Official Syllabus Item: 10, 40, 47 Mensuration

Perimeter & Area of Plane Figures and Field Book Surveying

Master perimeter and area of squares, rectangles, parallelograms, rhombuses, trapeziums, triangles (Heron's formula), pathways/borders, and Field Book land surveying for RIMC Maths 2026.

Overview & Fundamental Concept

Mensuration of two-dimensional plane figures and land surveying constitutes one of the heaviest scoring domains in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 10 ('Area and Perimeter'), 40 ('Perimeter & Area of Square, Rectangle, Parallelogram, Triangle, Arc Length circumference'), and 47 ('Area of figures which can be split up into the figures (field book)') of Chapter 8 in the official brochure, mensuration problems routinely carry between 25 to 40 marks across Part A (5 Marks) and Part B (10 Marks). RIMC examiners look for strict descriptive discipline: declaring geometric variables with units, drawing clear labeled boundary diagrams, writing out algebraic formulas prior to numerical substitution, demonstrating prime-factor decomposition in Heron's square-root evaluations rather than brute-force decimal multiplication, organizing field book survey data into structured tabular ledgers, and expressing final land areas in standard agricultural units (hectares and ares).

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Item 10 & 40	Perimeter & Area: Rectangles, squares, parallelograms, rhombuses, trapeziums, scalene triangles (Heron's formula), and pathways/borders.	Standard staple of Part A (5 Marks) and Part B (10 Marks) testing algebraic substitution and multi-step geometric partitioning.
Item 47	Field Book Land Surveying: Splitting irregular multi-sided polygons along a central chain survey baseline into right triangles and trapeziums.	Explicitly highlighted in brochure item 47. High-weightage 10-mark descriptive question requiring tabular ledger documentation.

1. Quadrilaterals: Formulas & Geometric Properties

A quadrilateral is a four-sided polygon. The fundamental quadrilateral types and their standard mensuration formulas are:

• Square (side $a$):
$$\text{Perimeter } P = 4a, \quad \text{Area } A = a^2 = \frac{d^2}{2}, \quad \text{Diagonal } d = a\sqrt{2}$$

• Rectangle (length $l$, breadth $b$):
$$\text{Perimeter } P = 2(l + b), \quad \text{Area } A = l \times b, \quad \text{Diagonal } d = \sqrt{l^2 + b^2}$$

• Parallelogram (base $b$, corresponding altitude $h$):
$$\text{Perimeter } P = 2(a + b), \quad \text{Area } A = b \times h$$Note: The area can be computed using either base and its respective altitude: $A = b_1 h_1 = b_2 h_2$.

• Rhombus (side $a$, diagonals $d_1$ and $d_2$):
Since the diagonals of a rhombus bisect each other at right angles ($90^\circ$):
$$\text{Area } A = \frac{1}{2} d_1 d_2, \quad \text{Side } a = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}, \quad \text{Perimeter } P = 4a$$

• Trapezium (parallel sides $a$ and $b$, perpendicular distance $h$):
$$\text{Area } A = \frac{1}{2}(a + b) \times h = \text{Average of parallel sides} \times \text{Altitude}$$

2. Triangles & Heron's Formula

Depending on the given side/altitude data, the area of a triangle $\triangle$ is determined using three primary approaches:

1. Base and Altitude: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b h$.
2. Equilateral Triangle (side $a$): $\text{Area} = \frac{\sqrt{3}}{4} a^2, \quad \text{Altitude } h = \frac{\sqrt{3}}{2} a$.
3. Scalene Triangle (Heron's Formula):
When all three side lengths $a, b,$ and $c$ are known without an altitude, calculate the semi-perimeter $s$:
$$s = \frac{a + b + c}{2}$$Then the area $\Delta$ is given by Heron's invariant formula:
$$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$$

Topper Arithmetic Tip: NEVER multiply the four numbers $s, (s-a), (s-b), (s-c)$ into a gargantuan product. Always break each term into its prime factors, pair the factors, and pull them cleanly out of the square root radical.

Secondary Triangle Relations:
• Inradius ($r$): $r = \frac{\Delta}{s}$ (radius of inscribed circle touching all three sides).
• Altitude to side $a$: $h_a = \frac{2\Delta}{a}$.

