Area

CHAPTER 7: AREA

COMPREHENSIVE NOTES FOR CLASS 8 MATHEMATICS (GANITA PRAKASH)

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7.1 RECTANGLE AND SQUARES

Understanding Area Through Unit Squares

**Definition of Area**: Area is the measure of the surface enclosed within a closed figure. We measure area by counting the number of **unit squares** (squares with side length 1 cm, 1 m, etc.) that can fit inside the region without overlapping.

**Key Insight**: Different shapes can be divided into unit squares or fractional parts of unit squares. By counting these, we find the total area.

**Real-life Context — Rangoli Powder**: When decorating rangoli (traditional Indian art form using colored powder), we need to know how much powder is required. The amount needed depends on the area being colored, not the perimeter. For example:

A rectangle 7 cm × 4 cm needs powder for 28 unit squares

A rectangle 8 cm × 3 cm needs powder for 24 unit squares

Even though the second rectangle appears long, it requires less powder

Area Formula for Rectangle

**Formula**: Area of a rectangle = **length × width**

**Why this works**: If you place a rectangle on a grid with 1 cm × 1 cm squares:

Each row contains 'width' number of unit squares

There are 'length' number of rows

Total = length × width unit squares

**Units**: Area is expressed in square units, written as:

Square centimeters (cm² or sq. cm)

Square meters (m² or sq. m)

Square kilometers (km² or sq. km)

**Example 1**: A rectangular park has length 50 m and width 30 m.

Area = 50 × 30 = 1500 m²

**Example 2 (Indian Context)**: A farmer's rectangular field is 60 m long and 40 m wide. He needs to apply fertilizer costing ₹5 per m².

Area = 60 × 40 = 2400 m²

Total cost = 2400 × 5 = ₹12,000

Area of Triangles from Rectangles

**Key Discovery**: When you draw a diagonal in a rectangle, it divides the rectangle into two **congruent (identical) triangles**.

Therefore: **Area of each triangle = (1/2) × Area of rectangle**

**Example**: A rectangle has length 7 cm and width 4 cm. Its diagonal creates two triangles.

Area of rectangle = 7 × 4 = 28 cm²

Area of each triangle = 28/2 = 14 cm²

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CRITICAL CONCEPT: Why Perimeter Cannot Measure Area

**Important Understanding**: Many students confuse perimeter with area. This is a fundamental error.

**Perimeter** = Length of the boundary (measured in cm, m, etc.)

**Area** = Surface enclosed (measured in cm², m², etc.)

Why Perimeter Fails as a Measure of Area

**Reason 1**: Two regions can have the **same perimeter but different areas**.

**Example**:

Rectangle A: length 10 cm, width 2 cm → Perimeter = 24 cm, Area = 20 cm²

Rectangle B: length 6 cm, width 6 cm → Perimeter = 24 cm, Area = 36 cm²

Same perimeter (24 cm) but different areas (20 cm² vs 36 cm²)

**Reason 2**: One region can have a **larger perimeter but smaller area** than another.

**Real-life Example (Indian Context)**: Two plots of land being sold:

Long narrow plot: 100 m × 5 m → Perimeter = 210 m, Area = 500 m²

Compact plot: 22 m × 22 m → Perimeter = 88 m, Area = 484 m²

The narrow plot has MORE perimeter but LESS area

This shows why farmers must understand area, not just boundary length

**Why This Matters**: When buying land for construction or farming, the area determines how much space you have. The perimeter only tells how much fencing is needed.

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TRIANGLES: COMPLETE TREATMENT

Developing the Triangle Area Formula

**How to find the area of a triangle**: Enclose the triangle in a rectangle using the triangle's base and height.

**Method for Right-angled Triangle**:

Consider triangle ABC where BC is the base. Draw a line parallel to BC through point A (call it line l). The rectangle BCDE is formed when we complete it with perpendiculars.

Key observations:

The diagonal of rectangle BCDE creates two congruent triangles

Triangle ABC is exactly half of the rectangle BCDE

Therefore: Area(△ABC) = (1/2) × Area(rectangle)

**The Height**: The **height** of a triangle is the perpendicular distance from a vertex to the opposite side (or the line containing that side). This perpendicular is drawn from the vertex to the base.

