Official Syllabus Item: 17, 40
Geometry & Mensuration
Circles: Properties, Chords, Circumference & Arc Length
Master circle terminology (radius, diameter, chord, secant, tangent, arc, sector, segment), circumference C = 2πr, semicircle perimeter, arc length l = (θ/360°) × 2πr, and wheel revolution problems for RIMC Maths 2026.
Overview & Fundamental Concept
Circles form a vital component of both plane geometry and mensuration in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 17 ('Circle') and 40 ('Perimeter & Area of Square, Rectangle, Parallelogram, Triangle, Arc Length circumference') of Chapter 8 in the official brochure, circle problems test candidate rigor across both Part A (5 Marks) and Part B (10 Marks). The RIMC descriptive evaluation standards demand precision: drawing accurate compass-and-pencil figures, distinguishing sharply between boundary length (circumference/perimeter) and interior regions (area/sector/segment), correctly accounting for straight bounding edges in composite curves (such as adding the diameter $2r$ in a semicircle perimeter), performing multi-unit conversions (e.g. converting cm of wheel radius to km/h of vehicular speed), and stating the value of $\pi = \frac{22}{7}$ unless explicitly specified as $3.1416$.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 17 | Circle: Radius, diameter, chord, secant, tangent, arc, sector, segment, concentric circles, and perpendicular chord bisector theorem. | Establishes foundational Euclidean circle terminology and chord properties used in 5-mark and 10-mark geometric proofs. |
| Item 40 | Circumference & Arc Length: Formulas for circumference C = 2πr, semicircle perimeter (πr + 2r), sector arc length l = (θ/360) × 2πr, and rotational mechanics. | Directly specified in syllabus item 40. Essential for perimeter calculations, wire rebending problems, and wheel revolution dynamics. |
1. Circle Anatomy & Core Geometric Elements
A circle is the locus of all coplanar points that are at a fixed distance (the radius) from a fixed central point (the centre). The essential geometric segments and lines associated with a circle are:
• Radius ($r$): A line segment joining the centre $O$ to any point on the boundary of the circle. All radii of a given circle are equal in length.
• Diameter ($d$): A line segment passing through the centre and connecting two points on the circle. It is the longest chord of the circle, and $d = 2r$.
• Chord: A straight line segment connecting any two distinct points on the circle.
• Secant: A straight line that intersects the circle at two distinct points, extending indefinitely in both directions.
• Tangent: A coplanar line that touches the circle at exactly one point (the point of contact). The tangent is always perpendicular to the radius drawn through the point of contact.
• Concentric Circles: Two or more coplanar circles that share the exact same centre but have different radii ($R > r$).
2. Arcs, Sectors, Segments & Semicircles
A continuous portion of the boundary curve of a circle is called an arc:
• Minor Arc & Major Arc: An arc smaller than a semicircle is a minor arc; an arc larger than a semicircle is a major arc.
• Central Angle ($\theta$): The angle subtended by an arc at the centre of the circle.
• Sector: The region of the circular plane enclosed by two radii and their intercepted arc. A minor sector corresponds to a central angle $\theta < 180^\circ$; a major sector corresponds to $(360^\circ - \theta)$.
• Segment: The region of the circular plane bounded by a chord and the corresponding arc intercepted by the chord.
• Semicircle: Exactly half of a circle, formed when a diameter divides the circle into two equal arcs. The angle subtended by a diameter at any point on the semicircle is always a right angle ($90^\circ$).
3. Metric Formulas: Circumference, Arc Length & Perimeters
Circumference of a Circle ($C$):
The total perimeter (curved boundary length) of a circle is given by:
$$C = 2\pi r = \pi d$$where $\pi = \frac{C}{d} \approx \frac{22}{7} \approx 3.14159$.
Arc Length ($l$):
For a circular sector of radius $r$ and central angle $\theta$ (in degrees):
$$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta}{180^\circ} \times \pi r$$
Perimeter of a Semicircular Region:
The boundary of a closed semicircular shape consists of the curved semicircular arc plus the straight diameter base:
$$P_{\text{semicircle}} = \pi r + 2r = r(\pi + 2) = r\left(\frac{22}{7} + 2\right) = \frac{36}{7}r$$
Perimeter of a Sector:
The complete boundary of a sector consists of the curved arc $l$ plus the two bounding radii:
$$P_{\text{sector}} = l + 2r = \left(\frac{\theta}{360^\circ} \times 2\pi r\right) + 2r$$
4. Wheel Revolutions & Chord Geometry Theorems
Rotational Mechanics of Wheels:
When a circular wheel rolls without slipping on a flat surface, the linear distance covered in exactly one full revolution equals its circumference:
$$\text{Distance in } 1 \text{ revolution} = C = 2\pi r = \pi d$$For a wheel completing $N$ revolutions covering total linear distance $D$:
$$D = N \times (2\pi r) \implies N = \frac{D}{2\pi r}$$$$\text{Speed of Vehicle} = \frac{\text{Revolutions per minute (RPM)} \times 2\pi r \times 60}{100000} \text{ km/h (with } r \text{ in cm)}$$
Fundamental Chord Theorems:
1. Perpendicular Chord Bisector Theorem: The perpendicular dropped from the centre of a circle to a chord bisects the chord. If $OM \perp AB$, then $AM = MB = \frac{AB}{2}$.
2. Chord-Radius Pythagoras Relation: In the right-angled $\triangle OMA$ formed by radius $r$, perpendicular distance $d_\perp = OM$, and half-chord $AM = \frac{L}{2}$:
$$r^2 = d_\perp^2 + \left(\frac{L}{2}\right)^2$$
3. Equidistant Chords Theorem: Chords of a circle that are equal in length are equidistant from the centre, and conversely, chords equidistant from the centre are equal in length.
