Perimeter and Area Class 7 Free Notes and Mind Map (Free PDF Download)

Leave a Comment / Others / By
perimeter

In our daily life, we see various geometric shapes beyond the common squares and rectangles, each requiring different methods to calculate their perimeter and area. When dealing with situations such as finding the perimeter of a boundary or calculating the area of a parallelogram-shaped piece of land, we need specific methods to find accurate measurements. This chapter shows approaches to calculate the area of different shapes, including the technique of converting a parallelogram into a rectangle while keeping the same area. Understanding these concepts is important because perimeter and area calculations vary for different geometric shapes, and learning these methods helps us solve practical problems involving measurement and space.

Area of a Parallelogram

  • A parallelogram’s area equals a rectangle’s area.
  • The rectangle’s length is the parallelogram’s base.
  • The rectangle’s breadth is the parallelogram’s height.
  • Formula: Area = base × height.
  • Any side can be the base of a parallelogram.
  • Height is the perpendicular from the opposite vertex.
  • Parallelograms with equal areas can have different perimeters.
  • Parallelograms with equal perimeters can have different areas.
  • Only base and height are needed for area calculation.

Area of a Triangle

  • A triangle’s area is related to a parallelogram.
  • Two identical triangles form a parallelogram.
  • Area of each triangle is half the parallelogram’s area.
  • Formula: Area = ½ × base × height.
  • Base is any side of the triangle.
  • Height is the perpendicular from the opposite vertex.
  • All triangles on the same base and height have equal areas.
  • Equal area triangles are not always congruent.
  • Congruent triangles always have equal areas.
  • Height can be outside in obtuse-angled triangles.

Circles

Circumference of a Circle

  • Circumference is the distance around a circle.
  • It cannot be measured directly with a ruler.
  • A string can measure the circumference.
  • Circumference is related to the diameter.
  • Ratio of circumference to diameter is constant.
  • This constant is called pi (π), approximately 22/7 or 3.14.
  • Formula: Circumference = π × diameter.
  • Since diameter = 2 × radius, Circumference = 2 × π × radius.
  • Circumference helps calculate lengths for circular objects.
  • It is used for fencing or decorating circular shapes.

Area of a Circle

  • Area is needed for circular regions like gardens.
  • A circle can be approximated as a parallelogram.
  • Dividing a circle into sectors forms a rectangle.
  • Rectangle’s breadth is the circle’s radius.
  • Rectangle’s length is half the circumference.
  • Formula: Area = π × radius².
  • Area calculation is vital for practical tasks.
  • Examples include fertilising gardens or polishing tabletops.

Imp Points

  • Parallelogram area = base × height.
  • Triangle area = ½ × base × height.
  • Circumference of a circle = π × diameter or 2 × π × radius.
  • Area of a circle = π × radius².
  • Equal area shapes can have different perimeters.
  • Pi (π) is a constant, approximately 22/7 or 3.14.

