Chapter Overview
2(lb + bh + hl)
Cuboid Surface Area
6l²
Cube Surface Area
2πr(r + h)
Cylinder Surface Area
l × b × h
Cuboid Volume
What You'll Learn
Perimeter and Area
Understanding perimeter as boundary distance and area as covered region for plane figures like triangles, rectangles, circles.
Area of Polygons
Splitting quadrilaterals and polygons into triangles and trapeziums to calculate areas.
Surface Area of Solids
Calculating total and lateral surface areas for cubes, cuboids, and cylinders.
Volume of Solids
Finding volumes for cubes, cuboids, and cylinders using formulas and unit cubes.
Historical Context
This chapter builds on basic plane figures to explore complex polygons and introduces 3D solids like cubes, cuboids, and cylinders. It emphasizes practical applications, such as calculating areas for fields or volumes for containers, with examples like trapezium-shaped fields and rhombus tiles.
Key Highlights
Key formulas include area of trapezium \( \frac{1}{2} h (a + b) \), rhombus \( \frac{1}{2} d_1 d_2 \), and cylinder volume \( \pi r^2 h \). Activities like dividing polygons and forming nets reinforce concepts.
Comprehensive Chapter Summary
1. Introduction to Mensuration
The chapter introduces perimeter as the distance around a closed plane figure and area as the region it covers. It reviews calculations for triangles, rectangles, circles, and borders in rectangles. It extends to quadrilaterals, polygons, and solids like cubes, cuboids, cylinders, with formulas for surface area and volume.
2. Area of a Polygon
Splitting Polygons
Quadrilaterals are split into triangles; polygons like pentagons into triangles and trapeziums using diagonals or perpendiculars. Example: Pentagon area = sum of triangle and trapezium areas.
Formula for trapezium: \( \frac{1}{2} h (a + b) \).
Examples and Try These
Activities divide polygons into parts. Example 1: Trapezium field area 480 m², height 15 m, one side 20 m, find other side (44 m).
Example 2: Rhombus area 240 cm², one diagonal 16 cm, find other (30 cm). Formula: \( \frac{1}{2} d_1 d_2 \).
Hexagon Division
Example 3: Hexagon side 5 cm divided into trapeziums or triangles/rectangle, area 64 cm² both ways.
3. Solid Shapes
Cuboid, Cube, Cylinder
Solids have faces: Cuboid (6 rectangular), Cube (6 squares), Cylinder (2 circular + curved). Activities identify congruent faces.
Right Circular Cylinder
Parallel congruent circular faces, perpendicular axis. Non-right cylinders exist but focus on right ones.
Activities
Cut boxes to observe faces; discuss why some shapes aren't cylinders.
4. Surface Area of Solids
Cuboid
Total: \( 2(lb + bh + hl) \). Lateral: \( 2h(l + b) \). Example: Room whitewashing.
Cube
Total: \( 6l^2 \). Activities form nets, calculate areas.
Cylinder
Curved: \( 2\pi r h \). Total: \( 2\pi r(r + h) \). Unrolling to rectangle.
5. Volume of Solids
Cuboid and Cube
Cuboid: \( l \times b \times h \). Cube: \( l^3 \). Use unit cubes.
Cylinder
Volume: \( \pi r^2 h \). Relation to capacity (1 L = 1000 cm³).
6. Examples and Applications
Examples calculate heights, volumes, costs for aquariums, rooms, pillars, tanks.
Questions and Answers from Chapter
Short Questions
Q1. What is the area of a trapezium with parallel sides 1 m and 1.2 m, height 0.8 m?
Answer: \( 0.88 \) m²
Q2. Find the length of the other parallel side of a trapezium with area 34 cm², one side 10 cm, height 4 cm.
Answer: 7 cm
Q3. Find the area of a rhombus with diagonals 7.5 cm and 12 cm.
Answer: 45 cm²
Q4. Find the side of a cube with surface area 600 cm².
Answer: 10 cm
Q5. Find the height of a cuboid with volume 900 cm³ and base area 180 cm².
Answer: 5 cm
Q6. Find the height of a cylinder with volume 1.54 m³ and base diameter 140 cm.