Illustrated Conceptual Diagrams & Visual Schematics
Comprehensive Circle Geometry Anatomy & Terminology
Circle Metrics: Radius ($r = d/2$), Circumference ($C = 2\pi r = \pi d$), Area ($A = \pi r^2$). Sector Area = $\frac{\theta}{360^\circ} \times \pi r^2$, Arc Length = $\frac{\theta}{360^\circ} \times 2\pi r$. Tangent is strictly perpendicular to radius at point of contact ($r \perp t$).
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: Figure & Dimensional Given Data
20% of marks
Draw a clean, compass-drawn circle. Mark centre $O$, radius $r$, diameter $d$, or chord $AB$. State explicit 'Given:' values with correct units (cm, m, km).
Step 2: Formula Statement & Substitution
25% of marks
Quote the exact formula algebraically before plugging in numbers (e.g., $P = \pi r + 2r$ or $D = N \times 2\pi r$). State the chosen value of $\pi$ (typically $\frac{22}{7}$).
Step 3: Dimensional Conversion & Arithmetic Execution
35% of marks
Show explicit unit conversions (e.g., $1\text{ km} = 1000\text{ m} = 100000\text{ cm}$). Perform neat fraction cancellations without skipping intermediary steps.
Step 4: Concluding Statement & Boxed Answer
20% of marks
Write out the final answer in full prose with standard metric units (e.g., 'Therefore, the wheel makes $5000$ revolutions.'). Enclose the magnitude in a prominent box.
Examiner Traps & Common Blunders (Marks Penalties)
Omitting the Straight Edge in Semicircle / Sector Perimeter Severe: 3 to 4 marks deducted out of 5 marks for fundamentally wrong perimeter formula.
Examiner Trap: Using $P = \pi r$ for the perimeter of a closed semicircle or $P = l$ for a sector, forgetting to add the straight boundaries ($2r$).
Diameter vs Radius Confusion in Wheel Revolutions Results in a factor-of-2 error across the entire calculation; maximum 1 mark awarded.
Examiner Trap: Plugging the given diameter $d$ directly into $2\pi r$ (computing $2\pi d$), thereby doubling the circumference.
Unit Inconsistency (km/h vs cm/s) Catastrophic order-of-magnitude error; 0 marks for arithmetic reduction.
Examiner Trap: Dividing a distance given in kilometres directly by a circumference in centimetres without converting units by multiplying by $100000$.
Assuming Angle in Semicircle without Proving Chord is Diameter Deduction of 2 marks in geometric proof deductions.
Examiner Trap: Stating an inscribed angle is $90^\circ$ without establishing that the chord passing through the ends is a diameter containing the centre.
Concentric Circles Annulus Width Error Loss of 2 to 3 marks in circular track word problems.
Examiner Trap: Computing the width of a circular path as $R + r$ or $\frac{R - r}{2}$ instead of the true width $w = R - r$.
Cadet Topper Mental Verification Shortcuts
Semicircle Perimeter $\frac{36}{7}r$ Shortcut
Remember that $P_{\text{semi}} = \pi r + 2r = \left(\frac{22}{7} + 2\right)r = \frac{36}{7}r$. If given $P = 72\text{ cm}$, immediately solve $\frac{36}{7}r = 72 \implies r = 14\text{ cm}$ in two seconds!
Quadrant Perimeter $\frac{25}{7}r$ Shortcut
For a quarter of a circle (quadrant): $P_{\text{quad}} = \frac{1}{2}\pi r + 2r = \left(\frac{11}{7} + 2\right)r = \frac{25}{7}r$. If $P = 25\text{ cm}$, $r = 7\text{ cm}$.
Speed from RPM Direct Master Formula
Vehicle Speed in $\text{km/h} = \frac{\text{RPM} \times 2 \times \frac{22}{7} \times r(\text{cm}) \times 60}{100000} = \frac{\text{RPM} \times 132 \times r}{35000}$.
Concentric Circles Circumference Difference Shortcut
The difference in circumference between two concentric circles depends ONLY on the track width $w = R - r$: $\Delta C = 2\pi R - 2\pi r = 2\pi(R - r) = 2\pi w$. If $\Delta C = 44\text{ m}$, width $w = \frac{44}{2\pi} = 7\text{ m}$ instantly!
Key Formulas & Conversion Cheat Sheet
Circle Circumference
$$C = 2\pi r = \pi d$$
Boundary length of full circle. $\pi \approx \frac{22}{7}$.
Arc Length of Sector
$$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta}{180^\circ} \times \pi r$$
$\theta$ is the central angle measured in degrees.
Perimeter of Semicircular Region
$$P_{\text{semi}} = \pi r + 2r = r\left(\frac{22}{7} + 2\right) = \frac{36}{7}r$$
Includes both curved arc ($\pi r$) and straight base ($2r$).
Perimeter of Sector
$$P_{\text{sector}} = l + 2r = \frac{\theta}{360^\circ}(2\pi r) + 2r$$
Sum of arc length and two bounding radii.
Wheel Revolutions Formula
$$N = \frac{\text{Total Distance Covered}}{\text{Circumference}} = \frac{D}{2\pi r}$$
Ensure $D$ and $r$ are expressed in identical units (e.g. centimetres).
Chord Perpendicular Distance (Pythagoras)
$$r^2 = d_\perp^2 + \left(\frac{L}{2}\right)^2 \iff L = 2\sqrt{r^2 - d_\perp^2}$$
$L$ is chord length, $d_\perp$ is perpendicular distance from centre.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Part A - Descriptive
5 Marks
The perimeter of a semicircular protractor is $72\text{ cm}$. Find its radius and diameter. (Take $\pi = \frac{22}{7}$)
Step-by-Step Model Solution:
Step 1: Formula for perimeter of a closed semicircular region
Let the radius of the semicircular protractor be $r\text{ cm}$.
A semicircular protractor is bounded by a curved semicircular arc of length $\pi r$ and a straight diameter of length $2r$.