General Triangle Area Formula

**Formula**: Area of a triangle = **(1/2) × base × height**

In symbols: A = (1/2) × b × h

where:

**base** = length of any side of the triangle

**height** = perpendicular distance from the opposite vertex to the base

**Critical Points**:

The height MUST be perpendicular to the base

Any side can be chosen as the base

The corresponding height changes based on which side is the base

Why This Formula Works for ALL Triangles

**For Acute-angled Triangle**: The foot of the perpendicular lies on the base itself.

**For Right-angled Triangle**: One side is already perpendicular to another, so the height equals the side.

**For Obtuse-angled Triangle**: The foot of the perpendicular lies outside the base. The formula still works!

**Proof for Obtuse Triangle**:

Consider triangle ABC where angle C is obtuse. Draw height h from A perpendicular to line BC (extended).

Let D be the foot of perpendicular on the extended line, dividing BC.

Area(△ABC) = Area(△ADC) - Area(△ADB)

= (1/2) × h × DC - (1/2) × h × DB

= (1/2) × h × (DC - DB)

= (1/2) × h × BC

The formula holds for all triangles!

Worked Examples

**Example 1**: Find the area of a triangle with base 8 cm and height 6 cm.

Area = (1/2) × 8 × 6 = (1/2) × 48 = 24 cm²

**Example 2 (Indian Context)**: A triangular agricultural field has base 40 m and height 25 m. The farmer wants to know the area to determine fertilizer needed.

Area = (1/2) × 40 × 25 = (1/2) × 1000 = 500 m²

**Example 3**: Using the formula in reverse — finding height when area is known.

A triangle has area 60 cm² and base 12 cm. Find the height.

60 = (1/2) × 12 × h

60 = 6h

h = 10 cm

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TRIANGLES WITH SAME BASE AND HEIGHT

Important Theorem

**Theorem**: All triangles with the same base and whose third vertex lies on a line parallel to that base have equal areas.

**Proof**: If BC is the base and vertex A lies on line l parallel to BC, then:

All triangles have the same base = BC

All triangles have the same height = perpendicular distance from l to BC

Therefore, all have the same area = (1/2) × BC × h

**Visual Understanding**: If you move point A along the line l, keeping the same base BC, the area never changes because the height remains constant.

**Example**:

Triangle with base 10 cm and height 5 cm has area = 25 cm²

Any other triangle with the same base 10 cm and height 5 cm also has area = 25 cm²

This is true regardless of where the third vertex is (as long as it maintains height 5 cm)

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MEDIANS AND EQUAL AREAS

Property of Medians

**Definition of Median**: A **median** of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

**Key Theorem**: A median of a triangle divides it into two triangles of equal area.

**Proof**: In triangle ABC, let M be the midpoint of BC. Draw median AM.

For triangle ABM:

Base = BM = BC/2

Height = h (perpendicular from A to BC)

Area(△ABM) = (1/2) × BM × h = (1/2) × (BC/2) × h

For triangle AMC:

Base = MC = BC/2

Height = h (same perpendicular from A to BC)

Area(△AMC) = (1/2) × MC × h = (1/2) × (BC/2) × h

Therefore: Area(△ABM) = Area(△AMC)

**Practical Insight**: If you draw all three medians in a triangle, they create 6 smaller triangles, all with equal area. This is called the **centroid property**.

Application: Diagonals of Rectangle

**Theorem**: The diagonals of a rectangle divide it into 4 triangles, all with equal area.

**Proof**: Let ABCD be a rectangle with diagonals AC and BD intersecting at O.

Consider triangles AOB and COD:

OB = OD (diagonals of rectangle bisect each other)

If OB is the base of △AOB and OD is the base of △COD

Both triangles have the same height from the base to their opposite vertices

Therefore, Area(△AOB) = Area(△COD)

By similar reasoning: Area(△BOC) = Area(△AOD) = Area(△AOB) = Area(△COD)

All four triangles have equal area, each being 1/4 of the rectangle's area.

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TRIANGLES BETWEEN PARALLEL LINES

Maximum Area Theorem

**Theorem**: Among all triangles with a fixed base BC and the third vertex on a line l parallel to BC, all triangles have the **same maximum area** (which equals (1/2) × base × height).

**Why**: Since all vertices lie on the same line parallel to the base, the height is constant for all such triangles.

Minimum Perimeter Theorem (Reflection Principle)

**Problem**: Given a line l and two points B and C on the same side of l, find a point A on l such that the perimeter of triangle ABC is minimum.

