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Area Exercise 7.4 Class 8 Maths – Chapter 7 NCERT Solutions

Class 8 NCERT Maths Term 2 Chapter 7 Area Exercise 7.4 detailed solutions, curated by the ALLEN expert team, are available on this webpage. These solutions can help students prepare well for their exams and support their academic learning. They are prepared according to the updated NCERT syllabus.Class 8 Maths Chapter 7 Area Exercise 7.4 mainly discusses the areas and perimeters of parallelograms. It also talks about Pythagoras, the comparison of rectangles and parallelograms, and the conversion of triangles and rectangles into figures of equal area by construction and dissection.

1.0Download Class 8 Maths Chapter 7 Ex 7.4 NCERT Solutions PDF

Download the detailed solutions for Exercise 7.4 for free and refer to them for better learning and revision.

Chapter 7 : Area Exercise 7.4

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.4 Class 8 Maths (Ganita Prakash Term 2)

1. Observe the parallelograms in the figure below. (i) What can we say about the areas of all these parallelograms? (ii) What can we say about their perimeters? Which figures appears to have the maximum perimeter and which has the minimum perimeter?

Sol. (i) (a) Area of parallelogram $=$ base × height $= 5 \times 3 = 15$ sq. units (b) Area of parallelogram $= 5 \times 3 = 15$ sq. units (c) Area of parallelogram $= 5 \times 3 = 15$ sq. units (d) Area of parallelogram $= 5 \times 3 = 15$ sq. units (e) Area of parallelogram $= 5 \times 3 = 15$ sq. units (f) Area of parallelogram $= 5 \times 3 = 15$ sq. units (g) Area of parallelogram $= 5 \times 3 = 15$ sq. units

All parallelograms have equal areas. (ii) The perimeters of these parallelograms are different even though their areas are the same.

Figure (d) has the minimum perimeter and Figure (g) has the maximum perimeter.

2. Find the area of the following parallelograms. (i)

(ii)

(iii)

(iv)

Sol. Area of the parallelogram $=$ base × height (i) Here, base $= 7 cm$ and height $= 4 cm$ Area of the parallelogram $= 7 cm \times 4 cm = 28 cm^{2}$ (ii) Here, base $= 5 cm$ and height $= 3 cm$ Area of the parallelogram $= 5 cm \times 3 cm = 15 cm^{2}$ (iii) Here, base $= 5 cm$ and height $= 4.8 cm$ Area of the parallelogram $= 5 cm \times 4.8 cm = 24 cm^{2}$ (iv) Here, base $= 2 cm$ and height $= 4.4 cm$ Area of the parallelogram $= 2 cm \times 4.4 cm = 8.8 cm^{2}$

3. Find QN.

Sol. In $△ PNQ, ∠ PNQ = 9 0^{\circ}$ $PN = 7.6 cm$ $PQ = 12 cm$ By Pythagoras theorem $PQ^{2} = PN^{2} + NQ^{2}$ $\Rightarrow (12)^{2} = (7.6)^{2} + NQ^{2}$ $\Rightarrow Q N^{2} = (12)^{2} - (7.6)^{2}$ $\Rightarrow QN^{2} = 144 - 57.76$ $\Rightarrow QN^{2} = 86.24$ $\Rightarrow QN = 9.28 cm$

4. Consider a rectangle and a parallelogram of the same side lengths: 5 cm and 4 cm . Which has the greater area? [Hint: Imagine constructing them on the same base.]

Sol. For rectangle: $ℓ = 5, w = 4$ , all angles $= 9 0^{\circ}$ ∴ Area $= ℓ \times w = 5 \times 4 = 20 cm^{2}$ For parallelogram: Base (b) $= 5 cm$ , one slanted side $= 4 cm$ Height will be less than 4 cm because the side is slanted.

Area $= 5 \times 4 < 20 cm^{2}$ Hence, the rectangle has a greater area than the parallelogram.

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Sol. Given: Triangle with area A Required: Rectangle with area $= 2 A$

Method 1:

If the triangle has base b and height h Area of triangle $= \frac{1}{2} \times b \times h = A$

To get a rectangle with an area of 2 A . Take length $= b$ , width $= h$

Area of a rectangle $= b \times h$ $= 2 \times (\frac{1}{2} \times b \times h) = 2 A$

Steps:

Measure the base and height of the given triangle.