Therefore, the total perimeter $P$ is given by:
$$P = \pi r + 2r = r(\pi + 2)$$
Step 2: Substitute given values and solve for $r$
Given that $P = 72\text{ cm}$ and $\pi = \frac{22}{7}$:
$$r\left(\frac{22}{7} + 2\right) = 72$$ $$r\left(\frac{22 + 14}{7}\right) = 72$$ $$r\left(\frac{36}{7}\right) = 72$$
Cross-multiplying to solve for $r$:
$$r = \frac{72 \times 7}{36} = 2 \times 7 = 14\text{ cm}$$
Step 3: Calculate the diameter
$$\text{Diameter } d = 2r = 2 \times 14\text{ cm} = 28\text{ cm}$$
$$\mathbf{\text{Conclusion: The radius is } 14\text{ cm and the diameter is } 28\text{ cm.}}$$
Let the radius of the semicircular protractor be $r\text{ cm}$.
A semicircular protractor is bounded by a curved semicircular arc of length $\pi r$ and a straight diameter of length $2r$.
Therefore, the total perimeter $P$ is given by:
$$P = \pi r + 2r = r(\pi + 2)$$
Step 2: Substitute given values and solve for $r$
Given that $P = 72\text{ cm}$ and $\pi = \frac{22}{7}$:
$$r\left(\frac{22}{7} + 2\right) = 72$$ $$r\left(\frac{22 + 14}{7}\right) = 72$$ $$r\left(\frac{36}{7}\right) = 72$$
Cross-multiplying to solve for $r$:
$$r = \frac{72 \times 7}{36} = 2 \times 7 = 14\text{ cm}$$
Step 3: Calculate the diameter
$$\text{Diameter } d = 2r = 2 \times 14\text{ cm} = 28\text{ cm}$$
$$\mathbf{\text{Conclusion: The radius is } 14\text{ cm and the diameter is } 28\text{ cm.}}$$
Rough Column Working:
P = 36/7 * r = 72
r = 72 * 7 / 36 = 14 cm
d = 28 cm.
P = 36/7 * r = 72
r = 72 * 7 / 36 = 14 cm
d = 28 cm.
Step Mark Allocation:
- Writing correct formula P = πr + 2r with diagram: 1.5 Marks
- Substituting π = 22/7 to get 36/7 * r = 72: 1.5 Marks
- Solving for r = 14 cm: 1 Marks
- Calculating diameter = 28 cm and boxed answer: 1 Marks
Examiner Scoring Key: The most common blunder is setting πr = 72, which yields r = 22.9 cm and loses 4 marks. Always include the straight diameter 2r in closed semicircular perimeters.
Part A - Descriptive
5 Marks
A sector of a circle of radius $10.5\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find:
(i) The length of the arc of the sector.
(ii) The total perimeter of the sector. (Take $\pi = \frac{22}{7}$)
(i) The length of the arc of the sector.
(ii) The total perimeter of the sector. (Take $\pi = \frac{22}{7}$)
Step-by-Step Model Solution:
Step 1: Identify given parameters
$$\text{Radius } r = 10.5\text{ cm} = \frac{21}{2}\text{ cm}$$ $$\text{Central Angle } \theta = 60^\circ$$
Step 2: (i) Calculate the length of the arc ($l$)
The formula for the arc length of a sector is:
$$l = \frac{\theta}{360^\circ} \times 2\pi r$$ Substituting the known values:
$$l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times \frac{21}{2}$$ $$l = \frac{1}{6} \times 2 \times \frac{22}{7} \times \frac{21}{2}$$ $$l = \frac{1}{6} \times 66 = 11\text{ cm}$$
$$\mathbf{\text{Length of arc } l = 11\text{ cm.}}$$
Step 3: (ii) Calculate the total perimeter of the sector
The total perimeter of the sector is bounded by the arc $l$ and the two radii $OA$ and $OB$ ($2r$):
$$P_{\text{sector}} = l + 2r$$ $$P_{\text{sector}} = 11\text{ cm} + 2(10.5\text{ cm}) = 11\text{ cm} + 21\text{ cm} = 32\text{ cm}$$
$$\mathbf{\text{Conclusion: (i) Arc length } = 11\text{ cm}; \; (ii) \text{ Total perimeter of sector } = 32\text{ cm.}}$$
$$\text{Radius } r = 10.5\text{ cm} = \frac{21}{2}\text{ cm}$$ $$\text{Central Angle } \theta = 60^\circ$$
Step 2: (i) Calculate the length of the arc ($l$)
The formula for the arc length of a sector is:
$$l = \frac{\theta}{360^\circ} \times 2\pi r$$ Substituting the known values:
$$l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times \frac{21}{2}$$ $$l = \frac{1}{6} \times 2 \times \frac{22}{7} \times \frac{21}{2}$$ $$l = \frac{1}{6} \times 66 = 11\text{ cm}$$
$$\mathbf{\text{Length of arc } l = 11\text{ cm.}}$$
Step 3: (ii) Calculate the total perimeter of the sector
The total perimeter of the sector is bounded by the arc $l$ and the two radii $OA$ and $OB$ ($2r$):
$$P_{\text{sector}} = l + 2r$$ $$P_{\text{sector}} = 11\text{ cm} + 2(10.5\text{ cm}) = 11\text{ cm} + 21\text{ cm} = 32\text{ cm}$$
$$\mathbf{\text{Conclusion: (i) Arc length } = 11\text{ cm}; \; (ii) \text{ Total perimeter of sector } = 32\text{ cm.}}$$
Rough Column Working:
l = 60/360 * 2 * 22/7 * 21/2
= 1/6 * 66 = 11 cm.
P = 11 + 2*(10.5) = 11 + 21 = 32 cm.
l = 60/360 * 2 * 22/7 * 21/2
= 1/6 * 66 = 11 cm.
