Area — Class 8 Maths | Preeti Kushwah Classes

📐 Introduction to Area

We already know how to find the area of rectangles, squares, triangles, parallelograms, and circles. In this chapter, we extend those skills to find the areas of more complex shapes: trapeziums, rhombuses, general quadrilaterals, and polygons. We will also get our first introduction to surface area and volume of 3D solids.

The key idea throughout this chapter is splitting complex shapes into simpler shapes whose areas we already know. A trapezium can be split into triangles and rectangles. A polygon can be split into triangles. Even the surface of a 3D object can be "unfolded" into flat 2D shapes.

🇮🇳 Indian Mathematical Heritage

Ancient Indian mathematicians were pioneers in area computation. The Sulbasutras (c. 800 BCE) gave detailed methods for constructing altars of specific areas. Aryabhata (476 CE) gave the formula for the area of a triangle as "half the base times the height" in his Aryabhatiya. Brahmagupta (628 CE) discovered the formula for the area of a cyclic quadrilateral, a result that was not known in the Western world until much later.

📜 Baudhayana (~800 BCE)

The Sulbasutras described how to construct squares equal in area to given rectangles, circles, and other shapes — essentially performing area transformations thousands of years before modern geometry.

🌟 Brahmagupta (628 CE)

Brahmagupta's formula for the area of a cyclic quadrilateral with sides a, b, c, d is: Area = √[(s−a)(s−b)(s−c)(s−d)] where s is the semi-perimeter. This generalises Heron's formula!

💫 Mahavira (c. 850 CE)

Mahavira's Ganitasarasangraha contains formulas for areas of various quadrilaterals and polygons, including practical methods for land measurement that farmers and builders used.

💡 Core Strategy: To find the area of any complex shape, split it into simpler shapes (triangles, rectangles, trapeziums) whose area formulas you know. Then add up the individual areas.

▵ Area of a Trapezium

A trapezium (called "trapezoid" in some countries) is a quadrilateral with exactly one pair of parallel sides. The parallel sides are called the bases, and the perpendicular distance between them is the height.

Trapezium ABCD with parallel sides a and b, and height h

🔢 Deriving the Formula

Draw a diagonal of the trapezium to split it into two triangles. Both triangles share the same height h (the distance between the parallel sides).

Triangle 1 has base = a (top parallel side), height = h
Triangle 2 has base = b (bottom parallel side), height = h

Area of trapezium = Area of Triangle 1 + Area of Triangle 2

= ½ × a × h + ½ × b × h

= ½ × (a + b) × h

Area of Trapezium = ½ × (a + b) × h where a and b are the parallel sides, and h is the height (perpendicular distance between them)

💡 Memory Trick: "Half the sum of parallel sides, times the height." Think of it as the area of a rectangle whose width is the average of the two parallel sides: average width × height.

Example 1: Find the area of a trapezium with parallel sides 8 cm and 12 cm, and height 5 cm.

Given: a = 8 cm, b = 12 cm, h = 5 cm

Formula: Area = ½ × (a + b) × h

Substitute: Area = ½ × (8 + 12) × 5 = ½ × 20 × 5 = 50 cm²

Example 2: The area of a trapezium is 105 cm². Its parallel sides are 9 cm and 12 cm. Find the height.

Given: Area = 105 cm², a = 9 cm, b = 12 cm, h = ?

Formula: 105 = ½ × (9 + 12) × h

Solve: 105 = ½ × 21 × h ⇒ 105 = 10.5h ⇒ h = 105 / 10.5 = 10 cm

📝 Note: A parallelogram is a special trapezium where both pairs of opposite sides are parallel. If a = b, then Area = ½ × (a + a) × h = a × h, which is the area of a parallelogram!

◆ Area of a Rhombus

A rhombus is a quadrilateral with all four sides equal. Its diagonals bisect each other at right angles. This special property gives us a neat area formula using the diagonals.

