NCERT Solutions For Class 7 Maths Chapter 9 Perimeter and Area Solutions

NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area introduce students to important basic concepts of geometry. In CBSE 7th Grade Mathematics, Chapter 9 Perimeter and Area plays an important role as it introduces students to key concepts that are useful for higher classes.

It is important to understand and get the basics of this chapter, so students should practice these solutions regularly.

With the help of NCERT Solutions for Class 7 Maths Chapter 9, students can understand every question in a simplistic manner. The step wise solutions are explained sequentially by professionals hence learning becomes easy and effective. With practice, the problem-solving skills are sharpened to give confidence in solving the tougher questions in the future.

NCERT Class 7 Maths Chapter 9 PDF

This NCERT Class 7 Maths Chapter 9 PDF is one of the most important parts of your Class 7 Maths syllabus. It teaches on how to find the perimeter and area of various forms of squares, and rectangles, triangles, parallelograms, and circles.

We have provided the free NCERT Class 7 Maths Chapter 9 PDF download link with complete solutions. Students can easily practice each question, check their answers, and build confidence. 

This chapter explains:

• Perimeter formulas of square, rectangle, triangle, parallelogram, circle
• Area formulas of different 2D shapes
• Application-based word problems
• Real-life examples of perimeter and area

NCERT Solutions Class 7 Maths Chapter 9 Perimeter and Area Textbook Exercises

Important Formulas Used in Class 7 Maths Chapter 9 Perimeter and Area

Here are some important formulas given for Class 7 students so that they can revise it properly before exams.

Important Topics Discussed in Class 7 Maths Chapter 9

Perimeter and Area is not just about numbers, but also about how shapes work in real life. This chapter understand that every boundary we see around us is a perimeter, and every surface we walk or sit on is an area.

Here are the important topics students found in this chapter:

• Perimeter of Simple Figures: Like squares, rectangles, triangles, and parallelograms. It tells us how to calculate the total boundary of a figure.
• Area of Simple Figures: Formulas to find how much space shapes like rectangles, squares, and parallelograms cover.
• Area of a Triangle: Different methods to calculate the area, including base × height formula.
• Circle: Perimeter and Area – Introduction to terms like circumference and the formula to find the area of a circle.
• Special Shapes: Perimeter and area of shapes made by combining rectangles and squares (like when we design a park or floor tiles).

Also Check:

• Class 7 Maths Sample Papers
• Class 7 Maths Revision Notes
• Class 7 Maths Worksheets
• Class 7 Maths Important Questions
• Class 7 Maths MCQ Questions

Perimeter and Area Class 7 Extra Questions with Answers

Q1. A rectangular park is 120 m long and 80 m wide. A path 5 m wide is built inside the boundary of the park. Find the area of the path.
Answer: Outer area = 120 × 80 = 9600 m²
Inner rectangle = (120 – 2×5) × (80 – 2×5) = 110 × 70 = 7700 m²
Path area = 9600 – 7700 = 1900 m²

Q2. A square playground has a side of 50 m. Four small square flowerbeds each of side 5 m are made at the corners. Find the area left for playing.
Answer: Total area = 50 × 50 = 2500 m²
Area of 4 small squares = 4 × (5 × 5) = 100 m²
Remaining area = 2500 – 100 = 2400 m²

Q3. A rectangular field is 60 m long and 40 m wide. If a wire fencing is done 3 times around it, find the total length of wire needed.
Answer: Perimeter = 2(60 + 40) = 200 m
For 3 rounds = 200 × 3 = 600 m

Q4. A square and a rectangle have the same perimeter of 96 m. If the length of rectangle is 30 m, find its breadth and compare areas of both shapes.
Answer: Square: side = 96 ÷ 4 = 24 m, Area = 576 m²
Rectangle: Perimeter = 96 → 2(30 + b) = 96 → b = 18 m → Area = 540 m²
Difference = 36 m²

Q5. A carpet covers the entire floor of a rectangular hall of length 15 m and breadth 12 m. If 1 m² of carpet costs ₹120, find the cost of carpeting.
Answer: Area = 15 × 12 = 180 m²
Cost = 180 × 120 = ₹21,600

Triangle & Parallelogram Based Questions

Q6. The base of a triangle is 20 cm, and height is 15 cm. Find the area. Also, if another triangle has the same area but base 25 cm, find its height.
Answer: Area = ½ × 20 × 15 = 150 cm²
For 2nd triangle: 150 = ½ × 25 × h → h = 12 cm

Q7. A parallelogram has base 18 cm and height 12 cm. Another parallelogram has same base but half the area. Find its height.
Answer: Area1 = 18 × 12 = 216 cm²
Area2 = 108 cm² = 18 × h → h = 6 cm

Q8. A triangle and a rectangle have equal area. The rectangle’s length = 20 cm, breadth = 12 cm. Triangle base = 24 cm. Find its height.
Answer: Rectangle area = 20 × 12 = 240 cm²
Triangle area = ½ × 24 × h = 240 → h = 20 cm

Q9. A parallelogram has area 180 cm². Its base is 15 cm. Find the height.
Answer: Area = base × height → 180 = 15 × h → h = 12 cm