3. Pathways, Verandahs & Central Cross-Roads

Problems involving pathways surrounding or intersecting rectangular lawns are ubiquitous in RIMC mensuration:

• Uniform Pathway Running OUTSIDE a Rectangle ($l \times b$, width $w$):
New outer dimensions: $L = l + 2w$ and $B = b + 2w$.
$$\text{Area of Outer Path} = (l + 2w)(b + 2w) - lb = 2w(l + b + 2w)$$
• Uniform Pathway Running INSIDE a Rectangle ($l \times b$, width $w$):
New inner dimensions: $L' = l - 2w$ and $B' = b - 2w$.
$$\text{Area of Inner Path} = lb - (l - 2w)(b - 2w) = 2w(l + b - 2w)$$
• Two Central Cross-Roads Intersecting at Right Angles (width $w$):
One road runs parallel to the length ($l \times w$), the second parallel to the breadth ($b \times w$). The central square intersection of area $w^2$ is shared by both roads:
$$\text{Total Area of Cross-Roads} = (l \times w) + (b \times w) - w^2 = w(l + b - w)$$$$\text{Remaining Area of Lawn} = lb - w(l + b - w) = (l - w)(b - w)$$

4. Field Book Surveying & Polygonal Decomposition

In cadastral land surveying, an irregular multi-sided field $ABCDEF...$ is surveyed using a single central straight survey baseline (or chain line), with perpendicular distances (called offsets) measured to vertices on the left and right sides.

Understanding Field Book Entries:
The surveyor's notebook records cumulative chainage distances from the initial base station (e.g. $A = 0$) up to the terminal station (e.g. $D = 180\text{ m}$). Offsets to the left and right indicate perpendicular heights from those chain line points to perimeter boundary points.

Decomposition & Tabular Ledger Presentation:
To earn full 10 marks in RIMC, partition the irregular figure into standard elementary shapes: Right-Angled Triangles (at corners) and Trapeziums (between offsets on the same side). Present your calculation in a formal surveyor ledger table:

$$\begin{array}{|c|c|c|c|c|c|} \hline \textbf{Fig.} & \textbf{Sub-Region} & \textbf{Shape} & \textbf{Base Length } (\Delta x) & \textbf{Offsets / Heights } (y_1, y_2) & \textbf{Area } (\text{m}^2) \\ \hline 1 & \triangle APB & \text{Rt. Triangle} & AP & PB & \frac{1}{2} \times AP \times PB \\ 2 & \text{Trap. } PBCQ & \text{Trapezium} & PQ = AQ - AP & PB \text{ and } QC & \frac{1}{2}(PB + QC) \times PQ \\ \hline \end{array}$$

Standard Land Units Conversion:
$$1\text{ Are} = 100\text{ m}^2, \quad 1\text{ Hectare (ha)} = 10,000\text{ m}^2 = 100\text{ Ares}, \quad 1\text{ km}^2 = 100\text{ ha}$$

Illustrated Conceptual Diagrams & Visual Schematics

2D Quadrilaterals & Plane Figures Area Geometry Map

Key Formulas: Parallelogram: $\text{Base} \times \text{Height}$. Rhombus: $\frac{1}{2} d_1 d_2$. Trapezium: $\frac{1}{2}(a+b)h$. Area of 4 Walls: $2h(l+b)$. Circular Ring Area: $\pi(R^2 - r^2)$.

Topper Strategy & Descriptive Step Breakdown

Mastering RIMC Mathematics Standard

In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.

Official Descriptive Marking Scheme & Step Allocation

In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:

Step 1: Labeled Layout Diagram & Given Dimensions 20% of marks

Draw an accurate pencil sketch of the polygon, path, or field book chain line. Mark all dimensions, offsets, stations ($A, B, C...$), and define variables with explicit units (m, cm).

Step 2: Partitioning & Formula Citations 25% of marks

State the area formulas algebraically before substituting numbers: $\text{Area} = \frac{1}{2}(a+b)h$, $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, or ledger formulas for each sub-region.

Step 3: Step-by-Step Ledger Computation & Prime Factoring 35% of marks

In field book questions, compute each sub-region in a clean tabular format. In Heron's formula, show prime factorization beneath the radical. Show all addition steps clearly.

Step 4: Unit Conversion & Boxed Final Summary 20% of marks

Convert the total square metres to hectares ($1\text{ ha} = 10,000\text{ m}^2$) or ares ($1\text{ are} = 100\text{ m}^2$) when requested. Conclude with a complete boxed statement.

Examiner Traps & Common Blunders (Marks Penalties)

Double Counting the Central Cross-Road Intersection 2 to 3 marks deducted for conceptual error in pathway mensuration.

Examiner Trap: Adding $(l \times w) + (b \times w)$ without subtracting the square intersection $w^2$ shared by both roads.

Multiplying Giant Numbers under Heron's Radical Arithmetic deduction of 2 to 4 marks; loss of precious exam time.