Rhombus ABCD with diagonals d₁ (AC) and d₂ (BD) meeting at O

🔢 Deriving the Formula

The diagonals of a rhombus divide it into 4 right-angled triangles. Since the diagonals bisect each other:

Each triangle has base = d₂/2 and height = d₁/2
Area of one triangle = ½ × (d₂/2) × (d₁/2) = d₁d₂/8
Total area = 4 × d₁d₂/8 = ½ × d₁ × d₂

Area of Rhombus = ½ × d₁ × d₂ where d₁ and d₂ are the lengths of the two diagonals

💡 Memory Trick: "Half the product of diagonals." This formula also works for any kite (a quadrilateral with two pairs of consecutive equal sides), since the diagonals of a kite are also perpendicular.

Example 3: Find the area of a rhombus whose diagonals are 16 cm and 12 cm.

Given: d₁ = 16 cm, d₂ = 12 cm

Formula: Area = ½ × d₁ × d₂ = ½ × 16 × 12 = 96 cm²

Example 4: The area of a rhombus is 120 cm² and one diagonal is 15 cm. Find the other diagonal.

Given: Area = 120 cm², d₁ = 15 cm, d₂ = ?

Solve: 120 = ½ × 15 × d₂ ⇒ 120 = 7.5 × d₂ ⇒ d₂ = 120 / 7.5 = 16 cm

Example 5: Find the area and side of a rhombus whose diagonals are 24 cm and 10 cm.

Area: = ½ × 24 × 10 = 120 cm²

Finding side: The diagonals bisect each other at right angles. Half-diagonals = 12 cm and 5 cm.

Using Pythagoras: side = √(12² + 5²) = √(144 + 25) = √169 = 13 cm

▭ Area of a General Quadrilateral

For a general quadrilateral (one that is neither a parallelogram, nor a trapezium, nor a rhombus), we use the strategy of splitting it into two triangles by drawing one diagonal.

Quadrilateral ABCD split into two triangles by diagonal AC = d

If we draw diagonal AC = d, then:

Triangle ABC has base d and height h₁ (perpendicular distance from B to AC)
Triangle ACD has base d and height h₂ (perpendicular distance from D to AC)

Area of Quadrilateral = ½ × d × (h₁ + h₂) where d is a diagonal, and h₁, h₂ are the perpendicular distances from the other two vertices to that diagonal

Example 6: In quadrilateral ABCD, diagonal AC = 20 cm. The perpendicular distances from B and D to AC are 6 cm and 8 cm respectively. Find the area.

Given: d = 20 cm, h₁ = 6 cm, h₂ = 8 cm

Area: = ½ × 20 × (6 + 8) = ½ × 20 × 14 = 140 cm²

📝 Note: This formula is very powerful! It works for any quadrilateral, whether convex or concave. You just need the length of one diagonal and the two perpendicular heights from the opposite vertices.

⬟ Area of a Polygon

To find the area of a polygon (a closed figure with 5 or more sides), we split it into triangles and/or trapeziums whose areas we can calculate, then add them up.

🔧 Method 1: Triangulation from One Vertex

Choose one vertex and draw diagonals to all non-adjacent vertices. This divides an n-sided polygon into (n − 2) triangles.

Pentagon ABCDE divided into 3 triangles (5 − 2 = 3) from vertex A

Area of pentagon = Area(△ABC) + Area(△ACD) + Area(△ADE)

🔧 Method 2: Using Trapeziums

For irregular polygons, another effective method is to draw perpendiculars from each vertex to a fixed base line. This creates a series of trapeziums (and possibly triangles at the ends). Calculate each trapezium area and add/subtract as needed.

Example 7: Find the area of a regular hexagon with side 6 cm.

Strategy: A regular hexagon can be divided into 6 equilateral triangles, each with side 6 cm.

Area of one equilateral triangle: = (√3/4) × side² = (√3/4) × 36 = 9√3 cm²

Area of hexagon: = 6 × 9√3 = 54√3 ≈ 93.53 cm²

Area of Regular Polygon = n × (½ × side × apothem) where n is the number of sides and the apothem is the perpendicular distance from centre to a side

🔧 Area of Composite Shapes

Composite shapes (also called combined shapes) are figures made by joining two or more simple shapes, or by cutting out a shape from a larger one.

➕ Addition Method

When two or more shapes are joined together, the total area = sum of individual areas.
Example: An L-shaped room = rectangle 1 + rectangle 2.