**Solution**: Use the **reflection principle**.

**Method**:

1. Reflect point C across line l to get point C'

2. Draw a straight line from B to C'

3. This line intersects l at point A

4. Triangle ABC (with this A) has minimum perimeter

**Why This Works**:

When C is reflected to C' across line l, the distance from any point A on l to C equals the distance from A to C' (mirror property)

AB + AC = AB + AC' (since AC = AC')

The sum AB + AC' is minimized when A lies on the straight line BC'

The shortest path between two points is a straight line

Therefore, the minimum perimeter path is B → A → C where A is on line BC'

**Verification**: To check if A lies on the perpendicular bisector of BC:

Draw perpendicular bisector of BC

Reflect principle gives a different point (unless B and C are equidistant from line l)

These are different methods giving different results

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APPLICATIONS OF TRIANGLE AREA FORMULA

Finding Unknown Sides Using Equal Areas

**Concept**: A triangle has fixed area. If we express this area using different bases and heights, we can find unknown measurements.

**Example**:

In triangle ABC:

Area = (1/2) × AX × BC = (1/2) × 5 × 6 = 15 cm²

The same triangle can also be written as:

Area = (1/2) × BY × AC = 15 cm²

If AC = 10 cm:

(1/2) × BY × 10 = 15

5 × BY = 15

BY = 3 cm

**Step-by-Step Approach**:

1. Calculate area using one base-height pair

2. Use the same area with a different base-height pair

3. Solve for the unknown

**Real-life Application (Indian Context)**: A triangular plot has area 1000 m². If we measure the distance from one corner to the opposite side (height) as 50 m, we can find the length of that side.

1000 = (1/2) × base × 50

base = 40 m

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AREA OF ANY POLYGON

General Approach: Breaking into Triangles

**Method**: Any polygon can be divided into triangles by drawing diagonals from one vertex to all other non-adjacent vertices.

**Process**:

1. Choose any vertex

2. Draw diagonals to all other non-adjacent vertices

3. This creates a set of triangles

4. Find the area of each triangle

5. Add all areas to get the polygon's area

**Why This Works**:

The triangles don't overlap

Together they completely cover the polygon

Total area of triangles = Total area of polygon

Area of Quadrilateral

**For a general quadrilateral ABCD**:

1. Draw diagonal AC

2. This divides ABCD into two triangles: △ABC and △ACD

3. Area(ABCD) = Area(△ABC) + Area(△ACD)

4. Calculate each triangle's area using the formula

5. Add them

**Example**: Quadrilateral ABCD where:

AC = 20 cm (diagonal)

Perpendicular distance from B to AC = 4 cm

Perpendicular distance from D to AC = 3 cm

Area(△ABC) = (1/2) × 20 × 4 = 40 cm²

Area(△ACD) = (1/2) × 20 × 3 = 30 cm²

Area(ABCD) = 40 + 30 = 70 cm²

Area of Pentagon and Higher Polygons

Same principle applies. A pentagon is divided into 3 triangles, a hexagon into 4 triangles, and so on.

**For an n-sided polygon**: It can be divided into (n-2) triangles.

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PARALLELOGRAM

Understanding Parallelogram Area Through Dissection

**Key Idea**: A parallelogram can be rearranged into a rectangle of the same area through **dissection** (cutting and rearranging pieces).

Derivation of Parallelogram Area Formula

**Process**:

Consider parallelogram ABCD where AB ∥ DC.

1. Draw perpendicular AX from A to line DC (extended if necessary). Call AX the **height** of the parallelogram.

2. Cut the parallelogram along AX into:

Triangle AXD

Trapezium ABCX

3. Rearrange: Slide triangle AXD to the right and place it next to trapezium ABCX to form rectangle ABYX.

**Verification that ABYX is a rectangle**:

∠AXB = 90° (we constructed AX ⊥ DC)

Since AB ∥ XC, interior angles sum to 180°, so ∠ABC = 90°

The third angle at Y is 90° by construction

**Verification that △AXD ≅ △BYC**:

BY = AX (opposite sides of rectangle ABYX)

∠BYC = ∠AXD = 90°

BC = AD (opposite sides of parallelogram ABCD)

By RHS (Right angle-Hypotenuse-Side) congruency, △AXD ≅ △BYC

**Area Relationship**:

Area(parallelogram ABCD) = Area(rectangle ABYX)

= AX × XY

= height × base

Since DX = CY (congruent triangles), we have:

DC = DX + XC + CY = DX + XC + DX = 2DX + XC

But more directly: DC = XY (by construction of the rectangle)

Parallelogram Area Formula

**Formula**: Area of a parallelogram = **base × height**

where:

**base** = length of any side

**height** = perpendicular distance from that side to the parallel opposite side

**Critical Notes**:

The height must be perpendicular to the base

Any side can be chosen as the base

The height changes if you choose a different base

Height is NOT the length of a slanted side

**Units**: Square units (cm², m², etc.)

Different Bases, Different Heights

**Important Truth**: The area remains the same regardless of which base you choose, but the corresponding height changes.