Construct a rectangle with these measurements as length and width.

Method 2:

Scaling method:

Take the rectangle.
Create a rectangle with base = (base of triangle) and height $=$ height of triangle
This rectangle automatically has twice the area of the triangle.

6. Give a method to obtain a rectangle of the same area as a given triangle.

Sol. Given: A triangle with base b and height h . Required: Rectangle with the same area Area of triangle $= \frac{1}{2} bh$

To get a rectangle with the same area Rectangle length $= \frac{b}{2}$ (half the triangle's base) Rectangle width $= h$ (same as the triangle's height) Area $= \frac{b}{2} \times h$ $= \frac{1}{2} \times b \times h$

7. An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

Sol. Given: Isosceles triangle ABC , where $AB =$ AC and AD is the altitude from A to BC . Method: Since the triangle is isosceles: AD is perpendicular to $BC . D$ is the midpoint of BC (property of an isosceles triangle). AD bisects the triangle into two congruent right triangles: $△ ADB$ and $△ ADC$ .

Dissection Process:

Step 1: The altitude AD divides the isosceles triangle into two congruent right triangles $△ ADB$ and $△ ADC$ . Step 2: Each of these right triangles can be cut into pieces. Step 3: Assembly: Take triangle $△ ADB$ and $△ ADC$ , rotate one triangle $18 0^{\circ}$ arrange them.

8. Give a method to convert a rectangle into an isosceles triangle by dissection.

Sol. Step-1: Take rectangle PQRS with length $ℓ$ and width w .

Step-2: Cut it along diagonal PR.

Step-3: Rotate $△ PQR$ and join with other $Δ$ in such way that PQ coincide with RS.

Allow triangle in required isosceles $Δ$ . 9. Which has greater area, an equilateral triangle or a square of the same side length as the triangle? Which has greater area two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons. Sol. Area of equilateral triangle $= \frac{3}{4} a^{2}$ Area of square $= a^{2}$ $\Rightarrow \frac{3}{4} a^{2} < a^{2}$ So, the area of a square is greater than the area of an equilateral triangle of the same side length. Area of two identical equilateral triangles $= \frac{3}{4} a^{2} + \frac{3}{4} a^{2} = \frac{2 3}{4} a^{2} = \frac{3}{2} a^{2}$ Area of square of side length $a = a^{2}$ Clearly, $\frac{3}{2} a^{2} < a^{2}$ So, the area of a square is greater than the area of two identical equilateral triangles.

3.0Key Concepts of Chapter 7 Area Exercise 7.4

Comparing the areas of parallelograms drawn on the same base and between the same parallel lines.
Observing how the perimeter changes while the area remains unchanged.
Relating the area of a parallelogram to the area of a rectangle with the same base and height.
Converting a parallelogram into a rectangle to understand its area formula.
Exploring different parallelograms formed on the same base and height.
Establishing the area formula of a parallelogram: Base × Height.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.4

Compares the areas and perimeters of different parallelograms.
Applies the base × height formula to calculate the area of parallelograms.
Uses the Pythagoras theorem to determine an unknown side of a triangle.
Compares the areas of a rectangle and a parallelogram with the same side lengths.
Converts a triangle into a rectangle with twice or the same area.
Explores dissection methods for converting an isosceles triangle into a rectangle and a rectangle into an isosceles triangle.
Compares the areas of an equilateral triangle and a square with the same side length.

NCERT Class 8 Maths Ch. 7 Area Other Exercises:-

Exercise 7.4

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 : Fractions in Disguise

Chapter 2 : The Baudhayana - The Pythagoras Theorem

Chapter 3 : Proportional Reasoning - 2

Chapter 4 : Exploring Some Geometric Themes

Chapter 5 : Tales by Dots and Lines

Chapter 6 : Algebra Play

Chapter 7 : Area

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Area Exercise 7.4 Class 8 Maths – Chapter 7 NCERT Solutions

Class 8 NCERT Maths Term 2 Chapter 7 Area Exercise 7.4 detailed solutions, curated by the ALLEN expert team, are available on this webpage. These solutions can help students prepare well for their exams and support their academic learning. They are prepared according to the updated NCERT syllabus.Class 8 Maths Chapter 7 Area Exercise 7.4 mainly discusses the areas and perimeters of parallelograms. It also talks about Pythagoras, the comparison of rectangles and parallelograms, and the conversion of triangles and rectangles into figures of equal area by construction and dissection.