P = 11 + 2*(10.5) = 11 + 21 = 32 cm.
Step Mark Allocation:
- Writing arc length formula and substituting values: 1.5 Marks
- Solving for arc length l = 11 cm: 1.5 Marks
- Writing sector perimeter formula P = l + 2r: 1 Marks
- Computing P = 32 cm and boxed answer: 1 Marks
Examiner Scoring Key: Writing 10.5 as 21/2 facilitates immediate cancellation with 7 and 2. Avoid cumbersome decimal long multiplication.
Part B - Rotational Dynamics
10 Marks
The diameter of the wheel of a car is $70\text{ cm}$.
(i) How many revolutions will the wheel make in travelling a distance of $22\text{ km}$?
(ii) If the car travels at a speed of $66\text{ km/h}$, find the number of revolutions made by the wheel per minute (RPM).
(i) How many revolutions will the wheel make in travelling a distance of $22\text{ km}$?
(ii) If the car travels at a speed of $66\text{ km/h}$, find the number of revolutions made by the wheel per minute (RPM).
Step-by-Step Model Solution:
Step 1: Calculate the circumference of the wheel
Given:
$$\text{Diameter of wheel } d = 70\text{ cm} \implies \text{Radius } r = \frac{70}{2} = 35\text{ cm}$$
The distance covered by the wheel in exactly one complete revolution is its circumference $C$:
$$C = \pi d = \frac{22}{7} \times 70 = 22 \times 10 = 220\text{ cm}$$
Step 2: (i) Calculate revolutions to travel $22\text{ km}$
Convert total distance $D$ from kilometres to centimetres:
$$1\text{ km} = 1000\text{ m} = 100000\text{ cm}$$ $$D = 22\text{ km} = 22 \times 100000 = 2,200,000\text{ cm}$$
Number of revolutions $N$ is given by:
$$N = \frac{\text{Total Distance}}{\text{Distance in } 1 \text{ revolution}} = \frac{D}{C}$$ $$N = \frac{2200000}{220} = 10000\text{ revolutions}$$
$$\mathbf{\text{The wheel makes } 10,000\text{ revolutions to travel } 22\text{ km.}}$$
Step 3: (ii) Calculate Revolutions Per Minute (RPM) at $66\text{ km/h}$
Given speed of car $= 66\text{ km/h}$.
First, determine the distance travelled by the car in $1$ minute:
$$\text{Distance in } 1 \text{ hour (60 min)} = 66\text{ km} = 66 \times 100000\text{ cm} = 6,600,000\text{ cm}$$ $$\text{Distance in } 1 \text{ minute} = \frac{6,600,000\text{ cm}}{60} = 110,000\text{ cm}$$
Now calculate the number of revolutions in $1$ minute:
$$\text{RPM} = \frac{\text{Distance in } 1 \text{ minute}}{C} = \frac{110000\text{ cm}}{220\text{ cm}}$$ $$\text{RPM} = \frac{110000}{220} = \frac{11000}{22} = 500\text{ revolutions per minute}$$
$$\mathbf{\text{Conclusion: (i) } 10,000\text{ revolutions; } \; (ii) \; 500\text{ revolutions per minute.}}$$
Given:
$$\text{Diameter of wheel } d = 70\text{ cm} \implies \text{Radius } r = \frac{70}{2} = 35\text{ cm}$$
The distance covered by the wheel in exactly one complete revolution is its circumference $C$:
$$C = \pi d = \frac{22}{7} \times 70 = 22 \times 10 = 220\text{ cm}$$
Step 2: (i) Calculate revolutions to travel $22\text{ km}$
Convert total distance $D$ from kilometres to centimetres:
$$1\text{ km} = 1000\text{ m} = 100000\text{ cm}$$ $$D = 22\text{ km} = 22 \times 100000 = 2,200,000\text{ cm}$$
Number of revolutions $N$ is given by:
$$N = \frac{\text{Total Distance}}{\text{Distance in } 1 \text{ revolution}} = \frac{D}{C}$$ $$N = \frac{2200000}{220} = 10000\text{ revolutions}$$
$$\mathbf{\text{The wheel makes } 10,000\text{ revolutions to travel } 22\text{ km.}}$$
Step 3: (ii) Calculate Revolutions Per Minute (RPM) at $66\text{ km/h}$
Given speed of car $= 66\text{ km/h}$.
First, determine the distance travelled by the car in $1$ minute:
$$\text{Distance in } 1 \text{ hour (60 min)} = 66\text{ km} = 66 \times 100000\text{ cm} = 6,600,000\text{ cm}$$ $$\text{Distance in } 1 \text{ minute} = \frac{6,600,000\text{ cm}}{60} = 110,000\text{ cm}$$
Now calculate the number of revolutions in $1$ minute:
$$\text{RPM} = \frac{\text{Distance in } 1 \text{ minute}}{C} = \frac{110000\text{ cm}}{220\text{ cm}}$$ $$\text{RPM} = \frac{110000}{220} = \frac{11000}{22} = 500\text{ revolutions per minute}$$
$$\mathbf{\text{Conclusion: (i) } 10,000\text{ revolutions; } \; (ii) \; 500\text{ revolutions per minute.}}$$
Rough Column Working:
C = 22/7 * 70 = 220 cm
22 km = 2,200,000 cm
N = 2,200,000 / 220 = 10,000 rev
Speed = 66 km/h = 1.1 km/min = 110,000 cm/min
RPM = 110,000 / 220 = 500.
C = 22/7 * 70 = 220 cm
22 km = 2,200,000 cm
N = 2,200,000 / 220 = 10,000 rev
Speed = 66 km/h = 1.1 km/min = 110,000 cm/min
RPM = 110,000 / 220 = 500.
