Official Syllabus Item: 20, 34 Geometry

Triangles: Classification, Properties, Congruence & Pythagorean Theorem

Master triangle classification (side-wise and angle-wise), angle sum theorem, exterior angle property, triangle inequality theorem, Pythagoras theorem, and congruence criteria for RIMC Maths 2026.

Overview & Fundamental Concept

Triangles constitute the fundamental structural unit of plane geometry in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 20 ('Plane Figures') and 34 ('Types of Angles Classification of Triangles - Sidewise, Angle wise Polygons') of Chapter 8 in the official brochure, triangle geometry appears prominently in both Part A (5 Marks) and Part B (10 Marks). RIMC examiners place extraordinary weight on rigorous Euclidean proofs: constructing auxiliary parallel lines, stating formal 'Given', 'To Prove', and 'Proof' sections, justifying every algebraic transition with recognized theorems (such as the Angle Sum Property, Exterior Angle Theorem, or Triangle Inequality), and applying congruence criteria (SSS, SAS, ASA, RHS) to deduce geometric equalities via CPCTC. Mastering both numerical computation and formal geometric proof writing is indispensable for scoring top-percentile marks.

Official Syllabus Coverage Breakdown
Syllabus Item Prescribed Scope Exam Relevance & Pedagogical Objective
Item 20 Plane Figures: Triangles as 3-sided closed rectilinear polygons, altitude, median, centroid, and circumcenter/incenter definitions. Establishes definitions of geometric segments and points of concurrency within plane figures.
Item 34 Classification of Triangles: Sidewise (Equilateral, Isosceles, Scalene) and Angle-wise (Acute, Right, Obtuse), Angle Sum Theorem, and Exterior Angle Property. Directly specified in syllabus item 34. Assesses geometric reasoning, algebraic systems, and congruence deductions.
1. Dual Classification of Triangles

A triangle is a three-sided closed plane polygon bounded by three straight line segments connecting three non-collinear points. Triangles are classified according to two independent criteria:

A. Classification Based on Length of Sides (Side-wise):
1. Equilateral Triangle: All three sides are equal in length ($a = b = c$). Consequently, all three interior angles are equal to $60^\circ$ each (equiangular).
2. Isosceles Triangle: At least two sides are equal in length ($a = b \ne c$). The angles opposite to the equal sides are also equal in measure (base angles theorem).
3. Scalene Triangle: All three sides have distinctly different lengths ($a \ne b \ne c$). All three interior angles have different measures.

B. Classification Based on Measure of Angles (Angle-wise):
1. Acute-angled Triangle: All three interior angles are strictly acute (each angle $< 90^\circ$).
2. Right-angled Triangle: Exactly one interior angle is a right angle ($= 90^\circ$). The side opposite the right angle is the hypotenuse (the longest side); the other two sides are legs or base and perpendicular.
3. Obtuse-angled Triangle: Exactly one interior angle is obtuse ($> 90^\circ$). The remaining two angles must be acute.

2. Medians, Altitudes & Points of Concurrency

Four special concurrent sets of lines define the key centers of a triangle:

• Median: A line segment joining any vertex of a triangle to the midpoint of the opposite side. Every triangle has exactly three medians. The three medians are concurrent at a single interior point called the Centroid ($G$). The centroid divides each median in the ratio $2 : 1$ from the vertex to the base.
• Altitude: The perpendicular line segment drawn from a vertex to the opposite side (or the line containing the opposite side). The three altitudes are concurrent at the Orthocenter ($H$). (The orthocenter lies inside an acute triangle, at the right-angle vertex of a right triangle, and outside an obtuse triangle).
• Perpendicular Bisector: A line perpendicular to a side at its midpoint. The three perpendicular bisectors meet at the Circumcenter ($O$), which is equidistant from all three vertices.
• Angle Bisector: A ray dividing an interior angle into two equal halves. The three internal angle bisectors meet at the Incenter ($I$), which is equidistant from all three sides.

3. Fundamental Triangle Theorems & Inequalities

Angle Sum Property of a Triangle:
The sum of the measures of the three interior angles of any triangle is strictly $180^\circ$ ($2$ right angles):
$$\angle A + \angle B + \angle C = 180^\circ$$

Exterior Angle Property:
If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles:
$$\angle \text{Ext} = \angle \text{Int.Opp}_1 + \angle \text{Int.Opp}_2$$
Furthermore, an exterior angle is always strictly greater than either of its interior opposite angles.

Triangle Inequality Theorem:
A triangle can be formed if and only if the sum of the lengths of any two sides is strictly greater than the length of the third side:
$$a + b > c, \quad b + c > a, \quad c + a > b$$
Topper Verification Rule: To check if three lengths can form a triangle, verify only that the sum of the two smaller sides is strictly greater than the longest side.
Equivalently, the difference between any two sides must be strictly less than the third side: $|a - b| < c < a + b$.

