NCERT Solutions for Class 8 Maths Ganit Prakash II Chapter 7

NCERT solutions for Class 8 Mathematics Chapter 7 Area – complete answers & explanations

Class 8 Mathematics Chapter 7 Area is an important chapter that helps students understand how to measure and compare the space occupied by different shapes. In this chapter, students learn about rectangles, squares, triangles, polygons, parallelograms, rhombuses, trapeziums, and real-life applications of area. These concepts are essential for solving practical problems related to land measurement, construction, and design. A clear understanding of area also strengthens mathematical reasoning and problem-solving skills required in higher classes. This blog provides clear and reliable NCERT solutions that help students learn the correct approach to answering questions step by step. Parents and teachers can use these solutions to guide students effectively during homework and revision. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance and build strong confidence in mathematics.

What this NCERT chapter covers?

1. This chapter introduces the concept of area and explains how to calculate the area of common shapes such as rectangles and squares.  
2. Students learn how different shapes can have the same area even if their perimeters are different.  
3. The chapter explains how to find the area of paths, crosspaths, and shaded regions using simple formulas.  
4. Learners explore how triangles and polygons can be divided into smaller shapes to calculate their total area easily.  
5. Important properties of parallelograms and rhombuses are discussed along with methods to calculate their area.  
6. Students understand how rearranging shapes can help find the area of complex figures.  
7. The chapter teaches how doubling the side of a square affects its area.  
8. Real-life examples are used to explain area measurement in everyday situations.  
9. Unit conversions between square centimetres, square metres, and other units are introduced.  
10. These concepts help students prepare for exams and improve their accuracy in solving geometry problems.  
11. The chapter builds a strong foundation for advanced geometry topics in higher classes.  

How to use these NCERT solutions?

1. First, students should carefully read each question from the worksheet and attempt to solve it on their own.  
2. After solving, they can compare their answers with the provided solutions to check correctness.  
3. Students should focus on understanding the method used rather than memorising the final answer.  
4. If a mistake occurs, they should review the steps and identify where the error happened.  
5. Practice similar questions regularly to strengthen problem-solving skills.  
6. Parents can support children by checking their work and encouraging them to explain their reasoning.  
7. Teachers can use these solutions as a guide for classroom discussion and revision sessions.  
8. The answers follow the worksheet order exactly, making it easy to match each solution with its question.  
9. Regular practice using these solutions helps students become confident in geometry concepts.  

Important tips & tricks for students

1. Always write the correct formula before starting the calculation.  
2. Carefully check the units used in the question and write the answer in the correct unit.  
3. Draw neat diagrams whenever required to understand the shape clearly.  
4. Avoid skipping calculation steps, as this can lead to mistakes in the final answer.  
5. Remember that shapes with the same area can have different perimeters.  
6. When a side length is doubled, the area increases four times, not twice.  
7. Read each question carefully to identify whether you need to find total area, shaded area, or remaining area.  
8. Practice unit conversions regularly to avoid confusion during exams.  
9. Use rough work to verify calculations before writing the final answer.  

NCERT solutions – complete answer key

Rectangle and Squares

Math talk

1. Many different ways are possible. Students should draw any 4 equal-area parts.

In-text answer

1. 28 cm², 24 cm²  
2. The rectangle of dimensions 7 cm × 4 cm requires more rangoli powder.

In-text answer

1. Area of each triangle = 14 cm²

Math talk

Example:

Region 1: 9 cm × 2 cm → Perimeter = 22 cm, Area = 18 cm²  
Region 2: 6 cm × 3 cm → Perimeter = 18 cm, Area = 18 cm²  

So, a larger perimeter need not mean a larger area.

Another example:

A long thin rectangle can have larger perimeter but smaller area than a compact square.

Figure it out

1. (i) (a) 5 in (b) 7 in  
  (ii) (a) 5 m (b) 10 m  

2. (i) Required measurements:  
Length and breadth of outer rectangle ABCD, and length and breadth of inner rectangle EFGH.

Example:

AB = 12 m, BC = 8 m  
EF = 8 m, FG = 5 m  

Area of path = Area of ABCD – Area of EFGH  
= (12 × 8) – (8 × 5)  
= 96 – 40  
= 56 m²

Formula:

Area of path = (outer length × outer breadth) – (inner length × inner breadth)

(ii) Yes, if the width of the path on each side is given along with the dimensions of the inner rectangle or outer rectangle.

Example:

Inner rectangle = 8 m × 5 m  
Width of path = 2 m  

Outer rectangle = (8 + 2 + 2) m × (5 + 2 + 2) m  
= 12 m × 9 m  

Area of path = (12 × 9) – (8 × 5)  
= 108 – 40  
= 68 m²

Formula:

If inner rectangle is l × b and path width is w,  
Area of path = (l + 2w)(b + 2w) – lb

(iii) No, the area of the path does not change.