1.0Download Class 8 Maths Chapter 7 Ex 7.4 NCERT Solutions PDF

Download the detailed solutions for Exercise 7.4 for free and refer to them for better learning and revision.

Chapter 7 : Area Exercise 7.4

Available soon

2.0Detailed NCERT Solutions for Area Exercise 7.4 Class 8 Maths (Ganita Prakash Term 2)

1. Observe the parallelograms in the figure below. (i) What can we say about the areas of all these parallelograms? (ii) What can we say about their perimeters? Which figures appears to have the maximum perimeter and which has the minimum perimeter?

Sol. (i) (a) Area of parallelogram $=$ base × height $= 5 \times 3 = 15$ sq. units (b) Area of parallelogram $= 5 \times 3 = 15$ sq. units (c) Area of parallelogram $= 5 \times 3 = 15$ sq. units (d) Area of parallelogram $= 5 \times 3 = 15$ sq. units (e) Area of parallelogram $= 5 \times 3 = 15$ sq. units (f) Area of parallelogram $= 5 \times 3 = 15$ sq. units (g) Area of parallelogram $= 5 \times 3 = 15$ sq. units

All parallelograms have equal areas. (ii) The perimeters of these parallelograms are different even though their areas are the same.

Figure (d) has the minimum perimeter and Figure (g) has the maximum perimeter.

2. Find the area of the following parallelograms. (i)

(ii)

(iii)

(iv)

Sol. Area of the parallelogram $=$ base × height (i) Here, base $= 7 cm$ and height $= 4 cm$ Area of the parallelogram $= 7 cm \times 4 cm = 28 cm^{2}$ (ii) Here, base $= 5 cm$ and height $= 3 cm$ Area of the parallelogram $= 5 cm \times 3 cm = 15 cm^{2}$ (iii) Here, base $= 5 cm$ and height $= 4.8 cm$ Area of the parallelogram $= 5 cm \times 4.8 cm = 24 cm^{2}$ (iv) Here, base $= 2 cm$ and height $= 4.4 cm$ Area of the parallelogram $= 2 cm \times 4.4 cm = 8.8 cm^{2}$

3. Find QN.

Sol. In $△ PNQ, ∠ PNQ = 9 0^{\circ}$ $PN = 7.6 cm$ $PQ = 12 cm$ By Pythagoras theorem $PQ^{2} = PN^{2} + NQ^{2}$ $\Rightarrow (12)^{2} = (7.6)^{2} + NQ^{2}$ $\Rightarrow Q N^{2} = (12)^{2} - (7.6)^{2}$ $\Rightarrow QN^{2} = 144 - 57.76$ $\Rightarrow QN^{2} = 86.24$ $\Rightarrow QN = 9.28 cm$

4. Consider a rectangle and a parallelogram of the same side lengths: 5 cm and 4 cm . Which has the greater area? [Hint: Imagine constructing them on the same base.]

Sol. For rectangle: $ℓ = 5, w = 4$ , all angles $= 9 0^{\circ}$ ∴ Area $= ℓ \times w = 5 \times 4 = 20 cm^{2}$ For parallelogram: Base (b) $= 5 cm$ , one slanted side $= 4 cm$ Height will be less than 4 cm because the side is slanted.

Area $= 5 \times 4 < 20 cm^{2}$ Hence, the rectangle has a greater area than the parallelogram.

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Sol. Given: Triangle with area A Required: Rectangle with area $= 2 A$

Method 1:

If the triangle has base b and height h Area of triangle $= \frac{1}{2} \times b \times h = A$

To get a rectangle with an area of 2 A . Take length $= b$ , width $= h$

Area of a rectangle $= b \times h$ $= 2 \times (\frac{1}{2} \times b \times h) = 2 A$

Steps:

Measure the base and height of the given triangle.

Construct a rectangle with these measurements as length and width.

Method 2:

Scaling method:

Take the rectangle.
Create a rectangle with base = (base of triangle) and height $=$ height of triangle
This rectangle automatically has twice the area of the triangle.