4. The Pythagorean Theorem & Congruence Criteria

Pythagorean Theorem: In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two perpendicular sides:
$$\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2 \iff c^2 = a^2 + b^2$$
Converse of Pythagoras Theorem: If in a triangle $c^2 = a^2 + b^2$, then the angle opposite to side $c$ is a right angle ($90^\circ$).
Pythagorean Triplets: Sets of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$. Generated by $(2m, m^2 - 1, m^2 + 1)$ for any integer $m > 1$. Standard triplets to memorize: $(3,4,5), (5,12,13), (7,24,25), (8,15,17), (9,40,41), (12,35,37), (20,21,29)$.

Congruence of Triangles ($\cong$): Two triangles are congruent if they are identical in shape and size (their corresponding sides and angles match). Valid criteria are:
1. SSS (Side-Side-Side): All three corresponding sides are equal.
2. SAS (Side-Angle-Side): Two sides and the included angle are equal.
3. ASA (Angle-Side-Angle): Two angles and the included side are equal.
4. AAS (Angle-Angle-Side): Two angles and a non-included side are equal.
5. RHS (Right angle-Hypotenuse-Side): Hypotenuse and one leg of a right triangle equal corresponding parts of another right triangle.
Warning: AAA proves similarity, not congruence! SSA (non-included angle) is mathematically invalid.

Illustrated Conceptual Diagrams & Visual Schematics
Right-Angled Triangle & Pythagoras Theorem Blueprint
A B (90°) C Perpendicular (p) Base (b) Hypotenuse (h) = √(p² + b²) Pythagorean Metric AC² = AB² + BC² h = √(p² + b²) p = √(h² - b²) b = √(h² - p²) Area of Right Triangle: Area = ½ × b × p Perimeter = p + b + h
Pythagoras Theorem: In any right-angled triangle $\triangle ABC$ right-angled at $B$: $\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2$ ($AC^2 = AB^2 + BC^2$). Essential standard RIMC triplets: $(3,4,5)$, $(5,12,13)$, $(7,24,25)$, $(8,15,17)$, $(9,40,41)$, $(20,21,29)$.

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: Given, To Prove / Find & Figure 20% of marks

Draw an accurate, ruler-and-pencil figure. Label all vertices ($A, B, C$). Mark right angles with square symbols and equal sides/angles with tick marks. Write explicit 'Given:' and 'To Prove / Find:' headers.

Step 2: Construction Statement 20% of marks

If an auxiliary line is needed (e.g., 'Through vertex $A$, draw line $XY \parallel BC$'), describe it precisely under 'Construction:' before commencing proof steps.

Step 3: Deductive Steps with Theorem Citations 40% of marks

State each equation followed by its justification in square brackets, e.g., '[Angle Sum Property of $\triangle ABC$]', '[By Pythagoras Theorem in right $\triangle ABD$]', '[By CPCTC]'.

Step 4: Calculation & Final Conclusion 20% of marks

State the final numerical magnitude with units (e.g., '$\text{Length of ladder} = 17\text{ m}$') in a prominent box, or terminate proofs with 'Hence Proved'.

Examiner Traps & Common Blunders (Marks Penalties)
Confusing Altitude with Median 2 to 4 marks deducted for incorrect geometric assumption in area/length calculations.
Examiner Trap: Assuming the altitude also bisects the base in a scalene triangle. An altitude bisects the base ONLY in an isosceles (between equal legs) or equilateral triangle.
Quoting SSA as a Valid Congruence Rule Complete disqualification of congruence proof; 0 marks awarded for deductive reasoning.
Examiner Trap: Using two sides and a non-included angle to claim two triangles are congruent without RHS conditions.
Partial Check of Triangle Inequality Zero marks for triangle existence questions.
Examiner Trap: Checking only one side combination (e.g., $10 + 15 > 6$) and concluding a triangle exists, without verifying whether $6 + 10 > 15$.
Applying Pythagoras to Non-Right Triangles Total loss of marks for that proof section.
Examiner Trap: Writing $a^2 + b^2 = c^2$ without establishing or being given that the angle between $a$ and $b$ is $90^\circ$.
Incorrect Exterior Angle Identification Disrupts entire angle-chasing chain; 3 to 5 mark penalty.
Examiner Trap: Considering an angle formed by extending both arms or drawing an internal ray as an exterior angle.
Cadet Topper Mental Verification Shortcuts
Centroid 2:1 Segment Rule
If $AD$ is a median meeting centroid $G$, then $AG = \frac{2}{3}AD$ and $GD = \frac{1}{3}AD$. If $GD = 4\text{ cm}$, immediately deduce $AG = 8\text{ cm}$ and $AD = 12\text{ cm}$.
Third Side Boundary Rule
For any two known sides $a$ and $b$ (with $a \ge b$), the integer length of the third side $c$ strictly satisfies: $(a - b + 1) \le c \le (a + b - 1)$.
Equilateral Triangle Altitude-to-Side Relation
In an equilateral triangle of side $a$, altitude $h = \frac{\sqrt{3}}{2}a$, and Area $= \frac{\sqrt{3}}{4}a^2$. This bypasses lengthy Pythagoras calculations.
Isosceles Right Triangle Ratio (45-45-90)
In an isosceles right triangle, side ratios are always $1 : 1 : \sqrt{2}$. If legs are $x$, hypotenuse is $x\sqrt{2}$. If hypotenuse is $H$, legs are $\frac{H}{\sqrt{2}}$.

