NCERT Class 8 Maths Area Exercise 7.5 Solutions

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

1.0Area Exercise 7.5 Class 8 Maths – Chapter 7 NCERT Solutions

NCERT Class 8 Maths Chapter 7 Exercise 7.5 detailed solutions are provided below. These detailed solutions are prepared by the ALLEN faculty team. The solutions are prepared to help students prepare for their exams in a well-structured manner and support their preparation for competitive exams.Exercise 7.5 is about the application of the area concepts on different geometric figures. There are problems on parallelograms, triangles, rectangles, figures with equal areas, constructions and comparison of areas.

2.0Download Class 8 Maths Chapter 7 Ex 7.5 NCERT Solutions PDF

Download the free PDF for Exercise 7.5 given below for better learning and to prepare well for exams.

Chapter 7 : Area Exercise 7.5

Available soon

3.0Detailed NCERT Solutions for Area Exercise 7.5 Class 8 Maths (Ganita Prakash Term 2)

1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm

Sol. Given, first diagonal $= 20 cm$ second diagonal $= 15 cm$

The area of a rhombus $= \frac{1}{2} \times$ (Product of diagonals) $= \frac{1}{2} \times$ First diagonal × second diagonal $= \frac{1}{2} \times 20 cm \times 15 cm = 150 cm^{2}$ Thus, the area of a rhombus is $150 cm^{2}$ 2. Give a method to convert a rectangle into a rhombus of equal area using dissection.

Sol. This is the reverse of the rhombus to rectangle dissection.

Method:

Given: Rectangle PQRS with length I and width W

Required:

Rhombus with the same area $= ℓ \times W$

Dissection Process:

Step 1: The rhombus will have diagonals $d_{1}$ and $d_{2}$ such that: $\frac{1}{2} \times d_{1} \times d_{2} = ℓ \times w$ So, $d_{1} \times d_{2} = 2 lw$ . Step 2: Choose convenient diagonal lengths: Let $d_{1} = 2 ℓ$ (twice the rectangle length)

Then $d_{2} = w$ (same as rectangle width) Check: $\frac{1}{2} \times 2 ℓ \times w = ℓ w$

Or

Let $d_{1} = 2 w$ (twice the rectangle width) Then $d_{2} = ℓ$ (same as rectangle length) Step 3: Dissection process (reverse of textbook method):

Divide the rectangle into two halves Mark the center point 0 . Cut and rotate pieces to form two isosceles triangles. Arrange these triangles to share a common diagonal. This creates a rhombus.

3. Find the area of the following figures:

(ii)

(iii)

(iv)

Sol. The area of the trapezium $= \frac{1}{2} \times$ (Sum of parallel sides) $\times$ (Distance between them) $= \frac{1}{2} \times (a + b) \times h$ (i) Here, $a = 10 ft, b = 7 ft$ and $h = 16 ft$ Area of trapezium $= \frac{1}{2} \times (10 + 7) \times 16 = 17 \times 8 = 136 ft^{2}$ (ii) Here, $a = 36 m, b$ $= 24 m$ and $h = 14 m$ Area of trapezium

$= \frac{1}{2} \times (36 + 24) \times 14 = 60 \times 7 = 420 m^{2}$

(iii) Here, $a = 14 in, b = 6 in$ and $h = 10 in$

Area of trapezium $= \frac{1}{2} \times (14 + 6) \times 10 = 20 \times 5 = 100 in^{2}$ (iv) Here, $a = 18 ft, b = 12 ft$ and $h = 8 ft$

Area of trapezium

$= \frac{1}{2} \times (18 + 12) \times 8 = 30 \times 4 = 120 ft^{2}$

Give a method to convert an isosceles trapezium to a rectangle using dissection. Sol. An isosceles trapezium has special properties that make dissection simple.

Properties of Isosceles Trapezium ABCD: $AB ∥ CD$ (parallel sides) $AD = BC$ (non-parallel sides are equal) $∠ A = ∠ B$ and $∠ D = ∠ C$ (base angles are equal)

Dissection Method:

Step-1: Take an isosceles trapezium, in which $AB ∥ CD$ & $AD = BC$ .