Examiner Trap: Computing $s(s-a)(s-b)(s-c)$ into a 7-digit number and attempting long-division square roots, frequently incurring arithmetic errors.

Using Cumulative Chainage as Segment Base in Field Books Severe: invalidates all trapezium sub-areas, leading to loss of 6 to 8 marks.

Examiner Trap: Using the total distance from the starting station (e.g. $140\text{ m}$) instead of the segment length between adjacent stations ($140 - 80 = 60\text{ m}$).

Assuming Non-Parallel Sides of a Trapezium are Altitudes Fatal conceptual error: 0 marks awarded for area calculation.

Examiner Trap: Multiplying $\frac{1}{2}(a + b)$ directly by an inclined slant leg rather than calculating the true perpendicular height $h$.

Incorrect Hectare / Are Metric Conversion 1 to 2 marks lost in the final conversion step.

Examiner Trap: Dividing by $1,000$ or $100,000$ instead of $10,000$ to convert square metres to hectares.

Cadet Topper Mental Verification Shortcuts

Rhombus Pythagorean Side Identity

If diagonals are $d_1$ and $d_2$, half-diagonals form legs of a right triangle with side $a$ as hypotenuse: $a = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$. If diagonals are $16$ and $12$, half-diagonals are $8$ and $6$, so side is $10\text{ cm}$ instantly!

Remaining Lawn in Cross-Roads Shortcut

When two cross-roads of width $w$ cross a rectangular lawn $l \times b$, the remaining lawn pieces can be pushed together to form a smaller rectangle of dimensions $(l - w) \times (b - w)$. The remaining area is simply $(l - w)(b - w)$!

Heron's Formula Prime Factorization Method

Write each term under the radical as prime factors. Example: $\sqrt{24 \times 14 \times 12 \times 2} = \sqrt{(2^3 \cdot 3) \times (2 \cdot 7) \times (2^2 \cdot 3) \times 2} = \sqrt{2^7 \cdot 3^2 \cdot 7}$. Extract integer pairs cleanly without long division.

Trapezium Altitude via Parallelogram-Triangle Split

For an irregular trapezium with known non-parallel sides, draw a line parallel to one leg from a top vertex to split the figure into a parallelogram and a triangle. Find the triangle's area via Heron's formula, then find height $h = \frac{2\Delta}{\text{base}}$.

Key Formulas & Conversion Cheat Sheet

Rectangle & Square Metrics

$$A_{\text{rect}} = lb, \quad d_{\text{rect}} = \sqrt{l^2 + b^2}, \quad A_{\text{sq}} = a^2 = \frac{d^2}{2}, \quad d_{\text{sq}} = a\sqrt{2}$$

Square diagonal $d = a\sqrt{2}$ enables direct area computation without finding $a$.

Parallelogram & Rhombus Area

$$A_{\text{parallelogram}} = b \times h, \quad A_{\text{rhombus}} = \frac{1}{2}d_1 d_2 = b \times h$$

Diagonals of a rhombus intersect at $90^\circ$ and bisect each other.

Trapezium Area

$$A = \frac{1}{2}(a + b)h$$

$a$ and $b$ are parallel sides, $h$ is perpendicular distance between them.

Heron's Formula for Scalene Triangles

$$\Delta = \sqrt{s(s - a)(s - b)(s - c)}, \quad \text{where } s = \frac{a + b + c}{2}$$

$s$ is semi-perimeter; inradius $r = \frac{\Delta}{s}$.

Pathway & Cross-Road Formulas

$$A_{\text{outer}} = 2w(l + b + 2w), \quad A_{\text{inner}} = 2w(l + b - 2w), \quad A_{\text{crossroads}} = w(l + b - w)$$

$w$ is uniform width; remaining lawn area $= (l - w)(b - w)$.

Field Book Survey Decomposition

$$A_{\text{triangle}} = \frac{1}{2} \times \Delta x \times y, \quad A_{\text{trapezium}} = \frac{1}{2}(y_1 + y_2) \times \Delta x$$

$\Delta x$ is segment length along chain baseline; $y$ are perpendicular offsets.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Part A - Descriptive 5 Marks

The diagonals of a rhombus are $24\text{ cm}$ and $10\text{ cm}$. Find:
(i) Its area.
(ii) The length of each side.
(iii) Its perimeter and altitude.