Key Formulas & Conversion Cheat Sheet

Angle Sum Property
$$\angle A + \angle B + \angle C = 180^\circ$$

Valid for all Euclidean planar triangles.

Exterior Angle Theorem
$$\angle ACD = \angle A + \angle B$$

Exterior angle equals the sum of its two interior opposite angles.

Triangle Inequality Rule
$$a + b > c, \quad |a - b| < c < a + b$$

Sum of any two sides must strictly exceed the third side.

Pythagorean Theorem
$$c^2 = a^2 + b^2 \iff \text{Hypotenuse} = \sqrt{a^2 + b^2}$$

Strictly for right-angled triangles where $c$ is opposite the $90^\circ$ angle.

Centroid Median Division
$$AG : GD = 2 : 1 \implies AG = \frac{2}{3}AD, \quad GD = \frac{1}{3}AD$$

$G$ is centroid, $D$ is midpoint of opposite side $BC$.

Equilateral Triangle Metric Formulas
$$h = \frac{\sqrt{3}}{2}a, \quad \text{Area} = \frac{\sqrt{3}}{4}a^2, \quad P = 3a$$

$a$ is side length, $h$ is altitude.

Solved Model Examples (Descriptive Format with Step Mark Distribution)
Part A - Descriptive 5 Marks
The angles of a triangle are in the ratio $2 : 3 : 4$. Find the measure of each angle. Classify the triangle both side-wise and angle-wise.
Step-by-Step Model Solution:
Step 1: Define unknown angles algebraically
Let the measures of the angles of the triangle be $2x, 3x,$ and $4x$, where $x$ is a common ratio multiplier.

Step 2: Apply the Angle Sum Property
The sum of all three interior angles of a triangle is strictly $180^\circ$ [Angle Sum Property of a Triangle]:
$$2x + 3x + 4x = 180^\circ$$ $$9x = 180^\circ$$ $$x = \frac{180^\circ}{9} = 20^\circ$$
Step 3: Calculate the measures of each angle
$$\angle 1 = 2x = 2(20^\circ) = 40^\circ$$ $$\angle 2 = 3x = 3(20^\circ) = 60^\circ$$ $$\angle 3 = 4x = 4(20^\circ) = 80^\circ$$
Step 4: Classify the triangle
• Angle-wise: Since all three angles ($40^\circ, 60^\circ, 80^\circ$) are strictly less than $90^\circ$, the triangle is an Acute-angled Triangle.
• Side-wise: Since all three angles have unequal measures, the sides opposite to them must also have unequal lengths [Sides opposite to unequal angles are unequal]. Therefore, the triangle is a Scalene Triangle.

$$\mathbf{\text{Conclusion: Angles are } 40^\circ, 60^\circ, 80^\circ; \; \text{The triangle is an Acute-angled Scalene Triangle.}}$$
Rough Column Working:
2x + 3x + 4x = 180
9x = 180 => x = 20
Angles: 40, 60, 80 deg.
All < 90 => Acute-angled.
All unequal => Scalene.
Step Mark Allocation:
  • Setting up 2x + 3x + 4x = 180 with theorem citation: 1.5 Marks
  • Solving for x = 20: 1 Marks
  • Computing all 3 angles accurately (40, 60, 80 deg): 1.5 Marks
  • Dual classification (Acute-angled and Scalene) with reasoning: 1 Marks
Examiner Scoring Key: Candidates often forget the second half of the question (dual classification). Always answer both side-wise and angle-wise classification.
Part A - Descriptive 5 Marks
(a) Can a triangle have sides of lengths $7\text{ cm}, 11\text{ cm},$ and $19\text{ cm}$? Justify.
(b) If two sides of a triangle are $8\text{ cm}$ and $13\text{ cm}$, between what two integer limits must the length of the third side lie?
Step-by-Step Model Solution:
Step 1: (a) Test for triangle existence using Triangle Inequality
By the Triangle Inequality Theorem, a triangle is possible if and only if the sum of the lengths of any two sides is strictly greater than the third side.
Given side lengths: $a = 7\text{ cm}, b = 11\text{ cm}, c = 19\text{ cm}$.
Check the sum of the two smaller sides:
$$a + b = 7\text{ cm} + 11\text{ cm} = 18\text{ cm}$$
Comparing with the third side $c = 19\text{ cm}$:
$$18\text{ cm} < 19\text{ cm} \implies a + b \ngtr c$$
Since the sum of two sides ($18\text{ cm}$) is less than the third side ($19\text{ cm}$), the triangle inequality is violated.
$$\mathbf{\text{Hence, a triangle with sides } 7\text{ cm}, 11\text{ cm}, \text{ and } 19\text{ cm cannot be formed.}}$$
Step 2: (b) Determine the range of the third side
Let the given sides be $a = 13\text{ cm}$ and $b = 8\text{ cm}$, and let the third side be $x\text{ cm}$.
By the Triangle Inequality Theorem, the length of the third side must be:
1. Strictly greater than the difference of the other two sides: $x > a - b$
$$x > 13 - 8 \implies x > 5\text{ cm}$$
2. Strictly less than the sum of the other two sides: $x < a + b$
$$x < 13 + 8 \implies x < 21\text{ cm}$$
Combining the two conditions:
$$5\text{ cm} < x < 21\text{ cm}$$
For integer lengths, the third side must be at least $6\text{ cm}$ and at most $20\text{ cm}$.