Step Mark Allocation:
- Calculating circumference of wheel C = 220 cm: 2 Marks
- Converting 22 km to 2,200,000 cm with explicit units: 2 Marks
- Solving for total revolutions N = 10,000: 2 Marks
- Converting 66 km/h to distance per minute = 110,000 cm/min: 2 Marks
- Calculating RPM = 500 and boxed final statements: 2 Marks
Examiner Scoring Key: Clearly exhibit the unit conversion: 1 km = 1,00,000 cm. Skipping dimensional conversions leads to severe mark deductions.
Part B - Chord Geometry & Pythagoras
10 Marks
In a circle with centre $O$, a chord $AB$ of length $16\text{ cm}$ is at a distance of $6\text{ cm}$ from the centre. Find the length of another chord $CD$ of the same circle which is at a distance of $8\text{ cm}$ from the centre.
Step-by-Step Model Solution:
Step 1: Figure, Given Data & Geometric Theorems
Let $O$ be the centre of the circle with radius $r\text{ cm}$.
Let chord $AB = 16\text{ cm}$.
Let $OM \perp AB$ be the perpendicular from centre $O$ to chord $AB$, where $OM = 6\text{ cm}$.
By the Perpendicular Chord Bisector Theorem, the perpendicular drawn from the centre of a circle to a chord bisects the chord:
$$AM = MB = \frac{AB}{2} = \frac{16\text{ cm}}{2} = 8\text{ cm}$$
Step 2: Calculate the Radius ($r$) using Pythagoras Theorem
Join $OA$. In right-angled $\triangle OMA$ (where $\angle OMA = 90^\circ$):
$$OA^2 = OM^2 + AM^2 \quad [\text{Pythagorean Theorem}]$$ $$r^2 = 6^2 + 8^2$$ $$r^2 = 36 + 64 = 100$$ $$r = \sqrt{100} = 10\text{ cm}$$
Therefore, the radius of the circle is $r = 10\text{ cm}$.
Step 3: Analyze the second chord $CD$
Let $CD$ be the second chord, and let $ON \perp CD$ with $ON = 8\text{ cm}$.
Join $OC$. Since $OC$ is a radius of the same circle, $OC = r = 10\text{ cm}$.
By the Perpendicular Chord Bisector Theorem, $N$ is the midpoint of $CD$:
$$CN = ND = \frac{CD}{2}$$
Step 4: Calculate the length of chord $CD$
In right-angled $\triangle ONC$ (where $\angle ONC = 90^\circ$):
$$OC^2 = ON^2 + CN^2 \quad [\text{Pythagorean Theorem}]$$ $$10^2 = 8^2 + CN^2$$ $$100 = 64 + CN^2$$ $$CN^2 = 100 - 64 = 36$$ $$CN = \sqrt{36} = 6\text{ cm}$$
Since $N$ bisects chord $CD$:
$$CD = 2 \times CN = 2 \times 6\text{ cm} = 12\text{ cm}$$
$$\mathbf{\text{Conclusion: The length of the second chord } CD \text{ is } 12\text{ cm.}}$$
Let $O$ be the centre of the circle with radius $r\text{ cm}$.
Let chord $AB = 16\text{ cm}$.
Let $OM \perp AB$ be the perpendicular from centre $O$ to chord $AB$, where $OM = 6\text{ cm}$.
By the Perpendicular Chord Bisector Theorem, the perpendicular drawn from the centre of a circle to a chord bisects the chord:
$$AM = MB = \frac{AB}{2} = \frac{16\text{ cm}}{2} = 8\text{ cm}$$
Step 2: Calculate the Radius ($r$) using Pythagoras Theorem
Join $OA$. In right-angled $\triangle OMA$ (where $\angle OMA = 90^\circ$):
$$OA^2 = OM^2 + AM^2 \quad [\text{Pythagorean Theorem}]$$ $$r^2 = 6^2 + 8^2$$ $$r^2 = 36 + 64 = 100$$ $$r = \sqrt{100} = 10\text{ cm}$$
Therefore, the radius of the circle is $r = 10\text{ cm}$.
Step 3: Analyze the second chord $CD$
Let $CD$ be the second chord, and let $ON \perp CD$ with $ON = 8\text{ cm}$.
Join $OC$. Since $OC$ is a radius of the same circle, $OC = r = 10\text{ cm}$.
By the Perpendicular Chord Bisector Theorem, $N$ is the midpoint of $CD$:
$$CN = ND = \frac{CD}{2}$$
Step 4: Calculate the length of chord $CD$
In right-angled $\triangle ONC$ (where $\angle ONC = 90^\circ$):
$$OC^2 = ON^2 + CN^2 \quad [\text{Pythagorean Theorem}]$$ $$10^2 = 8^2 + CN^2$$ $$100 = 64 + CN^2$$ $$CN^2 = 100 - 64 = 36$$ $$CN = \sqrt{36} = 6\text{ cm}$$
Since $N$ bisects chord $CD$:
$$CD = 2 \times CN = 2 \times 6\text{ cm} = 12\text{ cm}$$
$$\mathbf{\text{Conclusion: The length of the second chord } CD \text{ is } 12\text{ cm.}}$$
Rough Column Working:
AM = 16/2 = 8
r^2 = 6^2 + 8^2 = 100 => r = 10 cm
In 2nd triangle: CN^2 = 10^2 - 8^2 = 36 => CN = 6 cm
CD = 2 * 6 = 12 cm.
AM = 16/2 = 8
r^2 = 6^2 + 8^2 = 100 => r = 10 cm
In 2nd triangle: CN^2 = 10^2 - 8^2 = 36 => CN = 6 cm
CD = 2 * 6 = 12 cm.
Step Mark Allocation:
- Neat diagram, stating perpendicular bisector theorem AM = 8 cm: 2.5 Marks
- Using Pythagoras in triangle OMA to find radius r = 10 cm: 2.5 Marks
- Setting up Pythagoras in triangle ONC to find CN = 6 cm: 2.5 Marks
- Doubling CN to find chord CD = 12 cm with boxed answer: 2.5 Marks
Examiner Scoring Key: Candidates frequently stop after finding CN = 6 cm, forgetting to double it to get the full chord length CD. Read the requirement carefully.