➖ Subtraction Method

When a shape is cut out from a larger shape, the remaining area = area of larger shape − area of cut-out.
Example: A ring = area of outer circle − area of inner circle.

L-shaped figure split into simpler rectangles

Example 8: A rectangular garden 50 m × 30 m has a circular pond of radius 7 m in the middle. Find the area of the garden excluding the pond. (Use π = 22/7)

Area of garden: 50 × 30 = 1500 m²

Area of pond: πr² = (22/7) × 7 × 7 = 154 m²

Area excluding pond: 1500 − 154 = 1346 m²

Example 9: A path 2 m wide runs around the outside of a rectangular field 40 m × 25 m. Find the area of the path.

Outer dimensions: (40 + 2 + 2) × (25 + 2 + 2) = 44 m × 29 m

Outer area: 44 × 29 = 1276 m²

Inner area (field): 40 × 25 = 1000 m²

Area of path: 1276 − 1000 = 276 m²

📦 Surface Area of 3D Solids

The surface area of a 3D solid is the total area of all its outer faces. Think of it as the amount of wrapping paper needed to cover the solid completely. We study three basic solids: cube, cuboid, and cylinder.

📦 Surface Area of a Cuboid

Cuboid with length l, breadth b, and height h

A cuboid has 6 rectangular faces: top & bottom (l × b each), front & back (l × h each), left & right (b × h each).

Total Surface Area of Cuboid = 2(lb + bh + hl) Lateral Surface Area = 2h(l + b)

📦 Surface Area of a Cube

A cube is a special cuboid where l = b = h = a (all edges are equal). Each face is a square of side a.

Cube with edge length a

Total Surface Area of Cube = 6a² Lateral Surface Area = 4a²

📦 Surface Area of a Cylinder

Cylinder with radius r and height h

When we "unroll" a cylinder, the curved surface becomes a rectangle with width = circumference = 2πr and height = h.

Total Surface Area of Cylinder = 2πr(r + h) Curved (Lateral) Surface Area = 2πrh | Two circular bases = 2πr²

📦 Cuboid TSA

2(lb + bh + hl)
LSA = 2h(l + b)

📦 Cube TSA

6a²
LSA = 4a²

📦 Cylinder TSA

2πr(r + h)
CSA = 2πrh

Example 10: Find the total surface area of a cuboid 12 cm × 8 cm × 5 cm.

TSA = 2(lb + bh + hl) = 2(12×8 + 8×5 + 5×12) = 2(96 + 40 + 60) = 2 × 196 = 392 cm²

Example 11: Find the total surface area of a cube with edge 7 cm.

TSA = 6a² = 6 × 7² = 6 × 49 = 294 cm²

Example 12: A cylindrical pillar has radius 14 cm and height 40 cm. Find the cost of painting its curved surface at ₹2 per cm². (Use π = 22/7)

CSA = 2πrh = 2 × (22/7) × 14 × 40 = 2 × 22 × 2 × 40 = 3520 cm²

Cost = 3520 × 2 = ₹7040

🍂 Volume of 3D Solids

The volume of a 3D solid is the amount of space it occupies. We measure volume in cubic units (cm³, m³, litres, etc.). Remember: 1 litre = 1000 cm³ and 1 m³ = 1000 litres.

📦 Volume of a Cuboid

Volume of Cuboid = l × b × h where l is length, b is breadth, and h is height

📦 Volume of a Cube

Volume of Cube = a³ where a is the edge length

📦 Volume of a Cylinder

Volume of Cylinder = πr²h where r is the radius and h is the height

📦 Cuboid Volume

V = l × b × h
Think of stacking layers of unit cubes.

📦 Cube Volume

V = a³
All edges equal, so V = a × a × a.

📦 Cylinder Volume

V = πr²h
Base area (πr²) stacked h units high.

Example 13: A cuboidal water tank is 2.5 m long, 1.5 m wide, and 1.2 m deep. Find its capacity in litres.