Step 2: Draw AP & BQ perpendicular to DC . Here $△ ADP ≅ △ BCQ$ .

Step 3: Cut $△ ADP$ and join to the other side to get rectangle.

WXYZ is required rectangle.

Resulting Rectangle:

Length = CD + AP $= CD + \frac{( AB - CD )}{2}$ $= \frac{( AB + CD )}{2}$ Width $= h$ (height of trapezium) Area $= \frac{( AB + CD )}{2} \times h = \frac{1}{2} h (AB + CD)$ This matches the trapezium area formula

5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area:

Given the trapezium ABCD , how do we find the vertices of the rectangle EFGH ? [Hint: If $△ AHI ≅ △ DGI$ and $△ BEJ ≅ △ CFJ$ , then the trapezium and rectangle have equal areas.]

Sol. Given: Trapezium ABCD with $AB ∥ CD$ Required:

Find the positions of the vertices E, F, G and H to form a rectangle EFGH with equal area

Using the Hint:

If $Δ AHI ≅ Δ DGI$ and $Δ BEJ ≅ Δ CFJ$ , then areas are equal.

Method:

Step 1: The rectangle $EFG H$ should have: EF as one side (top side) GH is the opposite parallel side (bottom side)

EH and FG are the other pair of sides.

Step 2: Position the rectangle such that: Points I and J are strategically chosen on the trapezium.

Step 3: For congruency:

Mark I on side AD Mark J on side BC Choose positions such that: HI=GI (making $△ AHI ≅ △ DGI$ possible) $EJ = FJ$ (making $Δ BEJ ≅ △ CFJ$ possible) Step 4: The height of the rectangle $= h$ (height of the trapezium) Step 5: The length of rectangle $= \frac{( a + b )}{2}$ where $a$ and $b$ are parallel sides This ensures: Area of rectangle $= h \times \frac{( a + b )}{2}$ = Area of trapezium

Practical construction:

Draw the trapezium ABCD Calculate required rectangle length $= \frac{( AB + CD )}{2}$ Mark points H and E on AB such that the central portion has this length.

Draw perpendiculars to get rectangle EFGH.

Verify that the triangular pieces outside match those inside.

This construction beautifully demonstrates area conservation through dissection!

6. Using the idea of converting a trapezium into a rectangle of equal area and vice versa, construct a trapezium of area $144 cm^{2}$ . Sol. Area of the trapezium $= \frac{1}{2} \times (10 + 8) \times 16$ $= 144 cm^{2}$ Area of rectangle $= 16 \times 9 = 144 cm^{2}$

7. A regular hexagon is divided into a trapezium, an equilateral triangle and a rhombus, as shown. Find the ratio of their areas.

Sol. Here total area of hexagon $= 6 \times \frac{3}{4} a^{2} = \frac{3 3}{2} a^{2}$

Equilateral triangle:

Area $= \frac{3}{4} a^{2}$

Rhombus:

Area $= 2 \times \frac{3}{4} a^{2} = \frac{3}{2} a^{2}$

Trapezium:

Remaining Area = Total Area - Triangle Area - Rhombus Area $= \frac{6 3}{4} - \frac{3}{4} a^{2} - \frac{3}{2} a^{2}$ $= \frac{3 3}{4} a^{2}$

Ratio of area

= Triangle : Rhombus : Trapezium $= \frac{3}{4} a^{2} : \frac{3}{2} a^{2} : \frac{3 3}{4} a^{2} = 1 : 2 : 3$

8. ZYXW is a trapezium with $ZY ∥ WX . A$ is the midpoint of XY . Show that the area of the trapezium ZYXW is equal to the area of $Δ ZWB .$ .