Step-by-Step Model Solution:

Step 1: (i) Calculate the Area of the Rhombus
Given lengths of diagonals:
$$d_1 = 24\text{ cm} \quad \text{and} \quad d_2 = 10\text{ cm}$$ The area of a rhombus is given by the formula:
$$\text{Area} = \frac{1}{2} \times d_1 \times d_2$$ $$\text{Area} = \frac{1}{2} \times 24\text{ cm} \times 10\text{ cm} = 120\text{ cm}^2$$
$$\mathbf{\text{Area of the rhombus } = 120\text{ cm}^2.}$$
Step 2: (ii) Calculate the length of each side ($a$)
The diagonals of a rhombus bisect each other at right angles ($90^\circ$).
Let the diagonals intersect at point $O$. In right-angled $\triangle AOB$:
$$OA = \frac{d_1}{2} = \frac{24}{2} = 12\text{ cm}$$ $$OB = \frac{d_2}{2} = \frac{10}{2} = 5\text{ cm}$$
By the Pythagorean Theorem:
$$\text{Side } a = \sqrt{OA^2 + OB^2}$$ $$a = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}$$
$$\mathbf{\text{Length of each side } a = 13\text{ cm.}}$$
Step 3: (iii) Calculate the perimeter and altitude
• Perimeter: Since all four sides of a rhombus are equal in length:
$$P = 4a = 4 \times 13\text{ cm} = 52\text{ cm}$$
• Altitude ($h$): A rhombus is also a parallelogram, so its area can be written as $\text{base} \times \text{altitude}$:
$$\text{Area} = a \times h$$ $$120 = 13 \times h$$ $$h = \frac{120}{13} = 9\frac{3}{13}\text{ cm} \approx 9.23\text{ cm}$$
$$\mathbf{\text{Conclusion: (i) Area } = 120\text{ cm}^2; \; (ii) \text{ Side } = 13\text{ cm}; \; (iii) \text{ Perimeter } = 52\text{ cm}, \text{ Altitude } = \frac{120}{13}\text{ cm}.}$$

Rough Column Working:
Area = 1/2 * 24 * 10 = 120
Side = sqrt(12^2 + 5^2) = sqrt(169) = 13
Perimeter = 4 * 13 = 52
Altitude = 120 / 13 = 9 3/13 cm.

Step Mark Allocation:

Calculating Area = 120 sq cm with formula: 1.5 Marks
Using Pythagoras on half-diagonals to find side = 13 cm: 1.5 Marks
Calculating Perimeter = 52 cm: 1 Marks
Calculating Altitude = 120/13 cm and boxed answer: 1 Marks

Examiner Scoring Key: Candidates frequently forget that a rhombus is also a parallelogram and can have an altitude computed via Area = base * height.

Part B - Non-Right Trapezium 10 Marks

The parallel sides of a trapezium are $25\text{ cm}$ and $10\text{ cm}$, and its non-parallel sides are $14\text{ cm}$ and $13\text{ cm}$. Find the area of the trapezium.

Step-by-Step Model Solution:

Step 1: Geometric Construction & Figure Partitioning
Let $ABCD$ be the trapezium with parallel sides $AB = 25\text{ cm}$ and $CD = 10\text{ cm}$ ($AB \parallel CD$).
Let non-parallel sides be $AD = 13\text{ cm}$ and $BC = 14\text{ cm}$.
Construction: Through vertex $C$, draw a line $CE$ parallel to side $DA$, intersecting the longer base $AB$ at point $E$ (i.e., $CE \parallel DA$).

Step 2: Properties of the Parallelogram $AECD$
In quadrilateral $AECD$:
$$AE \parallel DC \quad [\text{Parts of parallel lines } AB \parallel CD]$$ $$CE \parallel DA \quad [\text{By Construction}]$$ Therefore, $AECD$ is a parallelogram.
In a parallelogram, opposite sides are equal:
$$AE = DC = 10\text{ cm}$$ $$CE = DA = 13\text{ cm}$$
Now determine the base of the remaining $\triangle CEB$:
$$EB = AB - AE = 25\text{ cm} - 10\text{ cm} = 15\text{ cm}$$
Step 3: Calculate the Area of $\triangle CEB$ using Heron's Formula
The sides of $\triangle CEB$ are $a = 15\text{ cm}, b = 14\text{ cm},$ and $c = 13\text{ cm}$.
Semi-perimeter $s$:
$$s = \frac{a + b + c}{2} = \frac{15 + 14 + 13}{2} = \frac{42}{2} = 21\text{ cm}$$
Applying Heron's Formula:
$$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$$ $$\Delta = \sqrt{21 \times (21 - 15) \times (21 - 14) \times (21 - 13)}$$ $$\Delta = \sqrt{21 \times 6 \times 7 \times 8}$$
Decompose into prime factors:
$$\Delta = \sqrt{(3 \times 7) \times (2 \times 3) \times 7 \times (2^3)}$$ $$\Delta = \sqrt{2^4 \times 3^2 \times 7^2} = 2^2 \times 3 \times 7 = 4 \times 21 = 84\text{ cm}^2$$
Step 4: Find the Altitude ($h$) of the Trapezium
Let $h$ be the altitude of $\triangle CEB$ from vertex $C$ to base $EB$.
$$\text{Area}(\triangle CEB) = \frac{1}{2} \times EB \times h$$ $$84 = \frac{1}{2} \times 15 \times h$$ $$h = \frac{84 \times 2}{15} = \frac{168}{15} = \frac{56}{5} = 11.2\text{ cm}$$
Step 5: Calculate the Total Area of Trapezium $ABCD$
$$\text{Area}(ABCD) = \frac{1}{2}(AB + CD) \times h$$ $$\text{Area}(ABCD) = \frac{1}{2}(25 + 10) \times 11.2 = \frac{1}{2} \times 35 \times 11.2 = 35 \times 5.6 = 196\text{ cm}^2$$
$$\mathbf{\text{Conclusion: The area of the trapezium is } 196\text{ cm}^2.}$$