$$\mathbf{\text{Conclusion: The third side must lie strictly between } 5\text{ cm and } 21\text{ cm (integer range: } 6\text{ cm to } 20\text{ cm).}}$$
Rough Column Working:
7 + 11 = 18 < 19 => Not possible!
Range: 13 - 8 = 5 < x < 13 + 8 = 21
Integers: 6 to 20 cm.
Step Mark Allocation:
  • Stating Triangle Inequality Theorem: 1 Marks
  • Showing 7 + 11 = 18 < 19 and concluding triangle is impossible: 1.5 Marks
  • Setting up difference and sum bounds (13 - 8 < x < 13 + 8): 1.5 Marks
  • Final bounds: 5 cm < x < 21 cm and boxed answer: 1 Marks
Examiner Scoring Key: Checking only the sum of the two smaller sides is the fastest and most rigorous proof of non-existence.
Part B - Euclidean Proof 10 Marks
Prove that the sum of the three angles of a triangle is $180^\circ$ ($2$ right angles).
Step-by-Step Model Solution:
Step 1: Given, To Prove & Figure
Given: A triangle $ABC$ with interior angles $\angle 1 = \angle BAC$, $\angle 2 = \angle ABC$, and $\angle 3 = \angle BCA$.
To Prove: $\angle 1 + \angle 2 + \angle 3 = 180^\circ$ (or $\angle A + \angle B + \angle C = 180^\circ$).

Step 2: Construction
Through the vertex $A$, draw a straight line $XY$ parallel to the base $BC$ (i.e., $XY \parallel BC$).
Let $\angle XAB = \angle 4$ and $\angle YAC = \angle 5$.

Step 3: Establish alternate interior angle equalities
Since line $XY \parallel BC$ and $AB$ is a transversal intersecting them at $A$ and $B$:
$$\angle 4 = \angle 2 \quad [\text{Alternate Interior Angles are Equal}] \quad \text{--- (Equation 1)}$$
Similarly, since line $XY \parallel BC$ and $AC$ is a transversal intersecting them at $A$ and $C$:
$$\angle 5 = \angle 3 \quad [\text{Alternate Interior Angles are Equal}] \quad \text{--- (Equation 2)}$$

Step 4: Use angles on a straight line
Since $XAY$ is a straight line, the sum of all angles formed on one side of a straight line at point $A$ is $180^\circ$ [Angles on a Straight Line]:
$$\angle 4 + \angle 1 + \angle 5 = 180^\circ \quad \text{--- (Equation 3)}$$
Step 5: Substitute Equations 1 and 2 into Equation 3
Replacing $\angle 4$ with $\angle 2$ and $\angle 5$ with $\angle 3$:
$$\angle 2 + \angle 1 + \angle 3 = 180^\circ$$ Rearranging the terms in alphabetical order:
$$\angle 1 + \angle 2 + \angle 3 = 180^\circ$$ $$\angle A + \angle B + \angle C = 180^\circ$$
$$\mathbf{\text{Hence Proved: The sum of the angles of a triangle is } 180^\circ.}$$
Rough Column Working:
Line XY || BC through A.
Angle 4 = Angle 2 (alt int)
Angle 5 = Angle 3 (alt int)
Angle 4 + 1 + 5 = 180 (straight line)
=> Angle 1 + 2 + 3 = 180 deg. Q.E.D.
Step Mark Allocation:
  • Given, To Prove, and neat labeled geometric diagram: 2 Marks
  • Construction statement: Draw line XY || BC through vertex A: 2 Marks
  • Stating alternate interior angle pairs with reasons (Eq 1 & 2): 3 Marks
  • Applying straight line angle sum and substituting to conclude proof: 3 Marks
Examiner Scoring Key: This is one of the most frequently tested 10-mark proofs in RIMC history. The auxiliary parallel line through vertex A must be explicitly stated.
Part B - Applied Pythagoras 10 Marks
A ladder $25\text{ m}$ long reaches a window of a building which is $20\text{ m}$ above the ground. If the foot of the ladder is pulled $8\text{ m}$ further away from the base of the wall, find the distance by which the top of the ladder slides down the wall.
Step-by-Step Model Solution:
Step 1: Model Initial Position of Ladder (Figure 1)
Let $AC$ represent the vertical wall and $BC$ represent the horizontal ground, meeting at right angle $\angle ACB = 90^\circ$.
Let $AB$ be the ladder in its initial position.
Given:
$$\text{Length of ladder } AB = 25\text{ m}$$ $$\text{Height of window } AC = 20\text{ m}$$ In right-angled $\triangle ACB$, applying the Pythagorean Theorem:
$$AB^2 = AC^2 + BC^2$$ $$25^2 = 20^2 + BC^2$$ $$625 = 400 + BC^2$$ $$BC^2 = 625 - 400 = 225$$ $$BC = \sqrt{225} = 15\text{ m}$$ Thus, the initial distance of the foot of the ladder from the base of the wall is $15\text{ m}$.