Volume = l × b × h = 2.5 × 1.5 × 1.2 = 4.5 m³

Capacity = 4.5 × 1000 = 4500 litres

Example 14: How many cubes of edge 3 cm can fit inside a box 18 cm × 12 cm × 9 cm?

Volume of box: 18 × 12 × 9 = 1944 cm³

Volume of one cube: 3³ = 27 cm³

Number of cubes: 1944 / 27 = 72 cubes

Example 15: A cylindrical vessel has inner radius 7 cm and height 20 cm. How many litres of milk can it hold? (Use π = 22/7)

Volume = πr²h = (22/7) × 7² × 20 = (22/7) × 49 × 20 = 22 × 7 × 20 = 3080 cm³

In litres: 3080 / 1000 = 3.08 litres

Example 16: A cuboid has volume 720 cm³. Its length is 15 cm and breadth is 8 cm. Find its height.

V = l × b × h ⇒ 720 = 15 × 8 × h ⇒ 720 = 120h ⇒ h = 720/120 = 6 cm

💡 Unit Conversions:
1 m³ = 1,000,000 cm³ | 1 litre = 1000 cm³ | 1 m³ = 1000 litres | 1 mL = 1 cm³

✏️ NCERT-Style Worked Examples

Problem 1: A cross-section of a canal is a trapezium with the bottom width 6 m, top width 16 m, and depth 4 m. If the canal is 100 m long, find the volume of earth dug out.

Step 1: Area of trapezoidal cross-section = ½ × (6 + 16) × 4 = ½ × 22 × 4 = 44 m²

Step 2: Volume = cross-sectional area × length = 44 × 100 = 4400 m³

Cross-section of the canal (trapezium)

Problem 2: The diagonals of a rhombus are 18 cm and 24 cm. Find its (i) area, (ii) perimeter, (iii) altitude when one side is taken as base.

(i) Area: = ½ × 18 × 24 = 216 cm²

(ii) Side: Half-diagonals = 9 cm and 12 cm. By Pythagoras: side = √(9² + 12²) = √(81 + 144) = √225 = 15 cm

Perimeter: = 4 × 15 = 60 cm

(iii) Altitude: Area = base × altitude ⇒ 216 = 15 × altitude ⇒ altitude = 216/15 = 14.4 cm

Problem 3: A room is 12 m long, 9 m wide, and 4 m high. It has 2 doors (2 m × 1.5 m each) and 3 windows (1.5 m × 1 m each). Find the cost of whitewashing the walls at ₹5 per m².

Step 1: Lateral surface area of walls:
= 2h(l + b) = 2 × 4 × (12 + 9) = 8 × 21 = 168 m²

Step 2: Area of 2 doors: 2 × (2 × 1.5) = 6 m²

Step 3: Area of 3 windows: 3 × (1.5 × 1) = 4.5 m²

Step 4: Area to be whitewashed: 168 − 6 − 4.5 = 157.5 m²

Step 5: Cost: 157.5 × 5 = ₹787.50

Problem 4: A field is in the shape of a trapezium with parallel sides 25 m and 10 m, and the non-parallel sides are 14 m and 13 m. Find the area.

Step 1: Drop perpendiculars from the ends of the shorter parallel side to the longer one. The base is divided into three parts: let the left part be x, the middle part = 10 m (= top side), and right part = 25 − 10 − x = 15 − x.

Step 2: Let height = h. Using Pythagoras in left triangle: x² + h² = 14² = 196. In right triangle: (15 − x)² + h² = 13² = 169.

Step 3: Subtract: x² − (15 − x)² = 196 − 169 = 27
x² − 225 + 30x − x² = 27 ⇒ 30x − 225 = 27 ⇒ 30x = 252 ⇒ x = 8.4

Step 4: h² = 196 − (8.4)² = 196 − 70.56 = 125.44 ⇒ h = √125.44 ≈ 11.2 m

Step 5: Area = ½(25 + 10) × 11.2 = ½ × 35 × 11.2 = 196 m²

Problem 5: A closed cylinder has total surface area 462 cm² and height 7 cm. Find its radius. (π = 22/7)