Part A - Descriptive Track
5 Marks
A circular running track has an inner circumference of $440\text{ m}$ and an outer circumference of $484\text{ m}$. Find the width of the track. (Take $\pi = \frac{22}{7}$)
Step-by-Step Model Solution:
Step 1: Formulate the concentric circles model
Let the inner radius of the track be $r\text{ m}$ and the outer radius be $R\text{ m}$.
The width of the circular track $w$ is the difference between the two radii:
$$w = R - r$$
Step 2: Express circumferences algebraically
$$\text{Outer Circumference } C_2 = 2\pi R = 484\text{ m}$$ $$\text{Inner Circumference } C_1 = 2\pi r = 440\text{ m}$$
Subtracting the inner circumference from the outer circumference:
$$C_2 - C_1 = 2\pi R - 2\pi r$$ $$484 - 440 = 2\pi(R - r)$$ $$44 = 2\pi w$$
Step 3: Solve for width $w$
$$44 = 2 \times \frac{22}{7} \times w$$ $$44 = \frac{44}{7} \times w$$ $$w = \frac{44 \times 7}{44} = 7\text{ m}$$
$$\mathbf{\text{Conclusion: The width of the circular track is } 7\text{ metres.}}$$
Let the inner radius of the track be $r\text{ m}$ and the outer radius be $R\text{ m}$.
The width of the circular track $w$ is the difference between the two radii:
$$w = R - r$$
Step 2: Express circumferences algebraically
$$\text{Outer Circumference } C_2 = 2\pi R = 484\text{ m}$$ $$\text{Inner Circumference } C_1 = 2\pi r = 440\text{ m}$$
Subtracting the inner circumference from the outer circumference:
$$C_2 - C_1 = 2\pi R - 2\pi r$$ $$484 - 440 = 2\pi(R - r)$$ $$44 = 2\pi w$$
Step 3: Solve for width $w$
$$44 = 2 \times \frac{22}{7} \times w$$ $$44 = \frac{44}{7} \times w$$ $$w = \frac{44 \times 7}{44} = 7\text{ m}$$
$$\mathbf{\text{Conclusion: The width of the circular track is } 7\text{ metres.}}$$
Rough Column Working:
2*pi*(R - r) = 484 - 440 = 44
44/7 * w = 44 => w = 7 m.
2*pi*(R - r) = 484 - 440 = 44
44/7 * w = 44 => w = 7 m.
Step Mark Allocation:
- Defining inner and outer radii and track width w = R - r: 1 Marks
- Setting up equation 2πR - 2πr = 44: 1.5 Marks
- Factoring 2π(R - r) = 44: 1 Marks
- Solving for w = 7 m and boxed answer: 1.5 Marks
Examiner Scoring Key: Factoring out 2π directly saves significant time over calculating R and r individually.
Part B - Wire Re-bending Mensuration
10 Marks
A wire is in the form of a square of side $22\text{ cm}$. If the same wire is straightened and then rebent into the shape of a circle, find:
(i) The radius of the circle.
(ii) The area of the square and the area of the circle.
(iii) Which shape encloses more area and by how much? (Take $\pi = \frac{22}{7}$)
(i) The radius of the circle.
(ii) The area of the square and the area of the circle.
(iii) Which shape encloses more area and by how much? (Take $\pi = \frac{22}{7}$)
Step-by-Step Model Solution:
Step 1: Determine the total length of the wire
Given that the wire is initially bent in the shape of a square of side $a = 22\text{ cm}$.
The total length of the wire equals the perimeter of the square:
$$\text{Length of wire } L = 4a = 4 \times 22\text{ cm} = 88\text{ cm}$$
Step 2: (i) Find the radius of the rebent circle
When the same wire is rebent into a circle, the circumference of the circle must equal the total length of the wire:
$$\text{Circumference of Circle } C = L = 88\text{ cm}$$ $$2\pi r = 88$$ $$2 \times \frac{22}{7} \times r = 88$$ $$\frac{44}{7} \times r = 88$$ $$r = \frac{88 \times 7}{44} = 2 \times 7 = 14\text{ cm}$$
$$\mathbf{\text{Radius of the circle } r = 14\text{ cm.}}$$
Step 3: (ii) Calculate the areas of both figures
• Area of the square:
$$\text{Area}_{\text{square}} = a^2 = 22^2 = 484\text{ cm}^2$$
• Area of the circle:
$$\text{Area}_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 14^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616\text{ cm}^2$$
Step 4: (iii) Compare the enclosed areas
$$\text{Difference in Area} = \text{Area}_{\text{circle}} - \text{Area}_{\text{square}}$$ $$\text{Difference in Area} = 616\text{ cm}^2 - 484\text{ cm}^2 = 132\text{ cm}^2$$
$$\mathbf{\text{Conclusion: (i) Radius } = 14\text{ cm}; \; (ii) \text{ Area}_{\text{sq}} = 484\text{ cm}^2, \text{ Area}_{\text{circ}} = 616\text{ cm}^2; \; (iii) \text{ Circle encloses } 132\text{ cm}^2 \text{ more area.}}$$
Given that the wire is initially bent in the shape of a square of side $a = 22\text{ cm}$.