Answer: 2 m
Q7. Find the area of a rhombus with side 5 cm and altitude 4.8 cm.
Answer: 24 cm²
Q8. Find the length of the other diagonal of a rhombus with side 5 cm, one diagonal 8 cm.
Answer: 6 cm
Q9. Find the area of a trapezium field with fence 120 m, sides 48 m, 17 m, 40 m (AB perpendicular).
Answer: 660 m²
Q10. Find the area of a quadrilateral field with diagonal 24 m, perpendiculars 8 m and 13 m.
Answer: 252 m²
Q11. Find the metal sheet required for a closed cylindrical tank with radius 7 m, height 3 m.
Answer: 440 m²
Q12. Find the perimeter of a rectangular sheet from a hollow cylinder lateral area 4224 cm², width 33 cm.
Answer: 322 cm
Q13. Find the area leveled by a road roller (diameter 84 cm, length 1 m) in 750 revolutions.
Answer: 1980 m²
Q14. Find the label area for a milk powder cylinder (diameter 14 cm, height 20 cm, label 2 cm from top/bottom).
Answer: 528 cm²
Q15. If each edge of a cube is doubled, how many times does the surface area increase?
Answer: 4 times
Medium Questions
Q1. Find the area of a trapezium field with area 10500 m², height 100 m, one side along river twice the road side.
Answer: Let road side = x m, river side = 2x m. \( \frac{1}{2} \times 100 \times (x + 2x) = 10500 \), so 150x = 10500, x = 70 m, river side = 140 m. (3 marks)
Q2. Find the total cost of polishing a floor with 3000 rhombus tiles (diagonals 45 cm, 30 cm) at ₹4/m².
Answer: Tile area = \( \frac{1}{2} \times 0.45 \times 0.3 = 0.0675 \) m², total = 202.5 m², cost = ₹810. (3 marks)
Q3. For two cuboidal boxes (60 cm × 50 cm × 50 cm and 50 cm × 60 cm × 50 cm), which requires less material?
Answer: Both have same surface area 2(60×50 + 50×50 + 50×60) = 13000 cm², equal material. (3 marks)
Q4. How many meters of 96 cm wide tarpaulin for 100 suitcases (80 cm × 48 cm × 24 cm)?
Answer: Surface area per suitcase = 2(80×48 + 48×24 + 24×80) = 9888 cm², total length = (988800 / 96) × 100 = 103000 m approx. (3 marks)
Q5. Rukhsar painted a cabinet 1 m × 2 m × 1.5 m except bottom. Find surface area covered.
Answer: 2(1×2 + 2×1.5 + 1.5×1) - 1×2 = 9 m². (3 marks)
Q6. Daniel paints a hall 15 m × 10 m × 7 m with 100 m² per can. How many cans?
Answer: Area = 2(15×7 + 10×7) + 15×10 = 640 m², cans = 7. (3 marks)
Q7. Compare two figures: cylinder (r=7 cm, h=7 cm) and cube (side=7 cm). Which has larger lateral area?
Answer: Cylinder lateral 308 cm², cube 196 cm², cylinder larger. (3 marks)
Q8. How many 6 cm cubes in a cuboid 60 cm × 54 cm × 30 cm?
Answer: (60/6) × (54/6) × (30/6) = 450. (3 marks)
Q9. Milk tank cylinder r=1.5 m, length 7 m. Quantity in liters?
Answer: Volume = π(1.5)² × 7 ≈ 49.5 m³ = 49500 L. (3 marks)
Q10. If cube edge doubled, how many times volume increases?
Answer: 8 times ( (2l)^3 / l^3 = 8 ). (3 marks)
Q11. Reservoir volume 108 m³ fills at 60 L/min. Hours to fill?
Answer: 108000 L / 60 = 1800 min = 30 hours. (3 marks)
Q12. Find volume of cylinder from paper 11 cm × 4 cm folded to height 4 cm.
Answer: Circumference = 11 cm, r = 11/(2π) ≈ 1.75 cm, V ≈ 38.5 cm³. (3 marks)
Q13. Compare volumes of cylinders A (d=7 cm, h=14 cm) and B (d=14 cm, h=7 cm).