Step 2: Model Second Position of Ladder (Figure 2)
The foot of the ladder is pulled $9\text{ m}$ further away from the wall to a new point $D$.
$$\text{New distance from wall } CD = BC + BD = 15\text{ m} + 9\text{ m} = 24\text{ m}$$
Let the top of the ladder slide down to point $E$ on the wall.
Since the ladder is rigid, its length remains unchanged:
$$\text{Length of ladder } ED = 25\text{ m}$$
In the new right-angled $\triangle ECD$, applying the Pythagorean Theorem:
$$ED^2 = EC^2 + CD^2$$ $$25^2 = EC^2 + 24^2$$ $$625 = EC^2 + 576$$ $$EC^2 = 625 - 576 = 49$$ $$EC = \sqrt{49} = 7\text{ m}$$
Step 3: Calculate the distance the ladder slides down
The distance the top of the ladder slides down is $AE$:
$$AE = AC - EC$$ $$AE = 20\text{ m} - 7\text{ m} = 13\text{ m}$$
$$\mathbf{\text{Conclusion: The top of the ladder slides down the wall by a distance of } 13\text{ metres.}}$$
Rough Column Working:
Initial: 25^2 - 20^2 = 225 => BC = 15 m
New base: 15 + 9 = 24 m
New height: 25^2 - 24^2 = 49 => EC = 7 m
Slide distance = 20 - 7 = 13 m.
Step Mark Allocation:
  • Diagram and right triangle formulation for position 1: 2.5 Marks
  • Applying Pythagoras to find initial base BC = 15 m: 2.5 Marks
  • Determining new base CD = 24 m and solving EC = 7 m: 3 Marks
  • Calculating slide distance AE = 20 - 7 = 13 m with boxed answer: 2 Marks
Examiner Scoring Key: Candidates frequently make the mistake of assuming the top slides down by the same amount the bottom was pulled (i.e. thinking it slides by 9 m). This receives zero credit. Both positions must be solved via Pythagoras.
Part A - Descriptive Congruence 5 Marks
In the given figure, line segment $AB$ is parallel to line segment $CD$, and $O$ is the midpoint of $AD$. Prove that:
(i) $\triangle AOB \cong \triangle DOC$
(ii) $O$ is also the midpoint of $BC$.
Step-by-Step Model Solution:
Step 1: Given Data & To Prove
Given: $AB \parallel CD$, and $O$ is the midpoint of $AD$ (which means $OA = OD$).
To Prove:
(i) $\triangle AOB \cong \triangle DOC$
(ii) $O$ is also the midpoint of $BC$ (i.e., $OB = OC$).

Step 2: Proof of Congruence (i)
In $\triangle AOB$ and $\triangle DOC$:
1. $\angle OAB = \angle ODC \quad [\text{Alternate Interior Angles, as } AB \parallel CD \text{ with transversal } AD]$
2. $OA = OD \quad [\text{Given, } O \text{ is the midpoint of } AD]$
3. $\angle AOB = \angle DOC \quad [\text{Vertically Opposite Angles are Equal}]$

Therefore, by ASA (Angle-Side-Angle) Congruence Criterion:
$$\triangle AOB \cong \triangle DOC$$
Step 3: Proof of Part (ii) via CPCTC
Since corresponding parts of congruent triangles are equal (CPCTC):
$$OB = OC \quad [\text{By CPCTC}]$$
Since $OB = OC$, it follows that $O$ is the midpoint of segment $BC$.