Rough Column Working:
EB = 25 - 10 = 15
Triangle sides: 15, 14, 13
s = 42/2 = 21
Delta = sqrt(21 * 6 * 7 * 8) = 84
1/2 * 15 * h = 84 => h = 11.2 cm
Trapezium Area = 1/2 * (25 + 10) * 11.2 = 35 * 5.6 = 196 cm^2.

Step Mark Allocation:

Construction of CE || DA to split figure into parallelogram and triangle: 2 Marks
Finding EB = 15 cm and setting up Heron's formula for triangle CEB: 2 Marks
Prime factorization to calculate triangle Area = 84 sq cm: 2 Marks
Solving for perpendicular height h = 11.2 cm: 2 Marks
Applying trapezium area formula to obtain 196 sq cm and boxed answer: 2 Marks

Examiner Scoring Key: This is a quintessential RIMC 10-mark problem. Drawing the auxiliary line CE to create a parallelogram and triangle is the only mathematically sound approach.

Part B - Heron's Triangle & Inradius 10 Marks

The sides of a triangular field are $21\text{ m}, 17\text{ m},$ and $10\text{ m}$.
(i) Find the area of the triangular field.
(ii) Find the length of the altitude drawn to the longest side.
(iii) Find the radius of the circle inscribed in this triangle (inradius).

Step-by-Step Model Solution:

Step 1: Calculate the semi-perimeter ($s$)
Given side lengths:
$$a = 21\text{ m}, \quad b = 17\text{ m}, \quad c = 10\text{ m}$$ Semi-perimeter $s$ is:
$$s = \frac{a + b + c}{2} = \frac{21 + 17 + 10}{2} = \frac{48}{2} = 24\text{ m}$$
Step 2: (i) Calculate the Area ($\Delta$) using Heron's Formula
Compute the three differences $(s - a), (s - b), (s - c)$:
$$s - a = 24 - 21 = 3\text{ m}$$ $$s - b = 24 - 17 = 7\text{ m}$$ $$s - c = 24 - 10 = 14\text{ m}$$
By Heron's Formula:
$$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$$ $$\Delta = \sqrt{24 \times 3 \times 7 \times 14}$$
Expressing each factor in prime factor form:
$$\Delta = \sqrt{(2^3 \times 3) \times 3 \times 7 \times (2 \times 7)}$$ $$\Delta = \sqrt{2^4 \times 3^2 \times 7^2}$$ $$\Delta = 2^2 \times 3 \times 7 = 4 \times 21 = 84\text{ m}^2$$
$$\mathbf{\text{Area of the triangular field } = 84\text{ m}^2.}$$
Step 3: (ii) Calculate the altitude to the longest side
The longest side is $a = 21\text{ m}$. Let the altitude to this side be $h_1$.
$$\text{Area} = \frac{1}{2} \times \text{base} \times h_1$$ $$84 = \frac{1}{2} \times 21 \times h_1$$ $$h_1 = \frac{84 \times 2}{21} = 4 \times 2 = 8\text{ m}$$
$$\mathbf{\text{Altitude to the longest side } = 8\text{ metres.}}$$
Step 4: (iii) Calculate the Inradius ($r$)
The area of any triangle can be expressed in terms of its inradius $r$ and semi-perimeter $s$ as:
$$\Delta = r \times s$$ $$84 = r \times 24$$ $$r = \frac{84}{24} = \frac{7}{2} = 3.5\text{ m}$$
$$\mathbf{\text{Conclusion: (i) Area } = 84\text{ m}^2; \; (ii) \text{ Altitude } = 8\text{ m}; \; (iii) \text{ Inradius } = 3.5\text{ m}.}$$

Rough Column Working:
s = 48/2 = 24
s - a = 3, s - b = 7, s - c = 14
Delta = sqrt(24 * 3 * 7 * 14) = sqrt(2^4 * 3^2 * 7^2) = 84
1/2 * 21 * h = 84 => h = 8 m
r = Delta / s = 84 / 24 = 3.5 m.

