is useful in calculating the area of parallelograms and triangles, as congruent triangles have equal areas. Thus, the area of the parallelogram is twice the area of either triangle formed by the diagonal.

Question 15:
Prove that the area of a triangle is half the product of its base and corresponding height. Use a diagram to illustrate your proof and explain each step clearly.
Answer:

To prove that the area of a triangle is half the product of its base and corresponding height, let's follow these steps:


Step 1: Draw the Diagram


Consider a triangle ABC with base BC and height AD perpendicular to BC.


Step 2: Construct a Parallelogram


Draw a line CE parallel to AB and a line BE parallel to AC, forming a parallelogram ABEC.


Step 3: Relate the Areas


The area of parallelogram ABEC is given by: base × height = BC × AD.


Step 4: Compare Triangle and Parallelogram


Since the parallelogram is made up of two congruent triangles (ABC and BEC), the area of ABC is half the area of the parallelogram.


Step 5: Derive the Formula


Thus, the area of triangle ABC = ½ × base × height = ½ × BC × AD.


Conclusion: This proves that the area of a triangle is indeed half the product of its base and corresponding height.

Question 16:
Prove that a diagonal of a parallelogram divides it into two congruent triangles. Support your answer with a labeled diagram and step-by-step reasoning.
Answer:

To prove that a diagonal of a parallelogram divides it into two congruent triangles, let us consider a parallelogram ABCD with diagonal AC.


Step 1: Draw the Diagram
Sketch parallelogram ABCD with points A, B, C, D in order. Draw diagonal AC connecting A to C.


Step 2: Identify the Triangles
The diagonal AC divides the parallelogram into two triangles: △ABC and △ADC.


Step 3: Prove Congruency Using Properties
In a parallelogram, opposite sides are equal and parallel. Thus:

  • AB = CD (Opposite sides of parallelogram)
  • AD = BC (Opposite sides of parallelogram)
  • AC = AC (Common side)

By the SSS (Side-Side-Side) congruency rule, △ABC ≅ △ADC.


Conclusion:
Since all corresponding sides are equal, the two triangles are congruent. Hence, the diagonal divides the parallelogram into two congruent triangles.


Value-Added Insight:
This property is useful in calculating the area of parallelograms and triangles, as congruent triangles have equal areas.

Question 17:
Prove that a diagonal of a parallelogram divides it into two congruent triangles. Also, find the area of each triangle if the base and height of the parallelogram are 12 cm and 8 cm respectively.
Answer:

To prove that a diagonal divides a parallelogram into two congruent triangles, consider parallelogram ABCD with diagonal AC.

Proof:
1. In triangles ABC and ADC:
- AB = CD (opposite sides of a parallelogram are equal).
- BC = AD (opposite sides of a parallelogram are equal).
- AC is common to both triangles.
2. By the SSS (Side-Side-Side) congruence rule, △ABC ≅ △ADC.

Area Calculation:
1. Area of parallelogram = base × height = 12 cm × 8 cm = 96 cm².
2. Since the diagonal divides the parallelogram into two equal areas,
Area of each triangle = 96 cm² ÷ 2 = 48 cm².

Question 18:
A farmer has a triangular field with sides 50 m, 78 m, and 112 m. He wants to divide it into two parts of equal area by building a fence parallel to the shortest side. Calculate the length of the fence required.
Answer:

Step 1: Find the area of the original triangle using Heron's formula.
1. Semi-perimeter, s = (50 + 78 + 112)/2 = 120 m.
2. Area = √[s(s-a)(s-b)(s-c)] = √[120(120-50)(120-78)(120-112)]
= √[120 × 70 × 42 × 8] = √2822400 = 1680 m².

Step 2: Divide the area into two equal parts.
1. Area of smaller triangle = 1680/2 = 840 m².
2. The smaller triangle is similar to the original triangle (since the fence is parallel).
So, (Area ratio) = (Side ratio)² ⇒ (840/1680) = (x/50)².
3. Solving, x² = 50² × 0.5 ⇒ x = 50 × √0.5 ≈ 35.36 m.