Sol. $∠ ZAY = ∠ BAX$ (Vertically opposite angles)

A Y = A X (∵ A

is mid point of

X Y)

∴ ∠ YZB = ∠ XBZ

( ∵ ZY || XB alternate interior angles are equal)

∠ ZYA = ∠ BXA

( ∵ they are alternate interior angles) So

△ ZAY ≅ △ BAX

By the AAS congruence. So, Area of

Δ ZAY =

Area of

Δ BAX

Thus, Area of trapezium ZYXW = Area of triangle ZWB

4.0Key Concepts of Chapter 7 Area Exercise 7.5

Comparing the area of a rectangle and a parallelogram having the same side lengths.
Finding an unknown side of a right-angled triangle using the Pythagoras theorem.
Constructing a rectangle with twice the area of a given triangle.
Constructing a rectangle with the same area as a given triangle using half of its base.
Converting an isosceles triangle into a rectangle using dissection.
Converting a rectangle into an isosceles triangle through cutting and rearranging.
Comparing the areas of an equilateral triangle and a square with the same side length.

5.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.5

Compares the areas of a rectangle and a parallelogram with the same side lengths.
Finds an unknown side of a right-angled triangle using the Pythagoras theorem.
Constructs a rectangle with twice the area of a given triangle.
Constructs a rectangle with the same area as a given triangle using half of its base.
Applies dissection to convert an isosceles triangle into a rectangle.
Uses cutting and rearrangement to convert 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 7.5

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

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

1.0Area Exercise 7.5 Class 8 Maths – Chapter 7 NCERT Solutions

NCERT Class 8 Maths Chapter 7 Exercise 7.5 detailed solutions are provided below. These detailed solutions are prepared by the ALLEN faculty team. The solutions are prepared to help students prepare for their exams in a well-structured manner and support their preparation for competitive exams.Exercise 7.5 is about the application of the area concepts on different geometric figures. There are problems on parallelograms, triangles, rectangles, figures with equal areas, constructions and comparison of areas.

2.0Download Class 8 Maths Chapter 7 Ex 7.5 NCERT Solutions PDF

Download the free PDF for Exercise 7.5 given below for better learning and to prepare well for exams.

Chapter 7 : Area Exercise 7.5

Available soon

3.0Detailed NCERT Solutions for Area Exercise 7.5 Class 8 Maths (Ganita Prakash Term 2)

1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm

Sol. Given, first diagonal $= 20 cm$ second diagonal $= 15 cm$

The area of a rhombus $= \frac{1}{2} \times$ (Product of diagonals) $= \frac{1}{2} \times$ First diagonal × second diagonal $= \frac{1}{2} \times 20 cm \times 15 cm = 150 cm^{2}$ Thus, the area of a rhombus is $150 cm^{2}$ 2. Give a method to convert a rectangle into a rhombus of equal area using dissection.

Sol. This is the reverse of the rhombus to rectangle dissection.

Method:

Given: Rectangle PQRS with length I and width W

Required:

Rhombus with the same area $= ℓ \times W$

Dissection Process:

Step 1: The rhombus will have diagonals $d_{1}$ and $d_{2}$ such that: $\frac{1}{2} \times d_{1} \times d_{2} = ℓ \times w$ So, $d_{1} \times d_{2} = 2 lw$ . Step 2: Choose convenient diagonal lengths: Let $d_{1} = 2 ℓ$ (twice the rectangle length)

Then $d_{2} = w$ (same as rectangle width) Check: $\frac{1}{2} \times 2 ℓ \times w = ℓ w$

Or

Let $d_{1} = 2 w$ (twice the rectangle width) Then $d_{2} = ℓ$ (same as rectangle length) Step 3: Dissection process (reverse of textbook method):

Divide the rectangle into two halves Mark the center point 0 . Cut and rotate pieces to form two isosceles triangles. Arrange these triangles to share a common diagonal. This creates a rhombus.