Step Mark Allocation:

Calculating semi-perimeter s = 24 m: 1.5 Marks
Heron's formula prime factoring to find Area = 84 sq m: 3.5 Marks
Calculating altitude to longest side h = 8 m: 2.5 Marks
Using Delta = r * s to solve inradius r = 3.5 m and boxed summary: 2.5 Marks

Examiner Scoring Key: Always write out prime factors under the radical. Showing the prime pairing method earns top marks for mathematical elegance.

Part B - Field Book Surveying 10 Marks

Find the area of the field $ABCDEF$ from the following field book notes (all measurements are in metres):
$$\begin{array}{rcc} & \textbf{To D} & \\ & 160 & \\ \text{To E (40)} & 120 & \\ & 80 & 60 \text{ to C} \\ \text{To F (50)} & 40 & \\ & \textbf{From A} & \end{array}$$
Also express the total area in hectares and ares.

Step-by-Step Model Solution:

Step 1: Understand the Survey Baseline and Offsets
The central chain baseline runs from starting station $A$ ($0\text{ m}$) to terminal station $D$ ($160\text{ m}$).
Let the perpendicular offset feet on the chain baseline $AD$ be denoted by $P_1, P_2, P_3$:
• At distance $40\text{ m}$ from $A$, offset to left is $F$ with length $P_1 F = 50\text{ m}$ ($P_1 = 40\text{ m}$).
• At distance $80\text{ m}$ from $A$, offset to right is $C$ with length $P_2 C = 60\text{ m}$ ($P_2 = 80\text{ m}$).
• At distance $120\text{ m}$ from $A$, offset to left is $E$ with length $P_3 E = 40\text{ m}$ ($P_3 = 120\text{ m}$).
• At distance $160\text{ m}$, the chain line terminates at vertex $D$ ($D = 160\text{ m}$).

Step 2: Partition the Field into Elementary Geometrical Sub-regions
The polygon $ABCDEF$ (assuming $B$ is on the baseline or offset, here on right with station $C$) splits into:
• Left Side:
1. Right $\triangle AP_1 F$ (Base $= AP_1 = 40\text{ m}$, Height $= P_1 F = 50\text{ m}$)
2. Trapezium $P_1 F E P_3$ (Parallel sides $= P_1 F = 50\text{ m}, P_3 E = 40\text{ m}$, Height $= P_1 P_3 = 120 - 40 = 80\text{ m}$)
3. Right $\triangle P_3 E D$ (Base $= P_3 D = 160 - 120 = 40\text{ m}$, Height $= P_3 E = 40\text{ m}$)
• Right Side:
4. Right $\triangle AP_2 C$ (Base $= AP_2 = 80\text{ m}$, Height $= P_2 C = 60\text{ m}$)
5. Right $\triangle P_2 C D$ (Base $= P_2 D = 160 - 80 = 80\text{ m}$, Height $= P_2 C = 60\text{ m}$)

Step 3: Tabular Ledger Calculation
$$\begin{array}{|c|c|c|c|c|c|} \hline \textbf{Part} & \textbf{Figure} & \textbf{Type} & \textbf{Base } (\text{m}) & \textbf{Heights/Offsets } (\text{m}) & \textbf{Area } (\text{m}^2) \\ \hline 1 & \triangle AP_1 F & \text{Rt. } \triangle & 40 & 50 & \frac{1}{2} \times 40 \times 50 = 1000 \\ 2 & \text{Trap. } P_1 F E P_3 & \text{Trapezium} & 120 - 40 = 80 & 50 \text{ and } 40 & \frac{1}{2}(50 + 40) \times 80 = 3600 \\ 3 & \triangle P_3 E D & \text{Rt. } \triangle & 160 - 120 = 40 & 40 & \frac{1}{2} \times 40 \times 40 = 800 \\ 4 & \triangle AP_2 C & \text{Rt. } \triangle & 80 & 60 & \frac{1}{2} \times 80 \times 60 = 2400 \\ 5 & \triangle P_2 C D & \text{Rt. } \triangle & 160 - 80 = 80 & 60 & \frac{1}{2} \times 80 \times 60 = 2400 \\ \hline \end{array}$$
Step 4: Sum Total Area and Convert to Land Units
$$\text{Total Area} = 1000 + 3600 + 800 + 2400 + 2400 = 10,200\text{ m}^2$$
• Conversion to Ares: Since $1\text{ Are} = 100\text{ m}^2$:
$$\text{Area in Ares} = \frac{10200}{100} = 102\text{ Ares}$$
• Conversion to Hectares: Since $1\text{ Hectare} = 10,000\text{ m}^2$:
$$\text{Area in Hectares} = \frac{10200}{10000} = 1.02\text{ Hectares}$$
$$\mathbf{\text{Conclusion: Total Area } = 10,200\text{ m}^2 = 102\text{ Ares } = 1.02\text{ Hectares.}}$$