Thus, the length of the fence required is approximately 35.36 meters.

Question 19:
Prove that a parallelogram and a rectangle on the same base and between the same parallels have equal areas. Support your answer with a diagram.
Answer:

To prove that a parallelogram and a rectangle on the same base and between the same parallels have equal areas, let us consider the following:


Given: A parallelogram ABCD and a rectangle ABEF lying on the same base AB and between the same parallels AB and CF.


Proof:


1. Both the parallelogram and the rectangle have the same base AB.


2. The height of the parallelogram is the perpendicular distance between AB and DC, which is equal to the height of the rectangle (distance between AB and EF).


3. Area of parallelogram = base × height = AB × h.


4. Area of rectangle = base × height = AB × h.


5. Hence, both have equal areas.


Diagram: (Draw parallelogram ABCD and rectangle ABEF sharing base AB with heights equal to h.)


Conclusion: Since both shapes have the same base and height, their areas are equal.

Question 20:
A farmer has a triangular field with sides 50 m, 65 m, and 65 m. He wants to divide it into two parts of equal area by building a fence parallel to the shortest side. Find the length of the fence required.
Answer:

To find the length of the fence parallel to the shortest side (50 m) that divides the triangular field into two equal areas, follow these steps:


Step 1: Calculate the area of the original triangle using Heron's formula.
1. Semi-perimeter, s = (50 + 65 + 65)/2 = 90 m.
2. Area = √[s(s - a)(s - b)(s - c)] = √[90(90 - 50)(90 - 65)(90 - 65)] = √[90 × 40 × 25 × 25] = 1500 m².


Step 2: Divide the area into two equal parts (750 m² each).
1. The fence divides the triangle into a smaller triangle and a trapezium, each with area 750 m².
2. The smaller triangle is similar to the original triangle.


Step 3: Use the property of similar triangles.
1. Area ratio = (Area of smaller triangle)/(Area of original triangle) = 750/1500 = 1/2.
2. Since area ratio = (length ratio)², length ratio = √(1/2) = 1/√2.
3. Therefore, the length of the fence = 50 × (1/√2) ≈ 35.36 m.


Conclusion: The required length of the fence is approximately 35.36 meters.

Question 21:
A farmer has a triangular field with sides 50 m, 65 m, and 65 m. He wants to divide it into two parts of equal area by building a fence parallel to the shortest side. Find the length of the fence required.
Answer:

To find the length of the fence parallel to the shortest side (50 m) that divides the triangular field into two parts of equal area, follow these steps:


Step 1: Verify the type of triangle
The sides 50 m, 65 m, and 65 m form an isosceles triangle with base 50 m and equal sides 65 m.


Step 2: Calculate the area of the original triangle
Using Heron's formula:
1. Semi-perimeter, s = (50 + 65 + 65)/2 = 90 m.
2. Area = √[s(s-a)(s-b)(s-c)] = √[90 × 40 × 25 × 25] = 1500 m².


Step 3: Divide the area equally
The fence must divide the area into two parts of 750 m² each.


Step 4: Use the Area Ratio Property
Since the fence is parallel to the base (50 m), the smaller triangle formed will be similar to the original triangle.
Area ratio = (750/1500) = 1/2.
Thus, the ratio of corresponding sides = √(1/2) = 1/√2.


Step 5: Find the length of the fence
Let the length of the fence be x.
Then, x/50 = 1/√2.
x = 50/√2 ≈ 35.36 m.


Conclusion: The required length of the fence is approximately 35.36 meters.