Questions and Answers

Exercise 9.1

  1. Find the area of each of the following parallelograms:
    • Specific measurements are needed for each parallelogram.
    • Use formula: Area = base × height.
    • Multiply given base and height values.
    • Ensure units are consistent (e.g., cm²).
    • Answer: Apply Area = base × height for each parallelogram with given values.
  2. Find the area of each of the following triangles:
    • Use formula: Area = ½ × base × height.
    • Identify base and height for each triangle.
    • Multiply base by height and divide by 2.
    • Ensure units are in square units (e.g., cm²).
    • Answer: Calculate Area = ½ × base × height for each triangle.
  3. Find the missing values:
    • a. Base = 20 cm, Area = 246 cm²
      • Area = base × height.
      • 246 = 20 × height.
      • Height = 246 ÷ 20 = 12.3 cm.
    • b. Height = 15 cm, Area = 154.5 cm²
      • Area = base × height.
      • 154.5 = base × 15.
      • Base = 154.5 ÷ 15 = 10.3 cm.
    • c. Height = 8.4 cm, Area = 48.72 cm²
      • Area = base × height.
      • 48.72 = base × 8.4.
      • Base = 48.72 ÷ 8.4 = 5.8 cm.
    • d. Base = 15.6 cm, Area = 16.38 cm²
      • Area = base × height.
      • 16.38 = 15.6 × height.
      • Height = 16.38 ÷ 15.6 = 1.05 cm.
    • Answer: a. Height = 12.3 cm, b. Base = 10.3 cm, c. Base = 5.8 cm, d. Height = 1.05 cm.
  4. Find the missing values:
    • Base = 15 cm, Area = 87 cm²
      • Area = ½ × base × height.
      • 87 = ½ × 15 × height.
      • Height = (87 × 2) ÷ 15 = 11.6 cm.
    • Height = 31.4 mm, Area = 1256 mm²
      • Area = ½ × base × height.
      • 1256 = ½ × base × 31.4.
      • Base = (1256 × 2) ÷ 31.4 = 80 mm.
    • Base = 22 cm, Area = 170.5 cm²
      • Area = ½ × base × height.
      • 170.5 = ½ × 22 × height.
      • Height = (170.5 × 2) ÷ 22 = 15.5 cm.
    • Answer: Height = 11.6 cm, Base = 80 mm, Height = 15.5 cm.
  5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
    • a. Area of the parallelogram PQRS
      • Area = base × height.
      • Base = SR = 12 cm, Height = QM = 7.6 cm.
      • Area = 12 × 7.6 = 91.2 cm².
    • b. QN, if PS = 8 cm
      • Area = base × height.
      • Area = 91.2 cm², Base = PS = 8 cm.
      • 91.2 = 8 × QN.
      • QN = 91.2 ÷ 8 = 11.4 cm.
    • Answer: a. Area = 91.2 cm², b. QN = 11.4 cm.
  6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm², AB = 35 cm, and AD = 49 cm, find the length of BM and DL.
    • Area = base × height.
    • For base AB = 35 cm: 1470 = 35 × DL.
    • DL = 1470 ÷ 35 = 42 cm.
    • For base AD = 49 cm: 1470 = 49 × BM.
    • BM = 1470 ÷ 49 = 30 cm.
    • Answer: BM = 30 cm, DL = 42 cm.
  7. Triangle ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm, and AC = 12 cm, find the area of triangle ABC. Also find the length of AD.
    • Area = ½ × base × height.
    • Use BC as base = 13 cm, AB as height = 5 cm.
    • Area = ½ × 13 × 5 = 32.5 cm².
    • For AD (height to BC): Area = ½ × BC × AD.
    • 32.5 = ½ × 13 × AD.
    • AD = (32.5 × 2) ÷ 13 = 5 cm.
    • Answer: Area = 32.5 cm², AD = 5 cm.
  8. Triangle ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC is 6 cm. Find the area of triangle ABC. What will be the height from C to AB i.e., CE?
    • Area = ½ × base × height.
    • Base = BC = 9 cm, Height = AD = 6 cm.
    • Area = ½ × 9 × 6 = 27 cm².
    • For height CE to base AB = 7.5 cm: Area = ½ × AB × CE.
    • 27 = ½ × 7.5 × CE.
    • CE = (27 × 2) ÷ 7.5 = 7.2 cm.
    • Answer: Area = 27 cm², CE = 7.2 cm.