Rough Column Working:
Part 1: 1/2 * 40 * 50 = 1000
Part 2: 1/2 * (50 + 40) * 80 = 3600
Part 3: 1/2 * 40 * 40 = 800
Part 4: 1/2 * 80 * 60 = 2400
Part 5: 1/2 * 80 * 60 = 2400
Total = 10200 m^2
102 ares = 1.02 ha.

Step Mark Allocation:

Neat sketch of survey baseline AD with labeled offset stations: 2 Marks
Correct identification of 5 geometric sub-regions: 2 Marks
Formal tabular ledger presentation with accurate calculations: 3 Marks
Summing total area = 10,200 sq m: 1.5 Marks
Converting to 102 Ares and 1.02 Hectares with boxed summary: 1.5 Marks

Examiner Scoring Key: Presenting the solution as a structured surveyor ledger table is mandatory for securing full 10 marks in RIMC Part B.

Part A - Descriptive Cross-Roads 5 Marks

A rectangular park of dimensions $60\text{ m} \times 40\text{ m}$ has two concrete crossroads of width $3\text{ m}$ each running through its center, one parallel to the length and the other parallel to the breadth. Find:
(i) The total area of the crossroads.
(ii) The cost of paving the crossroads at ₹$80\text{ per m}^2$.
(iii) The remaining area of the park for the lawn.

Step-by-Step Model Solution:

Step 1: Formulate the Crossroads Geometry
Given dimensions of rectangular park: $\text{Length } l = 60\text{ m}, \text{ Breadth } b = 40\text{ m}$.
Width of each concrete crossroad: $w = 3\text{ m}$.

Step 2: (i) Calculate the Total Area of Crossroads
• Area of road parallel to length $= l \times w = 60\text{ m} \times 3\text{ m} = 180\text{ m}^2$.
• Area of road parallel to breadth $= b \times w = 40\text{ m} \times 3\text{ m} = 120\text{ m}^2$.
• Area of central square intersection common to both roads $= w^2 = 3\text{ m} \times 3\text{ m} = 9\text{ m}^2$.
By the principle of inclusion-exclusion:
$$\text{Total Area of Crossroads} = (l \times w) + (b \times w) - w^2$$ $$\text{Total Area} = 180 + 120 - 9 = 300 - 9 = 291\text{ m}^2$$
$$\mathbf{\text{Total Area of Crossroads } = 291\text{ m}^2.}$$
Step 3: (ii) Calculate the cost of paving
$$\text{Cost} = \text{Area} \times \text{Rate} = 291\text{ m}^2 \times ₹80/\text{m}^2 = ₹23,280$$
$$\mathbf{\text{Cost of Paving } = ₹23,280.}$$
Step 4: (iii) Calculate the remaining lawn area
$$\text{Total Area of Park} = 60\text{ m} \times 40\text{ m} = 2400\text{ m}^2$$ $$\text{Remaining Area} = \text{Total Area} - \text{Crossroads Area} = 2400\text{ m}^2 - 291\text{ m}^2 = 2109\text{ m}^2$$
Verification via Topper Method:
$$\text{Remaining Area} = (l - w)(b - w) = (60 - 3)(40 - 3) = 57 \times 37 = 2109\text{ m}^2 \quad [\text{Verified}]$$
$$\mathbf{\text{Conclusion: (i) Crossroads Area } = 291\text{ m}^2; \; (ii) \text{ Paving Cost } = ₹23,280; \; (iii) \text{ Lawn Area } = 2109\text{ m}^2.}$$

Rough Column Working:
Road 1: 60 * 3 = 180
Road 2: 40 * 3 = 120
Common square: 3 * 3 = 9
Total road = 180 + 120 - 9 = 291 m^2
Cost = 291 * 80 = 23,280
Lawn = 57 * 37 = 2109 m^2.