The total length of the wire equals the perimeter of the square:
$$\text{Length of wire } L = 4a = 4 \times 22\text{ cm} = 88\text{ cm}$$
Step 2: (i) Find the radius of the rebent circle
When the same wire is rebent into a circle, the circumference of the circle must equal the total length of the wire:
$$\text{Circumference of Circle } C = L = 88\text{ cm}$$ $$2\pi r = 88$$ $$2 \times \frac{22}{7} \times r = 88$$ $$\frac{44}{7} \times r = 88$$ $$r = \frac{88 \times 7}{44} = 2 \times 7 = 14\text{ cm}$$
$$\mathbf{\text{Radius of the circle } r = 14\text{ cm.}}$$
Step 3: (ii) Calculate the areas of both figures
• Area of the square:
$$\text{Area}_{\text{square}} = a^2 = 22^2 = 484\text{ cm}^2$$
• Area of the circle:
$$\text{Area}_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 14^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616\text{ cm}^2$$
Step 4: (iii) Compare the enclosed areas
$$\text{Difference in Area} = \text{Area}_{\text{circle}} - \text{Area}_{\text{square}}$$ $$\text{Difference in Area} = 616\text{ cm}^2 - 484\text{ cm}^2 = 132\text{ cm}^2$$
$$\mathbf{\text{Conclusion: (i) Radius } = 14\text{ cm}; \; (ii) \text{ Area}_{\text{sq}} = 484\text{ cm}^2, \text{ Area}_{\text{circ}} = 616\text{ cm}^2; \; (iii) \text{ Circle encloses } 132\text{ cm}^2 \text{ more area.}}$$
Rough Column Working:
Perimeter = 4 * 22 = 88 cm
2 * 22/7 * r = 88 => r = 14 cm
Area sq = 22^2 = 484
Area circle = 22/7 * 14 * 14 = 616
Diff = 616 - 484 = 132 cm^2.
Perimeter = 4 * 22 = 88 cm
2 * 22/7 * r = 88 => r = 14 cm
Area sq = 22^2 = 484
Area circle = 22/7 * 14 * 14 = 616
Diff = 616 - 484 = 132 cm^2.
Step Mark Allocation:
- Calculating perimeter of square = 88 cm: 2 Marks
- Equating 2πr = 88 and solving for radius r = 14 cm: 2.5 Marks
- Calculating Area of square = 484 sq cm and Area of circle = 616 sq cm: 3 Marks
- Comparing areas and finding difference = 132 sq cm with conclusion: 2.5 Marks
Examiner Scoring Key: Classic isoperimetric theorem application: for a given fixed perimeter, a circle always encloses maximum area compared to any polygon.
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 circumference of a circle exceeds its diameter by $30\text{ cm}$. Find the radius of the circle. (Take $\pi = \frac{22}{7}$)
Model Descriptive Solution:
Let radius be $r$. Then diameter $d = 2r$ and circumference $C = 2\pi r$. Given $C - d = 30 \implies 2\pi r - 2r = 30 \implies 2r(\pi - 1) = 30 \implies 2r(\frac{22}{7} - 1) = 30 \implies 2r(\frac{15}{7}) = 30 \implies \frac{30}{7}r = 30 \implies r = 7\text{ cm}$.
Question 2
5 Marks
Find the perimeter of a quadrant of a circle whose radius is $14\text{ cm}$.
Model Descriptive Solution:
A quadrant has central angle $90^\circ$. Arc length $l = \frac{90^\circ}{360^\circ} \times 2\pi r = \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22\text{ cm}$. Total perimeter includes two radii: $P = l + 2r = 22 + 2(14) = 22 + 28 = 50\text{ cm}$.
Question 3
5 Marks
A bicycle wheel of diameter $56\text{ cm}$ makes $250$ complete rotations. What is the total distance covered in metres?
Model Descriptive Solution:
Circumference $C = \pi d = \frac{22}{7} \times 56 = 176\text{ cm}$. Total distance in 250 rotations $= 250 \times 176\text{ cm} = 44,000\text{ cm} = \frac{44000}{100} = 440\text{ metres}$.
Question 4
5 Marks
Two circles touch externally. The sum of their areas is $130\pi\text{ cm}^2$ and the distance between their centres is $14\text{ cm}$. Find the radii of the two circles.
Model Descriptive Solution:
For externally touching circles, distance between centres $r_1 + r_2 = 14\text{ cm}$. Sum of areas: $\pi r_1^2 + \pi r_2^2 = 130\pi \implies r_1^2 + r_2^2 = 130$. Using identity $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1 r_2 \implies 14^2 = 130 + 2r_1 r_2 \implies 196 - 130 = 2r_1 r_2 \implies 2r_1 r_2 = 66$. Then $(r_1 - r_2)^2 = (r_1^2 + r_2^2) - 2r_1 r_2 = 130 - 66 = 64 \implies r_1 - r_2 = 8$. Solving $r_1 + r_2 = 14$ and $r_1 - r_2 = 8$ gives $2r_1 = 22 \implies r_1 = 11\text{ cm}$ and $r_2 = 3\text{ cm}$.
Question 5
10 Marks
A horse is tied to a pole in a corner of a square grass field of side $30\text{ m}$ by means of a rope $14\text{ m}$ long. Find the area of the portion of the field over which the horse can graze, and find the ungrazed area.
Model Descriptive Solution:
The corner of a square field forms a right angle ($90^\circ$). The grazing area is a quadrant of a circle of radius $r = 14\text{ m}$. Grazing area $= \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 616 = 154\text{ m}^2$. Total area of square field $= 30 \times 30 = 900\text{ m}^2$. Ungrazed area $= 900 - 154 = 746\text{ m}^2$.
Question 6
10 Marks
A chord of length $30\text{ cm}$ is drawn in a circle of radius $17\text{ cm}$. Calculate its distance from the centre. Also calculate the length of another chord in the same circle whose distance from the centre is $15\text{ cm}$.
Model Descriptive Solution:
Half-chord $= \frac{30}{2} = 15\text{ cm}$. By Pythagoras theorem: $r^2 = d^2 + 15^2 \implies 17^2 = d^2 + 225 \implies 289 - 225 = d^2 \implies d^2 = 64 \implies d = 8\text{ cm}$. For second chord with $d_2 = 15\text{ cm}$: $17^2 = 15^2 + (L/2)^2 \implies 289 = 225 + (L/2)^2 \implies (L/2)^2 = 64 \implies L/2 = 8\text{ cm} \implies L = 16\text{ cm}$.