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 triangular field with base 12 m and height 5 m. Adjacent to it is a parallelogram-shaped plot with the same base and height.
Problem Interpretation: Compare their areas using the formula we studied.
Mathematical Modeling: Apply the area formulas for both shapes.
Answer:
Problem Interpretation: We compare areas of a triangle and parallelogram with identical base and height.
Mathematical Modeling: Area of triangle = ½ × base × height; Parallelogram = base × height.
Solution:
  • Triangle area = ½ × 12 × 5 = 30 m²
  • Parallelogram area = 12 × 5 = 60 m²
The parallelogram has double the area of the triangle, as per NCERT examples.
Question 2:
A kite is shaped like a parallelogram with diagonals 8 cm and 6 cm.
Problem Interpretation: Verify if its area matches the rhombus area formula.
Mathematical Modeling: Use the diagonal formula for quadrilaterals.
Answer:
Problem Interpretation: We check if a parallelogram’s area equals ½ × product of diagonals (like a rhombus).
Mathematical Modeling: Area = ½ × d₁ × d₂.
Solution:
  • Area = ½ × 8 × 6 = 24 cm²
This formula works here because the kite is a rhombus-type quadrilateral, confirming textbook derivations.
Question 3:
A farmer has a parallelogram-shaped field with base 50 m and height 30 m. He divides it into two triangular plots by fencing along one diagonal. Find the area of each triangular plot and justify your answer using the properties of parallelograms.
Answer:
Problem Interpretation

We need to find the area of two triangular plots formed by dividing a parallelogram along its diagonal.

Mathematical Modeling

We studied that a diagonal divides a parallelogram into two congruent triangles of equal area.

Solution
  • Area of parallelogram = base × height = 50 m × 30 m = 1500 m².
  • Area of each triangle = ½ × area of parallelogram = 750 m².
Question 4:
In ΔABC, D and E are midpoints of sides AB and AC respectively. Prove that area of ΔADE = ¼ area of ΔABC using the midsegment theorem.
Answer:
Problem Interpretation

We must prove the area relationship between ΔADE and ΔABC using midpoints.

Mathematical Modeling

Our textbook shows that the line joining midpoints divides the triangle into four congruent triangles.

Solution
  • By midsegment theorem, DE ∥ BC and DE = ½ BC.
  • ΔADE ~ ΔABC with a scale factor of ½, so area ratio = (½)² = ¼.
Question 5:
A farmer has a parallelogram-shaped field with base 120 m and height 80 m. He divides it into two triangular plots by drawing a diagonal. Find the area of each triangular plot and justify your answer using the properties of parallelograms.
Answer:
Problem Interpretation

We need to find the area of two triangular plots formed by dividing a parallelogram field using its diagonal.


Mathematical Modeling
  • Area of parallelogram = base × height = 120 m × 80 m = 9600 m²
  • Diagonal divides it into two congruent triangles with equal areas.

Solution

Area of each triangle = ½ × area of parallelogram = ½ × 9600 m² = 4800 m². Our textbook shows that diagonals of parallelograms create equal-area triangles.

Question 6:
A triangular park has sides 50 m, 120 m, and 130 m. Prove it is a right-angled triangle using the converse of Pythagoras theorem and calculate its area using the formula for right-angled triangles.
Answer:
Problem Interpretation

We must verify if the park is right-angled and find its area using the legs.


Mathematical Modeling
  • Check Pythagoras theorem: 50² + 120² = 2500 + 14400 = 16900 = 130²
  • Hence, it satisfies the converse of Pythagoras theorem.

Solution

Area = ½ × base × height = ½ × 50 m × 120 m = 3000 m². We studied that right-angled triangles follow this formula.