For parallelogram ABCD:

If AB is base: height = perpendicular distance from AB to DC

If AD is base: height = perpendicular distance from AD to BC

Area = AB × h₁ = AD × h₂

This means: AB × h₁ = AD × h₂

Worked Examples

**Example 1**: A parallelogram has base 12 cm and height 8 cm.

Area = 12 × 8 = 96 cm²

**Example 2**: A parallelogram has sides 10 cm and 8 cm. The height corresponding to the 10 cm base is 6 cm.

Area = 10 × 6 = 60 cm²

**Example 3 (Indian Context)**: A parallelogram-shaped field has:

One side = 50 m

Perpendicular height to that side = 20 m

Area = 50 × 20 = 1000 m²

**Example 4**: Finding height when area and base are known.

A parallelogram has area 120 cm² and base 15 cm. Find the height.

120 = 15 × h

h = 8 cm

Parallelogram vs Rectangle with Same Sides

**Key Comparison**: A rectangle and a parallelogram with the same base lengths have different areas (unless the parallelogram is actually a rectangle).

**Reason**: The rectangle has height = side length, while a non-rectangular parallelogram has height < side length (the height is the perpendicular, which is less than the slanted side).

**Consequence**: Among all parallelograms with the same base and side lengths, the rectangle has the maximum area.

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RHOMBUS

What is a Rhombus?

**Definition**: A rhombus is a special parallelogram where all four sides are equal in length.

**Properties**:

All sides equal

Opposite angles equal

Diagonals bisect each other at right angles (perpendicular)

Diagonals bisect the vertex angles

Area Formula for Rhombus (Using Diagonals)

Since a rhombus is a parallelogram, we can use: Area = base × height

However, rhombus has a special property: its diagonals are perpendicular. This gives us another formula.

**Derivation**:

Consider rhombus ABCD with diagonals AC and BD intersecting at O.

The diagonals divide the rhombus into 4 triangles: △AOB, △BOC, △COD, and △DOA.

Each triangle has:

One leg along a diagonal (which is the height)

Another leg along the other diagonal (which is the base)

For △AOB:

Base = AO = AC/2

Height = BO = BD/2

Area(△AOB) = (1/2) × (AC/2) × (BD/2) = (1/8) × AC × BD

By symmetry, all 4 triangles have equal area.

Total area of rhombus = 4 × (1/8) × AC × BD

= (1/2) × AC × BD

**Formula**: Area of rhombus = **(1/2) × d₁ × d₂**

where:

**d₁** = length of one diagonal

**d₂** = length of the other diagonal

**Proof Using Triangle Areas**:

The rhombus is divided by diagonal BD into two triangles: △ABD and △CBD.

In △ABD:

Base = BD

Height = AO (perpendicular distance from A to BD) = AC/2

Area(△ABD) = (1/2) × BD × (AC/2)

In △CBD:

Base = BD

Height = CO (perpendicular distance from C to BD) = AC/2

Area(△CBD) = (1/2) × BD × (AC/2)

Area(rhombus) = Area(△ABD) + Area(△CBD)

= (1/2) × BD × (AC/2) + (1/2) × BD × (AC/2)

= (1/2) × BD × AC

= (1/2) × d₁ × d₂

Worked Examples

**Example 1**: A rhombus has diagonals of length 8 cm and 6 cm.

Area = (1/2) × 8 × 6 = (1/2) × 48 = 24 cm²

**Example 2**: A rhombus has area 100 cm² and one diagonal is 16 cm. Find the other diagonal.

100 = (1/2) × 16 × d₂

100 = 8 × d₂

d₂ = 12.5 cm

**Example 3 (Indian Context)**: A rhombus-shaped park has diagonals 60 m and 80 m.

Area = (1/2) × 60 × 80 = (1/2) × 4800 = 2400 m²

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TRAPEZIUM

Definition of Trapezium

**Definition**: A trapezium is a quadrilateral with exactly one pair of parallel sides.

**Terminology**:

**Parallel sides** = bases (often called b₁ and b₂)

**Height** = perpendicular distance between the parallel sides

**Non-parallel sides** = legs

Deriving Trapezium Area Formula

**Method: Breaking into Simpler Figures**

Consider trapezium WXYZ where WX ∥ ZY.