3. Required measurement:  
Width of the crosspath.

Example:

Vertical strip width = 2 m  
Horizontal strip width = 2 m  
Plot dimensions = 14 m × 12 m  

Area of vertical strip = 14 × 2 = 28 m²  
Area of horizontal strip = 12 × 2 = 24 m²  
Overlapping square = 2 × 2 = 4 m²  

Area of crosspath = 28 + 24 – 4 = 48 m²

Formula:

If plot dimensions are L and B, and path width is w,  
Area of crosspath = Lw + Bw – w²

4. Area of spiral tube = 400 sq units  
Length of straight tube = 80 units

5. Region 1: Area becomes 4 times  
Region 2: Area becomes 4 times  
Region 3: Area becomes 4 times  

Reason:

When the side of a square is doubled, the area becomes 4 times.

6. Explanation:

Draw two perpendicular lines inside the square as shown, cut along them, and rearrange the 4 parts to form a larger square with a hole in the centre.

Triangles

In-text answer

1. ∆XDC and ∆YDC have equal areas.  
2. ∆XDC and ∆YBC have equal areas.  
3. Area of ∆XDC = 10 sq units

Figure it out

1. (i) 6 cm² (ii) 8 cm²  
  (iii) 6 cm²  

2. BY = 15/4 units = 3.75 units  

3. Area of ∆SUB = 48 sq units  

4. Explanation:

Draw a diagonal in the rectangle. One of the triangles formed has half the area of the rectangle and is the required triangle.

5. Explanation:

Take two identical copies of the triangle. Join them along a suitable side to form a rectangle of equal area.

6. (i) 49 sq units  
  (ii) 60 sq units  

7. Area of ∆XMN = 1/4 of the area of ∆XYZ

8. Explanation:

Reflect the water tank across the river and draw a straight line from the house to the reflected image. The point where this line meets the river gives the shortest path.

Area of any polygon

In-text answer

1. Quadrilateral ABCD can be divided into two triangles by joining BD.

Figure it out

1. Area of quadrilateral ABCD  
= Area of ∆ABC + Area of ∆ADC  
= 1/2 × 22 × 3 + 1/2 × 22 × 3  
= 33 + 33  
= 66 cm²

2. Area of shaded region  
= Area of rectangle – Area of top-left triangle – Area of top-right triangle  
= (18 × 10) – (1/2 × 10 × 6) – (1/2 × 8 × 10)  
= 180 – 30 – 40  
= 110 cm²

3. Required measurements:  
Side length and apothem or Perimeter and apothem

4. 1/4

5. Explanation:

Draw a diagonal to divide the quadrilateral into two triangles. Join the midpoints suitably to obtain a quadrilateral of half the area.

Parallelogram

Figure it out

1. (i) All the parallelograms have equal area.  
  (ii) Their perimeters are different.  

The most slanted figure has the maximum perimeter.  
The least slanted figure has the minimum perimeter.

2. (i) 28 cm²  
  (ii) 15 cm²  
  (iii) 24 cm²  
  (iv) 8.8 cm²  

3. Area of parallelogram = base × height  
= 12 × 6 = 72 cm²  

Also,

Area = 7.6 × QN  

So,

7.6 × QN = 72  
QN = 72/7.6  
QN ≈ 9.47 cm  

4. The rectangle has the greater area.

Rectangle area = 5 × 4 = 20 cm²  

A parallelogram with the same side lengths need not have included angle 90°, so its height is less than 4 cm.

Hence its area is less than 20 cm².

Rhombus

In-text answer

Area of rhombus = 1/2 × product of diagonals

Figure it out

1. Area = 1/2 × 20 × 15 = 150 cm²

2. Explanation:

Draw both diagonals of the rectangle. Cut and rearrange the pieces so that the figure obtained is a rhombus of the same area.

3. (i) 80 ft²  
  (ii) 336 m²  
  (iii) 84 in²  
  (iv) 120 ft²  

4. Explanation:

In an isosceles trapezium, cut off the two equal triangular parts at the sides and rearrange them with the central part to form a rectangle of equal area.

5. Explanation:

Construct the rectangle so that the triangles cut from the ends of the trapezium are congruent to the triangles added to the sides of the rectangle. This gives rectangle EFGH of equal area.

6. Explanation:

Take area = 1/2 × h × (sum of parallel sides) = 144 cm²

Example:

Let height = 12 cm and sum of parallel sides = 24 cm  

Then area = 1/2 × 12 × 24 = 144 cm²  

One possible trapezium:

Parallel sides 10 cm and 14 cm, height 12 cm  

7. Ratio = 1 : 1 : 1  

8. Area of trapezium ZYXW = Area of ∆ZWB

Areas in real life

In-text answer

1. Area of A4 sheet = 21 × 29.7 = 623.7 cm²  

2. (i) 5 in = 12.7 cm  
  (ii) 7.4 in = 18.796 cm  

3. (i) 5.08 cm = 2 in  
  (ii) 11.43 cm = 4.5 in  

4. 1 in² = 6.4516 cm²  

5. 10 in² = 64.516 cm²  

6. 161.29 cm² = 25 in²  

7. 1 ft² = 144 in²  

8. 1 km² = 10,00,000 m²

Why NCERT solutions help students?

NCERT solutions help students prepare confidently for exams by providing clear and accurate methods for solving questions. They improve concept understanding, reduce mistakes, and ensure that answers match the expected NCERT approach. Regular practice using these solutions builds strong problem-solving skills and boosts confidence in mathematics learning.

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