Question 7:
A farmer has a triangular field with base 12 m and height 8 m. Adjacent to it is a parallelogram-shaped plot with the same base and height.
Problem Interpretation: Compare their areas.
Mathematical Modeling: Use the area formulas.
Which shape has a larger area, and by how much?
Answer:
Problem Interpretation: We studied that area of a triangle = ½ × base × height, and area of a parallelogram = base × height.
Solution:
  • Triangle area = ½ × 12 × 8 = 48 m²
  • Parallelogram area = 12 × 8 = 96 m²
The parallelogram has a larger area by 48 m² (96 − 48).
Question 8:
In a trapezoid ABCD, AB || CD, and its diagonal AC divides it into two triangles.
Problem Interpretation: Prove both triangles have equal areas.
Mathematical Modeling: Use the property of triangles on the same base and between parallels.
Show the steps.
Answer:
Problem Interpretation: Our textbook shows triangles on the same base and between parallels have equal areas.
Solution:
  • Triangles ABC and ADC share base AC.
  • Since AB || CD, their heights are equal.
  • Area of ΔABC = Area of ΔADC = ½ × base × height.
Hence, both triangles are equal in area.
Question 9:
A farmer has a parallelogram-shaped field with base 120 m and height 80 m. He divides it into two triangular plots by drawing a diagonal. Find the area of each triangular plot. (Use formula: Area of Δ = ½ × base × height)
Answer:
Problem Interpretation

We need to find the area of two triangular plots formed by dividing a parallelogram.

Mathematical Modeling

Area of parallelogram = base × height = 120 × 80 = 9600 m². A diagonal divides it into two congruent triangles.

Solution

Area of each Δ = ½ × 9600 = 4800 m². Both triangles have equal area as per NCERT examples.

Question 10:
In ΔABC, D and E are midpoints of sides AB and AC. Prove that area of ΔADE = ¼ × area of ΔABC. (Hint: Use midpoint theorem and properties of similar triangles)
Answer:
Problem Interpretation

We must prove the area relationship using midpoint theorem.

Mathematical Modeling

By midpoint theorem, DE ∥ BC and DE = ½ BC. ΔADE ~ ΔABC with scale factor ½.

Solution

Area ratio of similar Δs = (½)² = ¼. Thus, area(ΔADE) = ¼ × area(ΔABC), as shown in our textbook.

Question 11:

A farmer has a triangular field with vertices at points A, B, and C. The coordinates of these points are A(0, 0), B(4, 0), and C(2, 3) respectively. The farmer wants to divide the field into two parts of equal area by drawing a line parallel to the side BC.

Find the coordinates of the point where this dividing line meets the side AB.

Answer:

To divide the triangular field into two equal areas, we need to find the midpoint of the side AB where the dividing line parallel to BC meets.


Step 1: Calculate the area of triangle ABC using the formula for the area of a triangle with coordinates.


Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|


Substituting the coordinates:


Area = (1/2) |0(0 - 3) + 4(3 - 0) + 2(0 - 0)| = (1/2) |0 + 12 + 0| = 6 square units.


Step 2: Since the dividing line is parallel to BC and divides the area into two equal parts, the smaller triangle formed will have an area of 3 square units.


Step 3: The ratio of the areas of the two similar triangles (original and the smaller one) is 1:2, so the ratio of their corresponding sides is 1:√2.


Step 4: The dividing line meets AB at a point D such that AD/AB = 1/√2.


Since AB is 4 units, AD = 4/√2 ≈ 2.828 units.


Step 5: The coordinates of D are (2.828, 0).


Thus, the dividing line meets AB at approximately (2.83, 0).

Question 12:

A parallelogram PQRS has vertices at P(1, 2), Q(4, 3), and R(6, 6). The fourth vertex S is unknown.

(a) Find the coordinates of vertex S.

(b) Calculate the area of the parallelogram PQRS.

Answer:

(a) Finding the coordinates of vertex S:


In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of PR is the same as the midpoint of QS.


Step 1: Find the midpoint of PR.


Midpoint of PR = ((1 + 6)/2, (2 + 6)/2) = (3.5, 4).


Step 2: Let the coordinates of S be (x, y). The midpoint of QS is ((4 + x)/2, (3 + y)/2).