1. Drop perpendiculars from W and X to line ZY, meeting at M and N respectively.

2. This creates:

Triangle WMZ (on the left)

Rectangle WXNM (in the middle)

Triangle XNY (on the right)

3. Area(trapezium) = Area(△WMZ) + Area(rectangle WXNM) + Area(△XNY)

**Calculation**:

Let:

MZ = x (part of base ZY)

WM = h (height of trapezium)

WX = a (one parallel side)

NY = y (another part of base ZY)

ZY = base (the other parallel side)

From the diagram: ZY = MZ + MN + NY = x + a + y

Therefore: x + y = ZY - a

Area(trapezium) = (1/2) × x × h + a × h + (1/2) × y × h

= h × [(1/2)x + a + (1/2)y]

= h × [(1/2)(x + y) + a]

= h × [(1/2)(ZY - a) + a]

= h × [(1/2)ZY + a - (1/2)a]

= h × [(1/2)ZY + (1/2)a]

= h × (1/2)(ZY + a)

= (1/2) × h × (ZY + a)

If we denote the parallel sides as b₁ and b₂:

**Formula**: Area of trapezium = **(1/2) × h × (b₁ + b₂)**

where:

**h** = height (perpendicular distance between parallel sides)

**b₁, b₂** = lengths of the two parallel sides

**Alternative Form**: Area = **h × (average of the two parallel sides)**

This means: Area = h × [(b₁ + b₂)/2]

Understanding the Formula

**Why the sum of bases?** The trapezium has two parallel sides of different lengths. The average of these two lengths, when multiplied by the height, gives the area.

**Geometric Insight**: Imagine the trapezium as being "filled" with horizontal slices. These slices average between the two base lengths, so their typical width is (b₁ + b₂)/2.

Worked Examples

**Example 1**: A trapezium has parallel sides of 8 cm and 12 cm, with height 5 cm.

Area = (1/2) × 5 × (8 + 12)

= (1/2) × 5 × 20

= (1/2) × 100

= 50 cm²

**Example 2**: A trapezium has area 60 m², one parallel side 10 m, and height 6 m. Find the other parallel side.

60 = (1/2) × 6 × (10 + b₂)

60 = 3 × (10 + b₂)

20 = 10 + b₂

b₂ = 10 m

**Example 3 (Indian Context)**: A trapezoidal agricultural field has:

One parallel side = 80 m

Other parallel side = 60 m

Height (perpendicular distance between them) = 40 m

Area = (1/2) × 40 × (80 + 60)

= (1/2) × 40 × 140

= 20 × 140

= 2800 m²

**Example 4**: Finding height when area and bases are known.

A trapezium has area 100 cm², parallel sides 12 cm and 18 cm. Find height.

100 = (1/2) × h × (12 + 18)

100 = (1/2) × h × 30

100 = 15h

h = 100/15 = 6.67 cm

Special Case: Trapezium Where Both Parallel Sides are Equal

If b₁ = b₂ = b, then:

Area = (1/2) × h × (b + b) = (1/2) × h × 2b = b × h

This is the area formula for a parallelogram! (Because a trapezium with equal parallel sides is a parallelogram.)

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COMMON ERRORS AND MISCONCEPTIONS

Error 1: Confusing Height with Slanted Side

**Mistake**: Using the length of a slanted side as the height.

**Correct**: Height must be perpendicular to the base.

**Example**:

WRONG: In a parallelogram with side 10 cm (slanted), using 10 cm as height

CORRECT: Measure the perpendicular distance; it will be less than 10 cm

Error 2: Using Perimeter to Estimate Area

**Mistake**: Assuming larger perimeter means larger area.

**Correct**: Perimeter and area are independent.

A 100 m × 1 m rectangle has perimeter 202 m and area 100 m²

A 20 m × 5 m rectangle has perimeter 50 m and area 100 m²

Same area, different perimeters!