$$\mathbf{\text{Hence Proved: } \triangle AOB \cong \triangle DOC \text{ and } O \text{ is the midpoint of } BC.}$$
Rough Column Working:
OA = OD (given)
Angle OAB = Angle ODC (alt int)
Angle AOB = Angle DOC (vert opp)
=> ASA congruence
=> OB = OC (CPCTC) => O is midpoint of BC. Q.E.D.
Step Mark Allocation:
  • Given and To Prove statements with labeled figure: 1 Marks
  • Listing three matching conditions with correct geometric reasons: 2 Marks
  • Stating ASA Congruence Criterion explicitly: 1 Marks
  • Deducing OB = OC via CPCTC and concluding proof: 1 Marks
Examiner Scoring Key: Ensure the criterion is correctly identified as ASA (the side OA is included between angle OAB and angle AOB).
Part B - Advanced Isosceles Geometry 10 Marks
In an isosceles triangle $ABC$, $AB = AC = 13\text{ cm}$ and base $BC = 10\text{ cm}$.
(i) Find the length of the altitude $AD$ drawn from vertex $A$ to base $BC$.
(ii) Calculate the area of $\triangle ABC$.
(iii) Find the length of the altitude drawn from vertex $B$ to side $AC$.
Step-by-Step Model Solution:
Step 1: Properties of Altitude in an Isosceles Triangle
In an isosceles triangle $ABC$ with $AB = AC$, the altitude $AD$ drawn from vertex $A$ to base $BC$ bisects the base.
Proof: In right $\triangle ADB$ and right $\triangle ADC$:
$$\text{Hypotenuse } AB = AC = 13\text{ cm} \quad [\text{Given}]$$ $$AD = AD \quad [\text{Common Side}]$$ $$\angle ADB = \angle ADC = 90^\circ \quad [AD \perp BC]$$ Therefore, $\triangle ADB \cong \triangle ADC$ by RHS Congruence Criterion.
Consequently, by CPCTC:
$$BD = DC = \frac{BC}{2} = \frac{10}{2} = 5\text{ cm}$$
Step 2: (i) Calculate the length of altitude $AD$
In right-angled $\triangle ADB$, applying the Pythagorean Theorem:
$$AB^2 = AD^2 + BD^2$$ $$13^2 = AD^2 + 5^2$$ $$169 = AD^2 + 25$$ $$AD^2 = 169 - 25 = 144$$ $$AD = \sqrt{144} = 12\text{ cm}$$
$$\mathbf{\text{Altitude } AD = 12\text{ cm.}}$$
Step 3: (ii) Calculate the Area of $\triangle ABC$
$$\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area}(\triangle ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 10\text{ cm} \times 12\text{ cm} = 60\text{ cm}^2$$
$$\mathbf{\text{Area of } \triangle ABC = 60\text{ cm}^2.}$$
Step 4: (iii) Calculate the altitude from vertex $B$ to side $AC$
Let $BE$ be the altitude drawn from vertex $B$ perpendicular to side $AC$ ($BE \perp AC$).
The area of the triangle can also be expressed using base $AC$ and altitude $BE$:
$$\text{Area}(\triangle ABC) = \frac{1}{2} \times AC \times BE$$ Substituting the known values ($AC = 13\text{ cm}$ and $\text{Area} = 60\text{ cm}^2$):
$$60 = \frac{1}{2} \times 13 \times BE$$ $$120 = 13 \times BE$$ $$BE = \frac{120}{13} = 9\frac{3}{13}\text{ cm} \approx 9.23\text{ cm}$$
$$\mathbf{\text{Conclusion: (i) } AD = 12\text{ cm}; \; (ii) \text{ Area } = 60\text{ cm}^2; \; (iii) \text{ Altitude } BE = \frac{120}{13}\text{ cm} \; (9\frac{3}{13}\text{ cm}).}$$
Rough Column Working:
BD = 10/2 = 5
AD^2 = 13^2 - 5^2 = 144 => AD = 12
Area = 1/2 * 10 * 12 = 60 cm^2
1/2 * 13 * BE = 60 => BE = 120/13 cm.
Step Mark Allocation:
  • Proving altitude bisects base using RHS congruence (BD = 5 cm): 2.5 Marks
  • Using Pythagoras theorem to find AD = 12 cm: 2.5 Marks
  • Calculating Area = 60 sq cm: 2.5 Marks
  • Equating area with base AC to find BE = 120/13 cm: 2.5 Marks
Examiner Scoring Key: Equating two different base-altitude pairs for the same triangle's area is a masterclass technique frequently rewarded in RIMC Part B.