Answer: A: 539 cm³, B: 2156 cm³, B greater. (3 marks)
Q14. Godown 60 m × 40 m × 30 m, boxes 0.8 m³. How many boxes?
Answer: 72000 / 0.8 = 90000. (3 marks)
Q15. Cylinder from paper 14 cm width, radius 20 cm. Volume?
Answer: π(20)² × 14 = 17600 cm³. (3 marks)
Long Questions
Q1. Explain how to find the area of a polygon by splitting it into triangles and trapeziums, with an example from the chapter.
Answer: Polygons can be divided using diagonals or perpendiculars. For a pentagon, draw diagonal AD and perpendiculars BF, CG: area = right ∆AFB + trapezium BFGC + right ∆CGD + ∆AED. Example: AD=8 cm, AH=6 cm, AG=4 cm, AF=3 cm, BF=2 cm, CH=3 cm, EG=2.5 cm. Area ∆AFB = \( \frac{1}{2} \times 3 \times 2 = 3 \) cm², trapezium FBCH = 3 × \( \frac{2+3}{2} = 7.5 \) cm², etc., total 21 cm². This method ensures accurate calculation by summing simpler shapes.
Q2. Describe the surface area calculation for a cuboid, including lateral area, with an example.
Answer: Total surface area = 2(lb + bh + hl). Lateral = 2h(l + b). Example: Aquarium 80 cm × 30 cm × 40 cm, cover base, sides, back: base 2400 cm², side 1200 cm² each, back 3200 cm², total 8000 cm². For rooms, lateral area excludes top/bottom.
Q3. Explain volume and capacity, with relations between units and an example.
Answer: Volume is space occupied; capacity is container quantity. 1 cm³ = 1 mL, 1 L = 1000 cm³, 1 m³ = 1000 L. Example: Milk tank r=1.5 m, h=7 m, volume π(1.5)²×7 ≈ 49.5 m³ = 49500 L.
Q4. Discuss finding area of a rhombus floor with tiles and cost calculation.
Answer: 3000 tiles, diagonals 45 cm, 30 cm. Area per tile \( \frac{1}{2} \times 45 \times 30 = 675 \) cm² = 0.0675 m², total 202.5 m², cost at ₹4/m² = ₹810.
Q5. Analyze painting cost for cylindrical pillars.
Answer: 24 pillars, r=28 cm=0.28 m, h=4 m. Curved area 2πrh ≈7.04 m² each, total 168.96 m², cost ₹8/m² = ₹1351.68.
Q6. Explain cylinder height from total surface area.
Answer: r=7 cm, area=968 cm². 2πr(h+r)=968, h=15 cm.
Q7. Discuss cuboid whitewashing cost including ceiling.
Answer: Room 12 m × 8 m × 4 m. Lateral 160 m² at ₹5=₹800, ceiling 96 m²=₹480, total ₹1280.
Q8. How to find boxes in godown?
Answer: 60×40×30=72000 m³, box 0.8 m³, 90000 boxes.
Q9. Explain cylinder from rolled paper volume.
Answer: Width 14 cm = h, r=20 cm, V=π(20)²×14=17600 cm³.
Q10. Compare cylinder volumes and surfaces.
Answer: A: r=3.5 cm, h=14 cm, V=539 cm³, SA=308 cm². B: r=7 cm, h=7 cm, V=1078 cm³, SA=308 cm². B larger volume, same SA.
Q11. Discuss octagonal surface area.
Answer: Regular octagon, side 11 m. Area = 2(1 + √2)s² ≈ 475.52 m².
Q12. Explain pentagonal park area methods.
Answer: Jyoti: Trapezium + triangle. Kavita: Two triangles + rectangle. Both yield same area.
Q13. Picture frame sections area.
Answer: Outer 24×28 cm, inner 16×20 cm, equal width. Each section area = 112 cm².
Q14. Discuss road roller area.
Answer: d=84 cm, length=1 m, 750 rev, area=1980 m².
Q15. Hollow cylinder sheet perimeter.
Answer: Area 4224 cm², width 33 cm, length=128 cm, perimeter=322 cm.