TSA: 2πr(r + h) = 462
2 × (22/7) × r × (r + 7) = 462

Simplify: (44/7) × r(r + 7) = 462 ⇒ r(r + 7) = 462 × 7/44 = 73.5

Solve: r² + 7r − 73.5 = 0. By trial: r = 7 gives 7(7+7) = 98 (too big). r = 5.25 gives too small. Let us check: (44/7) × r(r+7) = 462 ⇒ r(r+7) = 73.5. Trying r = 7: 7×14 = 98 ≠ 73.5. Trying r = 5.5: 5.5 × 12.5 = 68.75. Trying r = 6: 6 × 13 = 78. Close! Let us recalculate: 462 × 7 = 3234; 3234/44 = 73.5. So r² + 7r = 73.5. Using formula: r = (−7 + √(49 + 294))/2 = (−7 + √343)/2. √343 = 7√7 ≈ 18.52. So r ≈ (18.52 − 7)/2 ≈ 5.76 cm.

Alternative (if radius is a whole number): Checking: if TSA = 2 × (22/7) × r(r + 7) = 462, then r(r+7) = 462 × 7 / 44 = 73.5. Since 73.5 is not a perfect product, the radius is not a whole number. However, if the problem intends r = 7: TSA = 2×(22/7)×7×14 = 2×22×14 = 616 cm². So for TSA = 462 with h = 7, r ≈ 5.76 cm.

📊 Summary & Formula Sheet

▵ 2D Area Formulas

Shape	Formula	Variables
Trapezium	½ × (a + b) × h	a, b = parallel sides; h = height
Rhombus	½ × d₁ × d₂	d₁, d₂ = diagonals
General Quadrilateral	½ × d × (h₁ + h₂)	d = diagonal; h₁, h₂ = perpendicular heights
Equilateral Triangle	(√3/4) × a²	a = side
Regular Hexagon	(3√3/2) × a²	a = side

📦 Surface Area Formulas

Solid	TSA	LSA / CSA
Cuboid (l, b, h)	2(lb + bh + hl)	2h(l + b)
Cube (edge a)	6a²	4a²
Cylinder (r, h)	2πr(r + h)	2πrh

🍂 Volume Formulas

Solid	Volume	Key Fact
Cuboid	l × b × h	1 m³ = 1000 litres
Cube	a³	1 litre = 1000 cm³
Cylinder	πr²h	1 mL = 1 cm³

💡 Quick Recall:

Trapezium: "Half the sum of parallels, times the height"
Rhombus: "Half the product of diagonals"
Cuboid TSA: "Two of each pair: lb, bh, hl"
Cylinder TSA: "Two circles + one rectangle (when unrolled)"
Volume: Always "base area × height"

🧠 Multiple Choice Questions (10 MCQs)

Click on an option to see if your answer is correct. The correct option will turn green.