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
One of the acute angles of a right-angled triangle is $36^\circ$. Find the measure of the other acute angle.
Model Descriptive Solution:
In a right-angled triangle, one angle is $90^\circ$. By Angle Sum Property, the sum of all three angles is $180^\circ$. Therefore, the two acute angles are complementary: $\theta + 36^\circ = 90^\circ \implies \theta = 90^\circ - 36^\circ = 54^\circ$.
Question 2 5 Marks
An exterior angle of a triangle is $110^\circ$ and its two interior opposite angles are equal to each other. Find the measure of each angle of the triangle.
Model Descriptive Solution:
Let the two equal interior opposite angles be $x^\circ$ each. By Exterior Angle Theorem, $\text{Exterior Angle} = x + x \implies 2x = 110^\circ \implies x = 55^\circ$. The third interior angle forms a linear pair with the exterior angle: $180^\circ - 110^\circ = 70^\circ$. The angles of the triangle are $55^\circ, 55^\circ,$ and $70^\circ$.
Question 3 5 Marks
The lengths of two sides of a right-angled triangle containing the right angle are $12\text{ cm}$ and $16\text{ cm}$. Find the length of the hypotenuse.
Model Descriptive Solution:
By the Pythagorean Theorem, $\text{Hypotenuse}^2 = 12^2 + 16^2 = 144 + 256 = 400$. Hypotenuse $= \sqrt{400} = 20\text{ cm}$.
Question 4 5 Marks
Show that a triangle cannot have more than one right angle or more than one obtuse angle.
Model Descriptive Solution:
If a triangle has two right angles ($90^\circ$ each), their sum alone would be $90^\circ + 90^\circ = 180^\circ$, leaving $0^\circ$ for the third angle, which is impossible. Similarly, if a triangle had two obtuse angles ($> 90^\circ$ each), their sum would exceed $180^\circ$, violating the Angle Sum Property. Hence, a triangle can have at most one right angle or at most one obtuse angle.
Question 5 10 Marks
In $\triangle ABC$, $AD$ is the median to side $BC$. Prove that $AB + AC > 2AD$.
Model Descriptive Solution:
Extend median $AD$ to point $E$ such that $DE = AD$. Join $EC$. In $\triangle ADB$ and $\triangle EDC$: $AD = DE$ (by construction), $BD = CD$ ($AD$ is median), and $\angle ADB = \angle EDC$ (vertically opposite angles). By SAS Congruence, $\triangle ADB \cong \triangle EDC$. Hence, $AB = EC$ (CPCTC). Now in $\triangle AEC$, by the Triangle Inequality Theorem: $AC + EC > AE$. Since $EC = AB$ and $AE = AD + DE = 2AD$, we obtain: $AC + AB > 2AD$, or $AB + AC > 2AD$. Hence Proved.
Question 6 10 Marks
A tree of height $18\text{ m}$ is broken by the wind at a certain height above the ground such that its top touches the ground at a distance of $12\text{ m}$ from its foot. Find the height at which the tree broke.
Model Descriptive Solution:
Let total height of tree be $H = 18\text{ m}$. Let the tree break at height $x\text{ m}$ from the ground. Then the remaining broken part forms the hypotenuse: $\text{Hypotenuse} = (18 - x)\text{ m}$. The distance of top from foot is the base $= 12\text{ m}$. By Pythagoras Theorem: $(18 - x)^2 = x^2 + 12^2 \implies 324 - 36x + x^2 = x^2 + 144 \implies 324 - 144 = 36x \implies 180 = 36x \implies x = 5\text{ m}$. Hence, the tree broke at a height of $5\text{ metres}$ from the ground.
Question 7 5 Marks
The sides of a triangle are in the ratio $13 : 12 : 5$. If its perimeter is $60\text{ cm}$, prove that the triangle is right-angled.
Model Descriptive Solution:
Let sides be $13x, 12x,$ and $5x$. Perimeter $= 13x + 12x + 5x = 30x = 60\text{ cm} \implies x = 2$. Sides are $26\text{ cm}, 24\text{ cm},$ and $10\text{ cm}$. Longest side squared: $26^2 = 676$. Sum of squares of other two sides: $24^2 + 10^2 = 576 + 100 = 676$. Since $24^2 + 10^2 = 26^2$, by the Converse of the Pythagorean Theorem, the triangle is right-angled.
Question 8 5 Marks
In $\triangle ABC$, $\angle B = 35^\circ$ and $\angle C = 65^\circ$. The bisector of $\angle A$ meets $BC$ at $D$. Find $\angle ADB$ and $\angle ADC$.
Model Descriptive Solution:
In $\triangle ABC$, $\angle A = 180^\circ - (35^\circ + 65^\circ) = 180^\circ - 100^\circ = 80^\circ$. Since $AD$ bisects $\angle A$, $\angle BAD = \angle CAD = 40^\circ$. In $\triangle ABD$, $\angle ADB = 180^\circ - (\angle B + \angle BAD) = 180^\circ - (35^\circ + 40^\circ) = 105^\circ$. By linear pair, $\angle ADC = 180^\circ - 105^\circ = 75^\circ$.
Question 9 10 Marks
An equilateral triangle of side $12\text{ cm}$ has the same perimeter as an isosceles triangle whose base is $10\text{ cm}$. Compare their areas and determine which triangle has a larger area.
Model Descriptive Solution:
Perimeter of equilateral triangle $= 3 \times 12 = 36\text{ cm}$. Area of equilateral triangle $= \frac{\sqrt{3}}{4}(12)^2 = \frac{\sqrt{3}}{4}(144) = 36\sqrt{3} \approx 36 \times 1.732 = 62.35\text{ cm}^2$. For isosceles triangle: Perimeter $= 2a + 10 = 36 \implies 2a = 26 \implies a = 13\text{ cm}$ (equal sides). Half-base $= 5\text{ cm}$. Altitude of isosceles triangle $= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12\text{ cm}$. Area of isosceles triangle $= \frac{1}{2} \times 10 \times 12 = 60\text{ cm}^2$. Comparing: $62.35\text{ cm}^2 > 60\text{ cm}^2$. The equilateral triangle has a larger area by approximately $2.35\text{ cm}^2$.
Question 10 10 Marks
Points $D$ and $E$ are on sides $AB$ and $AC$ of an equilateral $\triangle ABC$ such that $AD = BE = CF$ in cyclic order. Prove that $\triangle DEF$ is also an equilateral triangle.
Model Descriptive Solution:
Given $\triangle ABC$ is equilateral: $AB = BC = CA = s$ and $\angle A = \angle B = \angle C = 60^\circ$. Given $AD = BE = CF = x$. Then $DB = s - x, EC = s - x, FA = s - x$. In $\triangle ADF, \triangle BED,$ and $\triangle CFE$: $AD = BE = CF = x$, $AF = BD = CE = (s - x)$, and included angles $\angle A = \angle B = \angle C = 60^\circ$. By SAS Congruence Criterion, $\triangle ADF \cong \triangle BED \cong \triangle CFE$. By CPCTC, corresponding third sides must be equal: $DF = ED = FE$. Since all three sides of $\triangle DEF$ are equal in length, $\triangle DEF$ is an equilateral triangle. Hence Proved.