Setting the midpoints equal:


(4 + x)/2 = 3.5 ⇒ 4 + x = 7 ⇒ x = 3.


(3 + y)/2 = 4 ⇒ 3 + y = 8 ⇒ y = 5.


Thus, the coordinates of S are (3, 5).


(b) Calculating the area of parallelogram PQRS:


Step 1: Use the shoelace formula for the area of a quadrilateral.


Area = (1/2) |x1y2 + x2y3 + x3y4 + x4y1 - (y1x2 + y2x3 + y3x4 + y4x1)|.


Substituting the coordinates in order P(1, 2), Q(4, 3), R(6, 6), S(3, 5):


Area = (1/2) |1*3 + 4*6 + 6*5 + 3*2 - (2*4 + 3*6 + 6*3 + 5*1)|


= (1/2) |3 + 24 + 30 + 6 - (8 + 18 + 18 + 5)|


= (1/2) |63 - 49| = (1/2) * 14 = 7 square units.


Thus, the area of parallelogram PQRS is 7 square units.

Question 13:

A farmer has a triangular field ABC with sides AB = 50 m, BC = 40 m, and AC = 30 m. He wants to divide the field into two parts of equal area by constructing a fence parallel to the side BC. Find the length of the fence required.

Answer:

To solve this problem, we will use the concept of similar triangles and area ratios.


Step 1: Verify if triangle ABC is a right-angled triangle.
Using the Pythagorean theorem: AC² + BC² = 30² + 40² = 900 + 1600 = 2500 = AB² (50²).
Since AC² + BC² = AB², triangle ABC is right-angled at C.

Step 2: Calculate the area of triangle ABC.
Area = (1/2) × base × height = (1/2) × 30 × 40 = 600 m².

Step 3: Let the fence be DE parallel to BC, dividing the area into two equal parts (300 m² each).
Triangles ADE and ABC are similar (by AA similarity).
Area ratio of ADE to ABC = (DE/BC)² = 300/600 = 1/2.
Thus, (DE/40)² = 1/2 ⇒ DE/40 = 1/√2 ⇒ DE = 40/√2 = 20√2 m.

Final Answer: The length of the fence required is 20√2 meters.
Question 14:

A parallelogram PQRS has vertices P(1, 2), Q(4, 6), R(7, 6), and S(a, b). Find the coordinates of point S and calculate the area of the parallelogram using the distance formula and area of parallelogram formula.

Answer:

To solve this, we will use the properties of a parallelogram and the distance formula.


Step 1: Find the coordinates of S.
In a parallelogram, the diagonals bisect each other. Let the midpoint of PR = midpoint of QS.
Midpoint of PR = ((1+7)/2, (2+6)/2) = (4, 4).
Midpoint of QS = ((4+a)/2, (6+b)/2).
Setting them equal: (4+a)/2 = 4 ⇒ a = 4; (6+b)/2 = 4 ⇒ b = 2.
Thus, S is (4, 2).

Step 2: Calculate the area using the distance formula.
Base PQ = √[(4-1)² + (6-2)²] = √(9 + 16) = 5 units.
Height is the perpendicular distance from R to PQ.
Equation of PQ: (y-2) = (6-2)/(4-1)(x-1) ⇒ y = (4/3)x + 2/3.
Distance from R(7,6) to PQ: |(4/3)(7) - 1(6) + (2/3)| / √[(4/3)² + 1] = |28/3 - 6 + 2/3| / (5/3) = 12/5 units.
Area = base × height = 5 × (12/5) = 12 square units.

Final Answer: The coordinates of S are (4, 2), and the area of the parallelogram is 12 square units.
Question 15:
A farmer has a triangular field with vertices at points A(2, 3), B(5, 7), and C(8, 3) on a coordinate plane. He wants to divide the field into two parts of equal area by drawing a line parallel to the base AC. Find the coordinates of the points where this dividing line meets the other two sides AB and BC.
Answer:

To solve this problem, we will use the concept of area division in triangles and the section formula.