Q1. The area of a trapezium with parallel sides 10 cm and 14 cm, and height 6 cm is:
- a) 60 cm²
- b) 72 cm²
- c) 84 cm²
- d) 144 cm²
✅ Answer: (b) 72 cm² — Area = ½ × (10 + 14) × 6 = ½ × 24 × 6 = 72 cm².
Q2. The area of a rhombus whose diagonals are 10 cm and 16 cm is:
- a) 160 cm²
- b) 26 cm²
- c) 80 cm²
- d) 40 cm²
✅ Answer: (c) 80 cm² — Area = ½ × 10 × 16 = 80 cm².
Q3. The total surface area of a cube with edge 5 cm is:
- a) 25 cm²
- b) 125 cm³
- c) 150 cm²
- d) 100 cm²
✅ Answer: (c) 150 cm² — TSA = 6a² = 6 × 25 = 150 cm².
Q4. The volume of a cuboid 8 cm × 6 cm × 5 cm is:
- a) 19 cm³
- b) 236 cm²
- c) 120 cm³
- d) 240 cm³
✅ Answer: (d) 240 cm³ — Volume = 8 × 6 × 5 = 240 cm³.
Q5. The curved surface area of a cylinder with radius 7 cm and height 10 cm is (π = 22/7):
- a) 440 cm²
- b) 880 cm²
- c) 220 cm²
- d) 308 cm²
✅ Answer: (a) 440 cm² — CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².
Q6. If the area of a trapezium is 180 cm², height is 12 cm, and one parallel side is 10 cm, the other parallel side is:
- a) 15 cm
- b) 20 cm
- c) 25 cm
- d) 30 cm
✅ Answer: (b) 20 cm — 180 = ½ × (10 + b) × 12 ⇒ 30 = 10 + b ⇒ b = 20 cm.
Q7. The lateral surface area of a cuboid 10 cm × 8 cm × 6 cm is:
- a) 480 cm²
- b) 376 cm²
- c) 216 cm²
- d) 160 cm²
✅ Answer: (c) 216 cm² — LSA = 2h(l + b) = 2 × 6 × (10 + 8) = 12 × 18 = 216 cm².
Q8. How many litres of water can a cuboidal tank 2 m × 1 m × 0.5 m hold?
- a) 100 litres
- b) 500 litres
- c) 1000 litres
- d) 2000 litres
✅ Answer: (c) 1000 litres — Volume = 2 × 1 × 0.5 = 1 m³ = 1000 litres.
Q9. The area of a quadrilateral with one diagonal 24 cm and the perpendicular distances from the other vertices to it being 5 cm and 7 cm is:
- a) 120 cm²
- b) 168 cm²
- c) 144 cm²
- d) 288 cm²
✅ Answer: (c) 144 cm² — Area = ½ × 24 × (5 + 7) = ½ × 24 × 12 = 144 cm².
Q10. The volume of a cylinder with radius 14 cm and height 10 cm is (π = 22/7):
- a) 880 cm³
- b) 4400 cm³
- c) 6160 cm³
- d) 3080 cm³
✅ Answer: (c) 6160 cm³ — V = πr²h = (22/7) × 196 × 10 = 22 × 28 × 10 = 6160 cm³.

✍️ NCERT Short Answer Questions

Q1. What is a trapezium? How is it different from a parallelogram?

A trapezium is a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs of parallel sides. Every parallelogram is a special case of a trapezium, but not vice versa.
Q2. Find the area of a trapezium with parallel sides 15 cm and 25 cm, and height 8 cm.

Area = ½ × (15 + 25) × 8 = ½ × 40 × 8 = 160 cm².
Q3. The diagonals of a rhombus are 20 cm and 48 cm. Find the side of the rhombus.

Half-diagonals = 10 cm and 24 cm. Side = √(10² + 24²) = √(100 + 576) = √676 = 26 cm.
Q4. What is the difference between total surface area and lateral surface area?

Total surface area (TSA) includes all faces of the solid. Lateral surface area (LSA) excludes the top and bottom faces — it includes only the "side" faces. For a cylinder, LSA is the curved surface area (CSA).
Q5. Find the volume of a cube whose total surface area is 384 cm².

6a² = 384 ⇒ a² = 64 ⇒ a = 8 cm. Volume = 8³ = 512 cm³.
Q6. A cuboid is 20 cm long, 15 cm wide, and 10 cm high. Find the length of its longest diagonal.

Longest diagonal = √(l² + b² + h²) = √(400 + 225 + 100) = √725 = 5√29 ≈ 26.93 cm.
Q7. Convert 2.5 m³ to litres.

1 m³ = 1000 litres, so 2.5 m³ = 2.5 × 1000 = 2500 litres.
Q8. The area of a rhombus is 210 cm² and one diagonal is 14 cm. Find the other diagonal.

210 = ½ × 14 × d₂ ⇒ d₂ = 420/14 = 30 cm.

📖 NCERT Long Answer Questions

Q1. The parallel sides of a trapezium are 40 cm and 20 cm. Its non-parallel sides are 25 cm and 13 cm. Find the area of the trapezium.

Solution: Let ABCD be the trapezium with AB = 20 cm (top), DC = 40 cm (bottom).

Drop perpendiculars from A and B to DC. Let AE = h and BF = h. Then EF = AB = 20 cm, so DE + FC = 40 − 20 = 20 cm. Let DE = x, FC = 20 − x.