Error 3: Forgetting Units

**Mistake**: Writing area as just "100" without units.

**Correct**: Always include unit squares.

Write as 100 cm², not just 100

Error 4: Wrong Base-Height Pair for Triangles

**Mistake**: Using unrelated base and height measurements.

**Correct**: The height must be perpendicular to the chosen base.

**Example**:

Triangle with base 8 cm

Height to that base is 6 cm

Area = (1/2) × 8 × 6 = 24 cm² ✓

NOT: (1/2) × 8 × (any other measurement)

Error 5: Assuming All Parallelograms with Same Sides Have Same Area

**Mistake**: Thinking all shapes with sides 5 cm and 8 cm have the same area.

**Correct**: The angle between sides matters. A thin slanted parallelogram has less area than a rectangle with the same sides.

Error 6: Trapezium Formula Confusion

**Mistake**: Using base × height instead of (1/2) × height × (sum of parallel sides).

**Correct**: Area = (1/2) × h × (b₁ + b₂)

The factor 1/2 appears because we're averaging two different bases.

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SPECIAL PROPERTIES AND RELATIONSHIPS

Dissection Property

**Definition**: **Dissection** is the process of cutting a figure into pieces and rearranging them to form a different figure of the same area.

**Applications**:

1. Converting parallelogram to rectangle

2. Converting rhombus to rectangle

3. Converting triangle to rectangle

4. Converting one shape to another while preserving area

**Historical Note**: Ancient Indian texts called **Śulba-Sūtras** contain many dissection problems. These were used in the construction of altars that had to maintain exact prescribed shapes and areas. Euclid's Elements also contains similar problems.

Median Property Extended

**Theorem**: All three medians of a triangle divide it into 6 triangles of equal area.

**Proof**: Each median divides the triangle into two equal-area triangles. When all three medians intersect, they create 6 smaller triangles around the centroid, all with area = (1/6) × Area(original triangle).

Area Doubling

**Pattern**: If you double each side length of a figure:

Area multiplies by 4 (not by 2!)

**Reason**: Area has two dimensions. If each dimension doubles, area = 2 × 2 = 4 times.

**Example**:

Original rectangle: 5 cm × 3 cm, Area = 15 cm²

Doubled rectangle: 10 cm × 6 cm, Area = 60 cm² = 15 × 4

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IMPORTANT FORMULAS SUMMARY

Rectangle

**Area = length × width**

**Area = l × w**

Triangle

**Area = (1/2) × base × height**

**Area = (1/2) × b × h**

Parallelogram

**Area = base × height**

**Area = b × h**

Rhombus

**Area = (1/2) × d₁ × d₂** (where d₁, d₂ are diagonals)

Also: **Area = base × height** (since rhombus is a parallelogram)

Trapezium

**Area = (1/2) × h × (b₁ + b₂)**

**Area = h × (sum of parallel sides)/2**

General Polygon

**Area = Sum of areas of triangles** (after dividing polygon into triangles)

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DIMENSIONAL ANALYSIS

Understanding Square Units

When we say "Area = 100 cm²", we mean:

100 unit squares, each 1 cm × 1 cm in size

Not "100 square centimeters of something"

Converting Between Units

1 m² = 10,000 cm² (because 1 m = 100 cm, and 100 × 100 = 10,000)

1 km² = 1,000,000 m²

1 hectare = 10,000 m² (used in agriculture in India)

**Example**: A field is 500 m². How many hectares?

500 m² ÷ 10,000 m²/hectare = 0.05 hectares

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REAL-LIFE APPLICATIONS (INDIAN CONTEXTS)

Agriculture and Land

1. **Calculating Fertilizer Needs**: A farmer's field is rectangular: 60 m × 40 m.

Area = 2400 m²

If fertilizer costs ₹10 per m², total cost = ₹24,000

2. **Irrigation Planning**: A trapezoidal field with parallel sides 50 m and 60 m, height 30 m.

Area = (1/2) × 30 × (50 + 60) = 1650 m²

Water needed = 1650 m² × irrigation requirement

Construction

3. **Tiling a Floor**: A rectangular room is 8 m × 6 m.

Area = 48 m²

If tiles are 0.5 m × 0.5 m (area = 0.25 m² each), need 48/0.25 = 192 tiles

4. **Painting Walls**: A parallelogram-shaped wall has base 10 m and height 5 m.

Area =