Exercise 9.2

  1. Find the circumference of the circles with the following radius: (Take π = 22/7)
    • a. 14 cm
      • Circumference = 2 × π × radius.
      • = 2 × 22/7 × 14 = 88 cm.
    • b. 28 mm
      • Circumference = 2 × 22/7 × 28 = 176 mm.
    • c. 21 cm
      • Circumference = 2 × 22/7 × 21 = 132 cm.
    • Answer: a. 88 cm, b. 176 mm, c. 132 cm.
  2. Find the area of the following circles, given that:
    • a. radius = 14 mm (Take π = 22/7)
      • Area = π × radius².
      • = 22/7 × 14 × 14 = 616 mm².
    • b. diameter = 49 m
      • Radius = 49 ÷ 2 = 24.5 m.
      • Area = 22/7 × 24.5 × 24.5 = 1886.5 m².
    • c. radius = 5 cm
      • Area = 22/7 × 5 × 5 = 78.57 cm².
    • Answer: a. 616 mm², b. 1886.5 m², c. 78.57 cm².
  3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)
    • Circumference = 2 × π × radius.
    • 154 = 2 × 22/7 × radius.
    • Radius = 154 × 7 / (2 × 22) = 24.5 m.
    • Area = π × radius² = 22/7 × 24.5 × 24.5 = 1886.5 m².
    • Answer: Radius = 24.5 m, Area = 1886.5 m².
  4. A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7)
    • Circumference = π × diameter = 22/7 × 21 = 66 m.
    • For 2 rounds: 2 × 66 = 132 m.
    • Cost = 132 × 4 = ₹ 528.
    • Answer: Rope length = 132 m, Cost = ₹ 528.
  5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
    • Area of larger circle = π × 4² = 3.14 × 16 = 50.24 cm².
    • Area of smaller circle = π × 3² = 3.14 × 9 = 28.26 cm².
    • Remaining area = 50.24 – 28.26 = 21.98 cm².
    • Answer: Remaining area = 21.98 cm².
  6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
    • Circumference = π × diameter = 3.14 × 1.5 = 4.71 m.
    • Cost = 4.71 × 15 = ₹ 70.65.
    • Answer: Lace length = 4.71 m, Cost = ₹ 70.65.
  7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
    • Perimeter = π × radius + diameter.
    • Assume diameter = 14 cm (from Example 10).
    • Radius = 7 cm.
    • Perimeter = (22/7 × 7) + 14 = 22 + 14 = 36 cm.
    • Answer: Perimeter = 36 cm.
  8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
    • Radius = 1.6 ÷ 2 = 0.8 m.
    • Area = π × 0.8² = 3.14 × 0.64 = 2.0096 m².
    • Cost = 2.0096 × 15 = ₹ 30.144.
    • Answer: Cost = ₹ 30.14.
  9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)
    • Circle: Circumference = 2 × π × radius = 44 cm.
    • Radius = 44 × 7 / (2 × 22) = 7 cm.
    • Area of circle = 22/7 × 7² = 154 cm².
    • Square: Perimeter = 44 cm.
    • Side = 44 ÷ 4 = 11 cm.
    • Area of square = 11 × 11 = 121 cm².
    • Circle’s area (154 cm²) > Square’s area (121 cm²).
    • Answer: Radius = 7 cm, Circle area = 154 cm², Square side = 11 cm, Circle encloses more area.
  10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed. Find the area of the remaining sheet. (Take π = 22/7)
    • Area of large circle = 22/7 × 14² = 616 cm².
    • Area of one small circle = 22/7 × 3.5² = 38.5 cm².
    • Area of two small circles = 2 × 38.5 = 77 cm².
    • Area of rectangle = 3 × 1 = 3 cm².
    • Remaining area = 616 – (77 + 3) = 536 cm².
    • Answer: Remaining area = 536 cm².
  11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
    • Area of square = 6 × 6 = 36 cm².
    • Area of circle = 3.14 × 2² = 12.56 cm².
    • Remaining area = 36 – 12.56 = 23.44 cm².
    • Answer: Remaining area = 23.44 cm².
  12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
    • Circumference = 2 × π × radius.
    • 31.4 = 2 × 3.14 × radius.
    • Radius = 31.4 ÷ (2 × 3.14) = 5 cm.
    • Area = π × 5² = 3.14 × 25 = 78.5 cm².
    • Answer: Radius = 5 cm, Area = 78.5 cm².
  13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)
    • Flower bed radius = 66 ÷ 2 = 33 m.
    • Path radius = 33 + 4 = 37 m.
    • Area of path = π × 37² – π × 33².
    • = 3.14 × (1369 – 1089) = 3.14 × 280 = 879.2 m².
    • Answer: Area of path = 879.2 m².
  14. A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
    • Garden area = 314 m².
    • Sprinkler area = π × 12² = 3.14 × 144 = 452.16 m².
    • Since 452.16 > 314, sprinkler covers entire garden.
    • Answer: Yes, the sprinkler waters the entire garden.
  15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
    • Assume outer radius = 10 cm, inner radius = 4 cm (from Example 14).
    • Outer circumference = 2 × π × 10 = 62.8 cm.
    • Inner circumference = 2 × π × 4 = 25.12 cm.
    • Answer: Outer = 62.8 cm, Inner = 25.12 cm.
  16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
    • Circumference = 2 × 22/7 × 28 = 176 cm = 1.76 m.
    • Rotations = 352 ÷ 1.76 = 200.
    • Answer: 200 rotations.
  17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)
    • In 1 hour, minute hand completes one revolution.
    • Circumference = 2 × π × 15 = 2 × 3.14 × 15 = 94.2 cm.
    • Answer: Tip moves 94.2 cm.

Download Free Mind Map from the link below

This mind map contains all important topics of this chapter

[Download PDF Here]

Visit our Main Page for Exploring Mind Maps on different topics

Exit mobile version