6. Give a method to obtain a rectangle of the same area as a given triangle.

Sol. Given: A triangle with base b and height h . Required: Rectangle with the same area Area of triangle $= \frac{1}{2} bh$

To get a rectangle with the same area Rectangle length $= \frac{b}{2}$ (half the triangle's base) Rectangle width $= h$ (same as the triangle's height) Area $= \frac{b}{2} \times h$ $= \frac{1}{2} \times b \times h$

7. An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

Sol. Given: Isosceles triangle ABC , where $AB =$ AC and AD is the altitude from A to BC . Method: Since the triangle is isosceles: AD is perpendicular to $BC . D$ is the midpoint of BC (property of an isosceles triangle). AD bisects the triangle into two congruent right triangles: $△ ADB$ and $△ ADC$ .

Dissection Process:

Step 1: The altitude AD divides the isosceles triangle into two congruent right triangles $△ ADB$ and $△ ADC$ . Step 2: Each of these right triangles can be cut into pieces. Step 3: Assembly: Take triangle $△ ADB$ and $△ ADC$ , rotate one triangle $18 0^{\circ}$ arrange them.

8. Give a method to convert a rectangle into an isosceles triangle by dissection.

Sol. Step-1: Take rectangle PQRS with length $ℓ$ and width w .

Step-2: Cut it along diagonal PR.

Step-3: Rotate $△ PQR$ and join with other $Δ$ in such way that PQ coincide with RS.

Allow triangle in required isosceles $Δ$ . 9. Which has greater area, an equilateral triangle or a square of the same side length as the triangle? Which has greater area two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons. Sol. Area of equilateral triangle $= \frac{3}{4} a^{2}$ Area of square $= a^{2}$ $\Rightarrow \frac{3}{4} a^{2} < a^{2}$ So, the area of a square is greater than the area of an equilateral triangle of the same side length. Area of two identical equilateral triangles $= \frac{3}{4} a^{2} + \frac{3}{4} a^{2} = \frac{2 3}{4} a^{2} = \frac{3}{2} a^{2}$ Area of square of side length $a = a^{2}$ Clearly, $\frac{3}{2} a^{2} < a^{2}$ So, the area of a square is greater than the area of two identical equilateral triangles.

3.0Key Concepts of Chapter 7 Area Exercise 7.4

Comparing the areas of parallelograms drawn on the same base and between the same parallel lines.
Observing how the perimeter changes while the area remains unchanged.
Relating the area of a parallelogram to the area of a rectangle with the same base and height.
Converting a parallelogram into a rectangle to understand its area formula.
Exploring different parallelograms formed on the same base and height.
Establishing the area formula of a parallelogram: Base × Height.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.4

Compares the areas and perimeters of different parallelograms.
Applies the base × height formula to calculate the area of parallelograms.
Uses the Pythagoras theorem to determine an unknown side of a triangle.
Compares the areas of a rectangle and a parallelogram with the same side lengths.
Converts a triangle into a rectangle with twice or the same area.
Explores dissection methods for converting an isosceles triangle into a rectangle and a rectangle into an isosceles triangle.
Compares the areas of an equilateral triangle and a square with the same side length.

Frequently Asked Questions

What is covered in Exercise 7.4 of Class 8 Maths Chapter 7?

How is the area of a parallelogram calculated?

Do parallelograms with the same base and height have equal areas?

Can parallelograms with equal areas have different perimeters?

How is a rectangle with the same area as a triangle obtained?

How can a rectangle be converted into an isosceles triangle?

Which has greater area, an equilateral triangle or a square with the same side length?

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Area Exercise 7.4 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.4 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.4 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.4

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.4

Frequently Asked Questions

What is covered in Exercise 7.4 of Class 8 Maths Chapter 7?

How is the area of a parallelogram calculated?

Do parallelograms with the same base and height have equal areas?

Can parallelograms with equal areas have different perimeters?

How is a rectangle with the same area as a triangle obtained?

How can a rectangle be converted into an isosceles triangle?

Which has greater area, an equilateral triangle or a square with the same side length?

Join ALLEN!

Area Exercise 7.4 Class 8 Maths – Chapter 7 NCERT Solutions

1.0Download Class 8 Maths Chapter 7 Ex 7.4 NCERT Solutions PDF

2.0Detailed NCERT Solutions for Area Exercise 7.4 Class 8 Maths (Ganita Prakash Term 2)

3.0Key Concepts of Chapter 7 Area Exercise 7.4

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.4