Step Mark Allocation:

Correct diagram and crossroad formula showing subtraction of central square: 1.5 Marks
Calculating total crossroad area = 291 sq m: 1.5 Marks
Calculating cost = ₹23,280: 1 Marks
Calculating remaining lawn area = 2109 sq m and boxed answer: 1 Marks

Examiner Scoring Key: Forgetting to subtract the central square intersection of 9 sq m loses 2 marks immediately.

Part B - Pathway & Tiling Mensuration 10 Marks

A hall $28\text{ m}$ long and $16\text{ m}$ wide is surrounded on all sides by an open verandah of uniform width $2\text{ m}$.
(i) Find the area of the verandah.
(ii) Find the number of square tiles of side $40\text{ cm}$ needed to pave the floor of the verandah.
(iii) If the cost of tiling is ₹$15\text{ per tile}$ plus ₹$25\text{ per m}^2$ for labor, find the total expenditure.

Step-by-Step Model Solution:

Step 1: Identify Initial and Outer Dimensions
Given dimensions of the hall:
$$\text{Length of hall } l = 28\text{ m}, \quad \text{Breadth of hall } b = 16\text{ m}$$ $$\text{Area of hall floor} = l \times b = 28\text{ m} \times 16\text{ m} = 448\text{ m}^2$$
Width of uniform verandah around the hall: $w = 2\text{ m}$.
Since the verandah surrounds the hall on all four sides, the outer dimensions are:
$$\text{Outer Length } L = l + 2w = 28\text{ m} + 2(2\text{ m}) = 28 + 4 = 32\text{ m}$$ $$\text{Outer Breadth } B = b + 2w = 16\text{ m} + 2(2\text{ m}) = 16 + 4 = 20\text{ m}$$
$$\text{Total Outer Area} = L \times B = 32\text{ m} \times 20\text{ m} = 640\text{ m}^2$$
Step 2: (i) Calculate the Area of the Verandah
$$\text{Area of Verandah} = \text{Total Outer Area} - \text{Area of Hall}$$ $$\text{Area of Verandah} = 640\text{ m}^2 - 448\text{ m}^2 = 192\text{ m}^2$$
$$\mathbf{\text{Area of the Verandah } = 192\text{ m}^2.}$$
Step 3: (ii) Calculate the Number of Square Tiles Required
Given side of each square tile $= 40\text{ cm}$.
Convert side to metres:
$$\text{Side of tile} = \frac{40}{100} = 0.4\text{ m} = \frac{2}{5}\text{ m}$$ $$\text{Area of one tile} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\text{ m}^2 = 0.16\text{ m}^2$$
The total number of tiles $N$ required is:
$$N = \frac{\text{Area of Verandah}}{\text{Area of one tile}} = \frac{192}{0.16} = \frac{19200}{16} = 1200\text{ tiles}$$
$$\mathbf{\text{Number of tiles required } = 1200\text{ tiles.}}$$
Step 4: (iii) Calculate Total Expenditure
• Cost of Tiles: $1200 \text{ tiles} \times ₹15/\text{tile} = ₹18,000$.
• Cost of Labor: $192\text{ m}^2 \times ₹25/\text{m}^2 = ₹4,800$.
$$\text{Total Expenditure} = ₹18,000 + ₹4,800 = ₹22,800$$
$$\mathbf{\text{Conclusion: (i) Area } = 192\text{ m}^2; \; (ii) \; 1200\text{ tiles}; \; (iii) \text{ Total Cost } = ₹22,800.}$$

Rough Column Working:
Outer: (28+4) * (16+4) = 32 * 20 = 640
Inner: 28 * 16 = 448
Verandah = 640 - 448 = 192 m^2
Tile area = 0.4 * 0.4 = 0.16 m^2
Tiles = 192 / 0.16 = 1200
Cost tiles = 1200 * 15 = 18,000
Labor = 192 * 25 = 4800
Total = 22,800.

Step Mark Allocation:

Finding outer dimensions (32 m x 20 m) and outer area = 640 sq m: 2.5 Marks
Calculating verandah area = 192 sq m: 2 Marks
Converting tile units and finding tile count = 1200: 2.5 Marks
Calculating tile cost (₹18,000) and labor cost (₹4,800): 1.5 Marks
Total expenditure ₹22,800 with boxed conclusions: 1.5 Marks

Examiner Scoring Key: Ensure the tile side is converted to metres (0.4 m) before computing its area, or convert the verandah area to square centimetres (192 * 10000 = 19,20,000 sq cm).

RIMC Practice & Drill Exercises

Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.

Question 1 5 Marks

The area of a square field is $0.5\text{ hectares}$. Find the length of its diagonal in metres.