3. Find the area of the following figures:

(ii)

(iii)

(iv)

Sol. The area of the trapezium $= \frac{1}{2} \times$ (Sum of parallel sides) $\times$ (Distance between them) $= \frac{1}{2} \times (a + b) \times h$ (i) Here, $a = 10 ft, b = 7 ft$ and $h = 16 ft$ Area of trapezium $= \frac{1}{2} \times (10 + 7) \times 16 = 17 \times 8 = 136 ft^{2}$ (ii) Here, $a = 36 m, b$ $= 24 m$ and $h = 14 m$ Area of trapezium

$= \frac{1}{2} \times (36 + 24) \times 14 = 60 \times 7 = 420 m^{2}$

(iii) Here, $a = 14 in, b = 6 in$ and $h = 10 in$

Area of trapezium $= \frac{1}{2} \times (14 + 6) \times 10 = 20 \times 5 = 100 in^{2}$ (iv) Here, $a = 18 ft, b = 12 ft$ and $h = 8 ft$

Area of trapezium

$= \frac{1}{2} \times (18 + 12) \times 8 = 30 \times 4 = 120 ft^{2}$

Give a method to convert an isosceles trapezium to a rectangle using dissection. Sol. An isosceles trapezium has special properties that make dissection simple.

Properties of Isosceles Trapezium ABCD: $AB ∥ CD$ (parallel sides) $AD = BC$ (non-parallel sides are equal) $∠ A = ∠ B$ and $∠ D = ∠ C$ (base angles are equal)

Dissection Method:

Step-1: Take an isosceles trapezium, in which $AB ∥ CD$ & $AD = BC$ .

Step 2: Draw AP & BQ perpendicular to DC . Here $△ ADP ≅ △ BCQ$ .

Step 3: Cut $△ ADP$ and join to the other side to get rectangle.

WXYZ is required rectangle.

Resulting Rectangle:

Length = CD + AP $= CD + \frac{( AB - CD )}{2}$ $= \frac{( AB + CD )}{2}$ Width $= h$ (height of trapezium) Area $= \frac{( AB + CD )}{2} \times h = \frac{1}{2} h (AB + CD)$ This matches the trapezium area formula

5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area:

Given the trapezium ABCD , how do we find the vertices of the rectangle EFGH ? [Hint: If $△ AHI ≅ △ DGI$ and $△ BEJ ≅ △ CFJ$ , then the trapezium and rectangle have equal areas.]

Sol. Given: Trapezium ABCD with $AB ∥ CD$ Required:

Find the positions of the vertices E, F, G and H to form a rectangle EFGH with equal area

Using the Hint:

If $Δ AHI ≅ Δ DGI$ and $Δ BEJ ≅ Δ CFJ$ , then areas are equal.

Method:

Step 1: The rectangle $EFG H$ should have: EF as one side (top side) GH is the opposite parallel side (bottom side)

EH and FG are the other pair of sides.

Step 2: Position the rectangle such that: Points I and J are strategically chosen on the trapezium.

Step 3: For congruency:

Mark I on side AD Mark J on side BC Choose positions such that: HI=GI (making $△ AHI ≅ △ DGI$ possible) $EJ = FJ$ (making $Δ BEJ ≅ △ CFJ$ possible) Step 4: The height of the rectangle $= h$ (height of the trapezium) Step 5: The length of rectangle $= \frac{( a + b )}{2}$ where $a$ and $b$ are parallel sides This ensures: Area of rectangle $= h \times \frac{( a + b )}{2}$ = Area of trapezium

Practical construction:

Draw the trapezium ABCD Calculate required rectangle length $= \frac{( AB + CD )}{2}$ Mark points H and E on AB such that the central portion has this length.

Draw perpendiculars to get rectangle EFGH.

Verify that the triangular pieces outside match those inside.

This construction beautifully demonstrates area conservation through dissection!

6. Using the idea of converting a trapezium into a rectangle of equal area and vice versa, construct a trapezium of area $144 cm^{2}$ . Sol. Area of the trapezium $= \frac{1}{2} \times (10 + 8) \times 16$ $= 144 cm^{2}$ Area of rectangle $= 16 \times 9 = 144 cm^{2}$

7. A regular hexagon is divided into a trapezium, an equilateral triangle and a rhombus, as shown. Find the ratio of their areas.