Step 1: Calculate the area of triangle ABC
Using the formula for the area of a triangle given its vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Substituting the coordinates:
Area = ½ |2(7 - 3) + 5(3 - 3) + 8(3 - 7)|
Area = ½ |8 + 0 - 32| = ½ × 24 = 12 square units.

Step 2: Find the area of the smaller triangle
Since the field is divided into two equal parts, the smaller triangle (similar to ABC) will have an area of 6 square units.

Step 3: Use the property of similar triangles
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let the dividing line be DE (parallel to AC).
Area of ΔADE / Area of ΔABC = (DE/AC)²
6/12 = (DE/AC)² ⇒ DE/AC = 1/√2.

Step 4: Find the coordinates of D and E
Using the section formula, D divides AB in the ratio 1 : (√2 - 1).
Coordinates of D:
x = (1×5 + (√2 - 1)×2) / (1 + √2 - 1) = (5 + 2√2 - 2) / √2 = (3 + 2√2)/√2.
y = (1×7 + (√2 - 1)×3) / √2 = (7 + 3√2 - 3)/√2 = (4 + 3√2)/√2.

Similarly, E divides BC in the same ratio.
Coordinates of E:
x = (1×8 + (√2 - 1)×5) / √2 = (8 + 5√2 - 5)/√2 = (3 + 5√2)/√2.
y = (1×3 + (√2 - 1)×7) / √2 = (3 + 7√2 - 7)/√2 = (-4 + 7√2)/√2.

Thus, the coordinates of D and E are calculated as above, ensuring the field is divided into two equal areas.

Question 16:
A parallelogram ABCD has vertices A(1, 2), B(4, 5), and D(2, -1). Find the coordinates of vertex C and verify that the diagonals of the parallelogram bisect each other.
Answer:

To solve this problem, we will use the properties of a parallelogram and the midpoint formula.


Step 1: Find the coordinates of vertex C
In a parallelogram, the opposite sides are equal and parallel. Therefore, vector AB = vector DC.
Vector AB = (4 - 1, 5 - 2) = (3, 3).
Let C be (x, y). Then, vector DC = (x - 2, y - (-1)) = (x - 2, y + 1).
Since AB = DC:
x - 2 = 3 ⇒ x = 5.
y + 1 = 3 ⇒ y = 2.
Thus, C is (5, 2).

Step 2: Find the midpoints of the diagonals
Diagonal AC has endpoints A(1, 2) and C(5, 2).
Midpoint of AC = ((1 + 5)/2, (2 + 2)/2) = (3, 2).

Diagonal BD has endpoints B(4, 5) and D(2, -1).
Midpoint of BD = ((4 + 2)/2, (5 + (-1))/2) = (3, 2).

Step 3: Verification
Since the midpoints of both diagonals coincide at (3, 2), the diagonals bisect each other, confirming that ABCD is a parallelogram.

Thus, the coordinates of C are (5, 2), and the diagonals bisect each other at (3, 2), satisfying the properties of a parallelogram.

Question 17:
A farmer has a triangular field with vertices at points A(2, 3), B(4, 7), and C(6, 2) on a coordinate plane. He wants to divide the field into two parts of equal area by drawing a line parallel to the base BC. Find the coordinates of the points where this dividing line meets the other two sides AB and AC.
Answer:

To divide the triangular field into two equal areas, we need to find the midpoints of sides AB and AC because a line parallel to the base and passing through the midpoint will divide the triangle into two regions of equal area.


Step 1: Find the midpoint of AB
Coordinates of A: (2, 3)
Coordinates of B: (4, 7)
Midpoint of AB (D) = ((2+4)/2, (3+7)/2) = (3, 5)

Step 2: Find the midpoint of AC
Coordinates of A: (2, 3)
Coordinates of C: (6, 2)
Midpoint of AC (E) = ((2+6)/2, (3+2)/2) = (4, 2.5)

Conclusion: The dividing line parallel to BC will pass through points D(3, 5) and E(4, 2.5).