In right ▵AED: x² + h² = 25² = 625 ... (i)
In right ▵BFC: (20 − x)² + h² = 13² = 169 ... (ii)

(i) − (ii): x² − (20 − x)² = 625 − 169 = 456
x² − 400 + 40x − x² = 456
40x − 400 = 456 ⇒ 40x = 856 ⇒ x = 21.4

From (i): h² = 625 − (21.4)² = 625 − 457.96 = 167.04
h = √167.04 ≈ 12.92 cm

Area = ½ × (40 + 20) × 12.92 = ½ × 60 × 12.92 ≈ 387.6 cm²
Q2. A closed rectangular box 80 cm × 60 cm × 40 cm is to be made from sheet metal. Find (i) the area of sheet required, (ii) the volume of the box, (iii) how many litres of water it can hold.

(i) Area of sheet = TSA = 2(lb + bh + hl)
= 2(80×60 + 60×40 + 40×80)
= 2(4800 + 2400 + 3200)
= 2 × 10400 = 20,800 cm²

(ii) Volume = 80 × 60 × 40 = 192,000 cm³

(iii) Capacity = 192,000 / 1000 = 192 litres
Q3. A field in the shape of a pentagon ABCDE has the following measurements: AC = 18 m, AF = 6 m (perpendicular from B to AC), AG = 8 m (perpendicular from E to AC), and the triangle ACD has base AC = 18 m with height from D = 10 m. Find the area of the field.

Solution: We split the pentagon into triangles using diagonal AC.

Area of ▵ABC = ½ × AC × (perp. from B) = ½ × 18 × 6 = 54 m²

Area of ▵ACD = ½ × AC × (perp. from D) = ½ × 18 × 10 = 90 m²

Area of ▵ACE = ½ × AC × (perp. from E) = ½ × 18 × 8 = 72 m²

Total area = 54 + 90 + 72 = 216 m²
Q4. A cylindrical roller 120 cm long has a radius of 42 cm. How many revolutions does it make to level a playground 264 m × 6.6 m? (Use π = 22/7)

Solution:
Step 1: Area covered in 1 revolution = CSA of cylinder
= 2πrh = 2 × (22/7) × 42 × 120 = 2 × 22 × 6 × 120 = 31,680 cm²

Convert to m²: 31,680 / 10,000 = 3.168 m²

Step 2: Area of playground = 264 × 6.6 = 1742.4 m²

Step 3: Number of revolutions = 1742.4 / 3.168 = 550 revolutions
Q5. The inner dimensions of a swimming pool are 50 m × 20 m × 2 m. The walls and floor are to be tiled. The tiles are 20 cm × 20 cm. Find the total number of tiles needed and the cost at ₹6 per tile.

Solution:
Area of floor: 50 × 20 = 1000 m²
Area of 4 walls (LSA): 2 × 2 × (50 + 20) = 4 × 70 = 280 m²
Total area to tile: 1000 + 280 = 1280 m²

Area of one tile: 0.2 × 0.2 = 0.04 m²
Number of tiles: 1280 / 0.04 = 32,000 tiles

Cost: 32,000 × 6 = ₹1,92,000

🌟 Fun Facts & Real-World Connections

🏟 Ancient Land Measurement

The word "geometry" comes from Greek: "geo" (earth) + "metron" (measure). Ancient Egyptians used area formulas to re-measure farmland after the annual flooding of the Nile!

🎯 Trapezium in Architecture

The famous Keystone in an arch is a trapezium shape! Its special shape locks all the other stones in place. The US Pentagon building is a regular pentagon — its area can be calculated using the polygon methods in this chapter.

🌎 Earth's Surface Area

Earth's total surface area is about 510 million km². About 71% (362 million km²) is water. To calculate the surface area of a sphere, we use 4πr² — a formula you will learn in Class 9!

📦 Packaging Science

Companies spend millions optimising the surface area of packaging boxes. A cube gives the least surface area for a given volume among all cuboids — that is why many products come in nearly cubic boxes!

🇮🇳 Sulbasutras Connection

The ancient Indian Sulbasutras contain instructions for building fire altars of specific shapes: falcons, tortoises, and chariots. The priests had to calculate the area of these complex shapes — essentially doing the same polygon decomposition we learn here!