Question 7
5 Marks
The ratio of the radii of two circles is $3 : 4$. Find the ratio of their circumferences and the ratio of their areas.
Model Descriptive Solution:
Let radii be $r_1 = 3x$ and $r_2 = 4x$. Ratio of circumferences: $\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2} = \frac{3}{4} = 3 : 4$. Ratio of areas: $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} = 9 : 16$.
Question 8
5 Marks
Prove that a diameter of a circle subtends a right angle at any point on the circumference of the circle.
Model Descriptive Solution:
Let $AB$ be a diameter with centre $O$. Let $P$ be any point on the circumference. Join $OP$. In $\triangle OAP$, $OA = OP$ (radii) $\implies \angle OAP = \angle OPA = x$. In $\triangle OBP$, $OB = OP$ (radii) $\implies \angle OBP = \angle OPB = y$. In $\triangle APB$, the sum of angles is $180^\circ$: $\angle PAB + \angle APB + \angle PBA = 180^\circ \implies x + (x + y) + y = 180^\circ \implies 2x + 2y = 180^\circ \implies 2(x + y) = 180^\circ \implies x + y = 90^\circ$. Since $\angle APB = x + y$, it follows that $\angle APB = 90^\circ$. Hence Proved.
Question 9
10 Marks
The minute hand of a clock is $10.5\text{ cm}$ long. Find the area swept by the minute hand between $9:00\text{ AM}$ and $9:20\text{ AM}$, and find the distance moved by the tip of the minute hand.
Model Descriptive Solution:
Time elapsed $= 20\text{ minutes}$. In 60 minutes, minute hand rotates $360^\circ$. In 20 minutes, $\theta = \frac{20}{60} \times 360^\circ = 120^\circ$. Radius $r = 10.5\text{ cm} = \frac{21}{2}\text{ cm}$. Area swept $= \frac{\theta}{360^\circ}\pi r^2 = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = \frac{1}{3} \times \frac{22}{7} \times \frac{441}{4} = 115.5\text{ cm}^2$. Distance moved by tip is arc length $l = \frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times \frac{21}{2} = \frac{1}{3} \times 66 = 22\text{ cm}$.
Question 10
10 Marks
An athletic track consists of two straight parallel segments of length $100\text{ m}$ each, connected by semicircular ends. The inner radius of the semicircular ends is $35\text{ m}$, and the track is $7\text{ m}$ wide throughout. Find the distance along the outer edge of the track in completing one full lap.
Model Descriptive Solution:
The outer edge consists of two straight sections of $100\text{ m}$ each, and two semicircular ends of outer radius $R = r + w = 35 + 7 = 42\text{ m}$. The two semicircular ends combine to form one complete circle of radius $R = 42\text{ m}$. Curved distance $= 2\pi R = 2 \times \frac{22}{7} \times 42 = 2 \times 22 \times 6 = 264\text{ m}$. Straight distance $= 2 \times 100\text{ m} = 200\text{ m}$. Total distance along outer edge $= 200\text{ m} + 264\text{ m} = 464\text{ metres}$.
High-Yield FAQs & Revision Notes
What is the difference between a chord, secant, and tangent?
A chord is a straight line segment whose endpoints both lie on the circle. A secant is a line that cuts through the circle at two points and extends infinitely. A tangent is a line that touches the circle at exactly one point and is perpendicular to the radius at that point.
Why is the perimeter of a semicircle not equal to πr?
The curved arc alone has length $\pi r$. However, a closed semicircular region also includes the straight diameter edge ($2r$) connecting the two endpoints. Hence, its perimeter is $P = \pi r + 2r$.
How many revolutions does a wheel make per kilometer?
Number of revolutions $= \frac{100,000\text{ cm}}{\text{Circumference in cm}} = \frac{100,000}{2\pi r}$.
What is the difference between a sector and a segment?
A sector is bounded by two radii and an arc (like a pizza slice). A segment is bounded by a chord and an arc.
What is the value of π to be used in RIMC exams?
Always use $\pi = \frac{22}{7}$ unless the question explicitly specifies $\pi = 3.14$ or $3.1416$.
What is the perpendicular chord bisector theorem?
The perpendicular dropped from the centre of a circle to any chord bisects that chord. This creates a right-angled triangle where $\text{Radius}^2 = \text{Perpendicular}^2 + (\frac{1}{2}\text{Chord})^2$.
Are concentric circles always parallel?
Concentric circles lie in the same plane with the same centre but different radii. The distance between them along any radial line is constant ($R - r$), forming a circular ring (annulus).
What is the relationship between the radius and the area of a circle?
Area is proportional to the square of the radius ($A = \pi r^2$). If the radius is doubled, the area increases by $2^2 = 4$ times.
Why does a circle enclose more area than a square of the same perimeter?
By the Isoperimetric Theorem, among all plane figures with a given fixed perimeter, the circle encloses the maximum possible area.
What is the central angle of a semicircle and a quadrant?
A semicircle has a central angle of $180^\circ$. A quadrant has a central angle of $90^\circ$.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 17, 40 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Checkpoint 1 | Remember: Perimeter of closed semicircle = πr + 2r = 36/7 * r. |
| Checkpoint 2 | Perimeter of a sector = arc length + 2r = (θ/360 * 2πr) + 2r. |
| Checkpoint 3 | Distance covered in 1 revolution = Circumference = 2πr = πd. |
| Checkpoint 4 | Ensure unit consistency in wheel problems: 1 km = 1,00,000 cm. |
| Checkpoint 5 | Chord Pythagoras relation: r^2 = (distance from centre)^2 + (half chord)^2. |
| Checkpoint 6 | Difference between concentric circumferences: ΔC = 2π(R - r) = 2πw. |