NCERT Solutions for Class 8 Maths Chapter 7 Area
NCERT Solutions for Class 8 Maths Chapter 7 Area
NCERT Solutions for Class 8 Maths Chapter 7 Area
This worksheet provides complete and accurate NCERT Solutions for Class 8 Mathematics Chapter 7 Area. This chapter helps students understand how to calculate the area of different shapes such as rectangles, squares, triangles, parallelograms, rhombuses, trapeziums, and polygons. It also explains how area is used in real-life situations like measuring land, designing paths, and comparing shapes. This chapter is important because it builds strong mathematical understanding and helps students solve practical geometry problems confidently. This worksheet follows the exact NCERT structure and provides reliable answers that support learning, revision, and exam preparation.
Chapter summary: stories, poems & themes
This chapter is concept-based and focuses on understanding how to measure and compare areas of different geometric shapes. Students learn how shapes can be divided into smaller parts such as triangles to calculate area easily. The chapter includes diagram-based and activity-based questions where students observe shapes, rearrange figures, and understand area relationships. It also introduces real-life applications of area measurement, such as calculating the area of paths, fields, and everyday objects. The main learning focus of this chapter is developing problem-solving skills using mathematical formulas and visual reasoning.
What this NCERT chapter covers?
• Understanding the concept of area and perimeter
• Calculating the area of rectangles and squares
• Finding the area of triangles and polygons
• Comparing areas of different shapes
• Learning formulas for parallelogram, rhombus, and trapezium
• Applying area concepts in real-life situations
• Developing logical reasoning and visual problem-solving skills
How to use these NCERT solutions?
Students should first attempt each question carefully by reading the problem and studying the diagram. After solving the question independently, they can check their answers using these NCERT Solutions to confirm accuracy. Parents and teachers can use these solutions to guide students step by step and explain mathematical methods clearly. The answers in this worksheet follow the exact NCERT order and structure, which makes revision easier and helps students build confidence in solving geometry problems.
Student tips & learning tricks
Students should always write the correct formula before starting any area calculation. It is important to read the question carefully and identify the shape correctly before applying the formula. Drawing neat diagrams and labeling measurements clearly helps avoid mistakes. Students should also check units carefully and ensure all measurements are in the same unit before calculating area. Practicing different types of problems regularly will improve speed and accuracy in exams.
Why NCERT solutions are important?
NCERT Solutions are designed according to the official curriculum and help students understand mathematical concepts clearly. Using NCERT-aligned solutions strengthens foundational learning and prepares students for school assessments and competitive exams. These solutions provide correct methods, improve problem-solving skills, and help students gain confidence in mathematics. Regular practice with accurate NCERT Solutions supports better academic performance and long-term learning success.
Complete answer key – NCERT solutions
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
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