Note: This method works because the area of the smaller triangle formed will be one-fourth of the original triangle, but since we are dividing the area into two equal parts, the line must pass through the midpoints of the non-base sides.
Question 18:
A parallelogram ABCD has vertices A(1, 2), B(4, 3), and D(2, 5). Find the coordinates of vertex C and calculate the area of the parallelogram using the coordinate geometry method.
Answer:

To find the coordinates of vertex C and the area of parallelogram ABCD, we follow these steps:


Step 1: Find the coordinates of C
In a parallelogram, the diagonals bisect each other. Let the intersection point of diagonals be O.
Coordinates of A: (1, 2)
Coordinates of B: (4, 3)
Coordinates of D: (2, 5)
O is the midpoint of BD:
O = ((4+2)/2, (3+5)/2) = (3, 4)
Since O is also the midpoint of AC:
Let C be (x, y)
(1 + x)/2 = 3 → x = 5
(2 + y)/2 = 4 → y = 6
So, C is (5, 6)

Step 2: Calculate the area
Using the shoelace formula for quadrilateral ABCD:
Area = 1/2 |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)|
Substitute the coordinates in order (A → B → C → D → A):
Area = 1/2 |(1*3 + 4*6 + 5*5 + 2*2) - (2*4 + 3*5 + 6*2 + 5*1)|
Area = 1/2 |(3 + 24 + 25 + 4) - (8 + 15 + 12 + 5)|
Area = 1/2 |56 - 40| = 1/2 * 16 = 8 square units

Note: The shoelace formula is a straightforward method to find the area of any polygon when the coordinates of its vertices are known.
Question 19:
A farmer has a triangular field ABC with base BC = 40 m and height AD = 20 m. He divides the field into two parts by drawing a line segment PQ parallel to BC, such that the area of the smaller triangle APQ is one-fourth of the area of the original triangle ABC. Find the length of PQ.
Answer:

To find the length of PQ, we follow these steps:


Step 1: Calculate the area of the original triangle ABC.
Area of ABC = (1/2) × base × height = (1/2) × 40 m × 20 m = 400 m².

Step 2: Determine the area of the smaller triangle APQ.
Given, area of APQ = (1/4) × area of ABC = (1/4) × 400 m² = 100 m².

Step 3: Since PQ is parallel to BC, triangles APQ and ABC are similar by the AA (Angle-Angle) similarity criterion.
The ratio of their areas is the square of the ratio of their corresponding sides.
So, (Area of APQ / Area of ABC) = (PQ / BC)².
Substituting the values: (100 / 400) = (PQ / 40)².
This simplifies to (1/4) = (PQ / 40)².

Step 4: Solve for PQ.
Taking the square root of both sides: √(1/4) = PQ / 40.
Thus, (1/2) = PQ / 40.
Therefore, PQ = 40 × (1/2) = 20 m.

The length of PQ is 20 meters.

Question 20:
A parallelogram ABCD has sides AB = 12 cm and AD = 8 cm. The height corresponding to side AB is 6 cm. Find the area of the parallelogram and the height corresponding to side AD.
Answer:

To solve the problem, we follow these steps:


Step 1: Calculate the area of the parallelogram using the given height corresponding to side AB.
Area = base × height = AB × height corresponding to AB = 12 cm × 6 cm = 72 cm².

Step 2: Now, find the height corresponding to side AD.
Since the area of the parallelogram remains the same, we can write:
Area = AD × height corresponding to AD.
Substituting the known values: 72 cm² = 8 cm × height corresponding to AD.

Step 3: Solve for the height corresponding to AD.
height corresponding to AD = 72 cm² / 8 cm = 9 cm.

The area of the parallelogram is 72 cm², and the height corresponding to side AD is 9 cm.

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