Q10. A triangular park has base 120 m and height 80 m. Around it, a road of 2 m width is constructed. Find the area of road.
Answer: Area of park = ½ × 120 × 80 = 4800 m²
Approx outer triangle base = 124 m, height ≈ 84 m
Outer area ≈ ½ × 124 × 84 = 5208 m²
Road area ≈ 5208 – 4800 = 408 m²

Circle Based Questions

Q11. Find the area of a circular garden whose radius is 21 m. Use π = 22/7.
Answer: Area = πr² = (22/7) × 21 × 21 = 1386 m²

Q12. A circular pond has a diameter of 28 m. Find its circumference.
Answer: Circumference = πd = (22/7) × 28 = 88 m

Q13. A circular park has radius 14 m. A path of width 2 m is built around it. Find the area of the path.
Answer: Outer radius = 16 m → Area = π(16²) = 804.25 × 2 ≈ 804.25?? (Recalc properly)
Wait I’ll correct in explanation:
Area outer = π × 16² = (22/7) × 256 = 804.57
Area inner = (22/7) × 14² = 616
Path area = 804.57 – 616 ≈ 188.57 m²

Q14. A wire of length 176 m is bent into a circle. Find its radius.
Answer: Circumference = 176 → 2πr = 176 → r = 176 ÷ (2 × 22/7) = 28 m

Q15. Find the cost of fencing a circular garden of radius 17.5 m at ₹120 per metre.
Answer: Circumference = 2πr = 2 × 22/7 × 17.5 = 110 m
Cost = 110 × 120 = ₹13,200

Mixed Word Problems

Q16. A rectangular sheet of paper is 40 cm by 30 cm. A circle of maximum radius is cut out. Find the area of paper left.
Answer: Circle radius = 15 cm → Area circle = 706.5 cm²
Rectangle area = 1200 cm²
Remaining = 1200 – 706.5 = 493.5 cm²

Q17. A garden in shape of rectangle is 50 m × 40 m. A square pond of side 20 m is dug inside. Find area left for plantation.
Answer: Rectangle area = 2000 m²
Square pond = 400 m²
Remaining = 1600 m²

Q18. A circular path runs around a square park of side 70 m. If width of path is 7 m, find area of path.
Answer: Outer radius = 35 + 7 = 42 m, inner = 35 m
Path area = π(42² – 35²) = (22/7) × (1764 – 1225) = (22/7) × 539 = 1694 m²

Q19. A rectangular hall is 24 m × 15 m. A carpet of 2 m width is laid along inside boundary. Find carpet area.
Answer: Outer area = 360 m²
Inner = (24 – 4) × (15 – 4) = 20 × 11 = 220 m²
Carpet area = 360 – 220 = 140 m²

Q20. A farmer wants to fence his rectangular land of 120 m × 90 m with 4 strands of wire. Find total wire length.
Answer: Perimeter = 2(120 + 90) = 420 m
Total wire = 420 × 4 = 1680 m

Challenging Conceptual Problems

Q21. The perimeter of a rectangle is 96 cm. If its length is double of breadth, find its area.
Answer: Let breadth = x, length = 2x
Perimeter = 2(2x + x) = 6x = 96 → x = 16, length = 32
Area = 32 × 16 = 512 cm²

Q22. The length of rectangle is 4 cm more than breadth. If perimeter = 64 cm, find its dimensions and area.
Answer: Let breadth = x, length = x + 4
Perimeter = 2(x + x+4) = 64 → 2(2x+4) = 64 → 2x+4 = 32 → x=14
Length = 18, breadth = 14, area = 252 cm²

Q23. A rope of length 44 m is bent to form a square. Find the area of square.
Answer: Side = 44 ÷ 4 = 11 m, Area = 121 m²

Q24. The perimeter of a square and a rectangle are equal. If square’s side = 14 cm and rectangle length = 18 cm, find breadth of rectangle.
Answer: Square perimeter = 56 cm → Rectangle perimeter = 56
2(18 + b) = 56 → b = 10 cm

Q25. A triangular park has sides 50 m, 60 m and 70 m. Find its area using Heron’s formula.
Answer: s = (50+60+70)/2 = 90
Area = √[90(90–50)(90–60)(90–70)] = √[90×40×30×20]
= √(2160000) = 1469.7 m²

Q26. A rectangular field is twice as long as it is broad. If area = 392 m², find its perimeter.
Answer: Let breadth = x, length = 2x
Area = 2x × x = 2x² = 392 → x²=196 → x=14, length=28
Perimeter=2(28+14)=84 m

Q27. A path of 2 m wide runs inside a square field of side 50 m. Find the area of path.
Answer: Outer = 2500 m²
Inner = (50–4)² = 46²=2116
Path=384 m²

Q28. The diameter of a wheel is 84 cm. Find distance covered in 100 revolutions.
Answer: Circumference = πd = (22/7)×84=264 cm
100 rev=264×100=26400 cm=264 m

Q29. A circular park of radius 20 m has a path of width 2 m running around it. Find cost of cementing path at ₹150 per m².
Answer: Outer=π×22²=1519.76, Inner=π×20²=1256
Path=263.76, Cost≈₹39,564

Q30. The perimeter of a semicircular garden is 72 m. Find its radius.
Answer: Perimeter = πr + 2r = 72
r(π+2) =72 → r=72/(22/7+2)=72/(36/7)=14 m

Chapter Wise NCERT Solutions for Class 7 Maths Free PDF