High-Yield FAQs & Revision Notes

Can a triangle have two right angles?
No. If a triangle had two right angles ($90^\circ + 90^\circ = 180^\circ$), the sum of two angles alone would equal $180^\circ$, leaving $0^\circ$ for the third angle. By the Angle Sum Property, the sum of all three angles must equal $180^\circ$.
What is the difference between an altitude and a median?
An altitude is the perpendicular line segment drawn from a vertex to the opposite side. A median is a line segment joining a vertex to the midpoint of the opposite side. They only coincide in equilateral triangles or from the vertex between equal sides in an isosceles triangle.
Why is SSA not a valid congruence criterion?
Given two sides and a non-included angle, two non-congruent triangles can often be constructed (one acute, one obtuse) sharing those exact measurements. Only RHS (where the angle is $90^\circ$) guarantees congruence.
How do I test if three side lengths can form a triangle?
Simply check whether the sum of the two smaller side lengths is strictly greater than the longest side length ($a + b > c$). If it is greater, the triangle is valid; otherwise, it is impossible.
What is the Centroid of a triangle?
The centroid is the point where the three medians of a triangle intersect. It always lies inside the triangle and divides each median in the ratio $2 : 1$ measured from the vertex.
What is the Orthocenter of a triangle?
The orthocenter is the point where the three altitudes of a triangle intersect. It is located inside an acute triangle, at the right-angle vertex in a right triangle, and outside an obtuse triangle.
What is the Incenter and Circumcenter?
The Incenter is the intersection of internal angle bisectors and is equidistant from all three sides. The Circumcenter is the intersection of perpendicular bisectors of sides and is equidistant from all three vertices.
What is the formula for the area and height of an equilateral triangle?
For side $a$, the altitude is $h = \frac{\sqrt{3}}{2}a$, and the area is $\Delta = \frac{\sqrt{3}}{4}a^2$.
Does CPCTC require full justification in RIMC answers?
Yes. First prove the two triangles are congruent using a valid criterion (SSS, SAS, ASA, AAS, or RHS), state the congruence symbol with vertices in corresponding order, and then deduce side/angle equality citing '[By CPCTC]'.
How to generate Pythagorean triplets from any integer?
For any integer $m > 1$, the three numbers $2m, m^2 - 1,$ and $m^2 + 1$ form a Pythagorean triplet because $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 20, 34 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 Always state 'Given:', 'To Prove:', and draw a neat pencil figure with tick marks for equal sides.
Checkpoint 2 Verify triangle existence quickly: sum of two smaller sides > longest side.
Checkpoint 3 Remember the 5 valid congruence criteria: SSS, SAS, ASA, AAS, RHS. Never use SSA or AAA.
Checkpoint 4 In right triangles, use Pythagoras: Hypotenuse^2 = Base^2 + Perpendicular^2.
Checkpoint 5 Median joins vertex to midpoint; Centroid divides medians in 2:1 ratio.
Checkpoint 6 Altitude in isosceles triangle bisects base and splits triangle into two congruent RHS triangles.