📊 Volume Fun

An Olympic swimming pool holds about 2,500,000 litres (2500 m³) of water. That is about 2500 cubes of 1 m edge! If it were a cylinder of the same volume, you could calculate its dimensions using V = πr²h.

✅ Revision Checklist

Make sure you can confidently do each of these before the exam:

✅ State and derive the area formula for a trapezium
✅ State and derive the area formula for a rhombus using diagonals
✅ Find the area of a general quadrilateral by splitting into triangles
✅ Find the area of a polygon by triangulation or trapezium method
✅ Solve composite/combined shape problems using addition and subtraction
✅ Calculate TSA and LSA of cuboid, cube, and cylinder
✅ Calculate volume of cuboid, cube, and cylinder
✅ Convert between m³, cm³, and litres
✅ Solve real-world problems involving cost of painting, tiling, fencing
✅ Find missing dimensions given area, surface area, or volume

📐 Area

📜 Baudhayana (~800 BCE)

🌟 Brahmagupta (628 CE)

💫 Mahavira (c. 850 CE)

Example 1: Find the area of a trapezium with parallel sides 8 cm and 12 cm, and height 5 cm.

Example 2: The area of a trapezium is 105 cm². Its parallel sides are 9 cm and 12 cm. Find the height.

Example 3: Find the area of a rhombus whose diagonals are 16 cm and 12 cm.

Example 4: The area of a rhombus is 120 cm² and one diagonal is 15 cm. Find the other diagonal.

Example 5: Find the area and side of a rhombus whose diagonals are 24 cm and 10 cm.

Example 6: In quadrilateral ABCD, diagonal AC = 20 cm. The perpendicular distances from B and D to AC are 6 cm and 8 cm respectively. Find the area.

Example 7: Find the area of a regular hexagon with side 6 cm.

➕ Addition Method

➖ Subtraction Method

Example 8: A rectangular garden 50 m × 30 m has a circular pond of radius 7 m in the middle. Find the area of the garden excluding the pond. (Use π = 22/7)

Example 9: A path 2 m wide runs around the outside of a rectangular field 40 m × 25 m. Find the area of the path.

📦 Cuboid TSA

📦 Cube TSA

📦 Cylinder TSA

Example 10: Find the total surface area of a cuboid 12 cm × 8 cm × 5 cm.

Example 11: Find the total surface area of a cube with edge 7 cm.

Example 12: A cylindrical pillar has radius 14 cm and height 40 cm. Find the cost of painting its curved surface at ₹2 per cm². (Use π = 22/7)

📦 Cuboid Volume

📦 Cube Volume

📦 Cylinder Volume

Example 13: A cuboidal water tank is 2.5 m long, 1.5 m wide, and 1.2 m deep. Find its capacity in litres.

Example 14: How many cubes of edge 3 cm can fit inside a box 18 cm × 12 cm × 9 cm?

Example 15: A cylindrical vessel has inner radius 7 cm and height 20 cm. How many litres of milk can it hold? (Use π = 22/7)

Example 16: A cuboid has volume 720 cm³. Its length is 15 cm and breadth is 8 cm. Find its height.

Problem 1: A cross-section of a canal is a trapezium with the bottom width 6 m, top width 16 m, and depth 4 m. If the canal is 100 m long, find the volume of earth dug out.

Problem 2: The diagonals of a rhombus are 18 cm and 24 cm. Find its (i) area, (ii) perimeter, (iii) altitude when one side is taken as base.

Problem 3: A room is 12 m long, 9 m wide, and 4 m high. It has 2 doors (2 m × 1.5 m each) and 3 windows (1.5 m × 1 m each). Find the cost of whitewashing the walls at ₹5 per m².

Problem 4: A field is in the shape of a trapezium with parallel sides 25 m and 10 m, and the non-parallel sides are 14 m and 13 m. Find the area.

Problem 5: A closed cylinder has total surface area 462 cm² and height 7 cm. Find its radius. (π = 22/7)

🏟 Ancient Land Measurement

🎯 Trapezium in Architecture

🌎 Earth's Surface Area

📦 Packaging Science

🇮🇳 Sulbasutras Connection

📊 Volume Fun

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