Sol. Here total area of hexagon $= 6 \times \frac{3}{4} a^{2} = \frac{3 3}{2} a^{2}$

Equilateral triangle:

Area $= \frac{3}{4} a^{2}$

Rhombus:

Area $= 2 \times \frac{3}{4} a^{2} = \frac{3}{2} a^{2}$

Trapezium:

Remaining Area = Total Area - Triangle Area - Rhombus Area $= \frac{6 3}{4} - \frac{3}{4} a^{2} - \frac{3}{2} a^{2}$ $= \frac{3 3}{4} a^{2}$

Ratio of area

= Triangle : Rhombus : Trapezium $= \frac{3}{4} a^{2} : \frac{3}{2} a^{2} : \frac{3 3}{4} a^{2} = 1 : 2 : 3$

8. ZYXW is a trapezium with $ZY ∥ WX . A$ is the midpoint of XY . Show that the area of the trapezium ZYXW is equal to the area of $Δ ZWB .$ .

Sol. $∠ ZAY = ∠ BAX$ (Vertically opposite angles)

A Y = A X (∵ A

is mid point of

X Y)

∴ ∠ YZB = ∠ XBZ

( ∵ ZY || XB alternate interior angles are equal)

∠ ZYA = ∠ BXA

( ∵ they are alternate interior angles) So

△ ZAY ≅ △ BAX

By the AAS congruence. So, Area of

Δ ZAY =

Area of

Δ BAX

Thus, Area of trapezium ZYXW = Area of triangle ZWB

4.0Key Concepts of Chapter 7 Area Exercise 7.5

Comparing the area of a rectangle and a parallelogram having the same side lengths.
Finding an unknown side of a right-angled triangle using the Pythagoras theorem.
Constructing a rectangle with twice the area of a given triangle.
Constructing a rectangle with the same area as a given triangle using half of its base.
Converting an isosceles triangle into a rectangle using dissection.
Converting a rectangle into an isosceles triangle through cutting and rearranging.
Comparing the areas of an equilateral triangle and a square with the same side length.

5.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.5

Compares the areas of a rectangle and a parallelogram with the same side lengths.
Finds an unknown side of a right-angled triangle using the Pythagoras theorem.
Constructs a rectangle with twice the area of a given triangle.
Constructs a rectangle with the same area as a given triangle using half of its base.
Applies dissection to convert an isosceles triangle into a rectangle.
Uses cutting and rearrangement to convert 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 7.5 of Class 8 Maths Chapter 7?

How is the area of a parallelogram calculated in Exercise 7.5?

How is the unknown side of the triangle found?

How can a rectangle with twice the area of a triangle be constructed?

How can a rectangle with the same area as a triangle be obtained?

How is an isosceles triangle converted into a rectangle?

What comparison of areas is included in Exercise 7.5?

Join ALLEN!

1.0Area Exercise 7.5 Class 8 Maths – Chapter 7 NCERT Solutions

2.0Download Class 8 Maths Chapter 7 Ex 7.5 NCERT Solutions PDF

3.0Detailed NCERT Solutions for Area Exercise 7.5 Class 8 Maths (Ganita Prakash Term 2)

4.0Key Concepts of Chapter 7 Area Exercise 7.5

5.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.5

Frequently Asked Questions

What is covered in Exercise 7.5 of Class 8 Maths Chapter 7?

How is the area of a parallelogram calculated in Exercise 7.5?

How is the unknown side of the triangle found?

How can a rectangle with twice the area of a triangle be constructed?

How can a rectangle with the same area as a triangle be obtained?

How is an isosceles triangle converted into a rectangle?

What comparison of areas is included in Exercise 7.5?

Join ALLEN!

1.0Area Exercise 7.5 Class 8 Maths – Chapter 7 NCERT Solutions

2.0Download Class 8 Maths Chapter 7 Ex 7.5 NCERT Solutions PDF

3.0Detailed NCERT Solutions for Area Exercise 7.5 Class 8 Maths (Ganita Prakash Term 2)

4.0Key Concepts of Chapter 7 Area Exercise 7.5

5.0Benefits of NCERT Solutions Class 8 Maths Chapter 7 Exercise 7.5