The building blocks of mensuration | Kerala PSC Overseer Gr.2
Mensuration is the branch of mathematics dealing with measurement of geometric figures — calculating their area, volume, and perimeter.
What is a Polygon?
A polygon is a simple closed figure made entirely of straight line segments.
Sides: the line segments that form it
Vertex: meeting point of two adjacent sides
Diagonal: a segment joining two non-adjacent vertices
If a polygon has n sides → it has n vertices and n internal angles
📌 Note: A regular polygon has all sides equal AND all interior angles equal. Equilateral triangle and square are the simplest examples.
Key Formulas — Regular Polygon
📌 Formula: • Sum of interior angles = (n − 2) × 180° • One interior angle = (n − 2) × 180° / n • Perimeter = n × side
Polygons by Number of Sides
Sides (n)
Name
Sum of Angles
Each Angle (regular)
3
Triangle
180°
60°
4
Quadrilateral
360°
90°
5
Pentagon
540°
108°
6
Hexagon
720°
120°
7
Heptagon
900°
≈128.57°
8
Octagon
1080°
135°
9
Nonagon
1260°
140°
10
Decagon
1440°
144°
⚠️ PSC TRAP: The interior angle of a REGULAR HEXAGON is exactly 120°, and the sum of all six is 720°. PSC frequently asks both — never confuse the SUM with EACH angle.
Pentagon & Hexagon — Standard Areas
📌 Formula: • Regular pentagon area = 1.7205 a² (a = side) • Regular hexagon area = (3√3 / 2) a² ≈ 2.598 a² • Hexagon flat-to-flat distance = √3 × side
A triangle is a 3-sided polygon. Sum of internal angles = 180°. The vertex opposite the base is the apex; lines from the midpoint of a side to the opposite vertex are medians; the medians meet at the centroid which divides each median in a 2:1 ratio.
📌 Formula — Triangle Area: • Area = ½ × base × height • If two sides a, b and included angle θ → Area = ½ a b sin θ • Heron's formula → Area = √(s(s−a)(s−b)(s−c)), where s = (a+b+c)/2
Triangle Types — by Sides
Type
Property
Area Formula
Scalene
All sides different
Heron's formula
Isosceles
Two sides equal
½ b × √(a² − b²/4)
Equilateral
All sides equal, each angle 60°
(√3 / 4) a²
Triangle Types — by Angle
Type
Angle Condition
Acute-angled
All angles < 90°
Right-angled
One angle = 90°
Obtuse-angled
One angle > 90°
Right Isosceles
90°, 45°, 45°
⚠️ PSC TRAP: The CENTROID divides each median in 2:1 ratio — the longer part is towards the vertex, the shorter is towards the midpoint of the side. Reverse direction = wrong.
Quadrilaterals — Properties & Areas
A quadrilateral has 4 sides, 4 vertices, and 4 internal angles summing to 360°.
Shape
Property
Area
Perimeter
Rectangle
Opposite sides equal
l × b
2(l + b)
Square
All sides equal
a² = d²/2
4a
Parallelogram
Opposite sides parallel & equal
b × h
2(a + b)
Rhombus
All sides equal
½ × d₁ × d₂
4a
Trapezium
One pair parallel sides
½ (a + b) h
sum of all 4 sides
⚠️ PSC TRAP — UK vs US Trapezium: NIMI / Indian PSC follows the UK definition: Trapezium = ONE pair of parallel sides. Trapezoid = ZERO parallel sides. The US system is the exact opposite. Read the question carefully if it quotes a US source.
📐 WORKED EXAMPLE — Square Diagonal: Side a = 10 cm. Diagonal d = a√2 = 10√2 ≈ 14.14 cm Area = a² = 100 cm² OR d²/2 = 200/2 = 100 cm² ✓
⭕ Circles & Ellipses
Curved 2-D figures and their measurements
Circle Fundamentals
A circle is the path traced by a point moving in a plane at a fixed distance (radius r) from a fixed point (centre).
Chord: a segment joining any two points on the circle
Diameter (d) = 2r — the longest chord
Circumference (C) = π d = 2πr
Area = πr² = (π/4) d²
π ≈ 3.14159 or 22/7 for clean numerics
📌 Formula — Circle parts: • Arc length L = r θ (θ in radians) or (2 π r θ) / 360 (θ in degrees) • Sector area = ½ r² θ (radians) or (π r² θ) / 360 (degrees) • Semicircle area = π r² / 2; Semicircle perimeter = π r + 2 r
🧠 Memory Trick:"π radians = 180°" — convert any θ between systems by multiplying by π/180 or 180/π.
📐 WORKED EXAMPLE — Semicircle perimeter: Diameter = 10 cm → r = 5 cm. Perimeter = π r + 2 r = 5 π + 10 = 5(3.14) + 10 = 25.7 cm
📐 WORKED EXAMPLE — Sector area: r = 6 cm, θ = 60°. Area = (π × 36 × 60) / 360 = 6 π cm² ≈ 18.85 cm²
Segment vs Sector
Region
Bounded By
Sector
Two RADII and an arc (pie slice)
Segment
A CHORD and an arc
⚠️ PSC TRAP: SECTOR uses two radii (slice of a pie). SEGMENT uses a chord. Don't swap them — PSC loves this confusion.
Ellipse — Quick Reference
An ellipse is the locus of a point such that the sum of distances from two fixed points (foci) is constant. Major axis = 2a, Minor axis = 2b.
📌 Formula: • Perimeter ≈ π (a + b) (approximate) • Area = π a b = (π / 4) × Major × Minor • Eccentricity (ellipse) < 1 | Parabola = 1 | Hyperbola > 1
Circle (e = 0)
All points equidistant
Ellipse (e < 1)
Two foci
Parabola (e = 1)
One focus
🧊 Solids — Volumes & Surface Areas
3-D shapes — the heart of mensuration numericals
Cube & Cuboid
A cube has 6 equal square faces, 8 vertices, 12 edges. A cuboid (rectangular prism) has 6 rectangular faces.
Solid
LSA / CSA
TSA
Volume
Cube (side a)
4a²
6a²
a³
Cuboid (l, b, h)
2(l + b) h
2(lb + bh + lh)
l × b × h
Cylinder (r, h)
2 π r h
2 π r (r + h)
π r² h
Cone (r, h, l)
π r l
π r (r + l)
⅓ π r² h
Sphere (r)
—
4 π r²
⁴⁄₃ π r³
Hemisphere (r)
2 π r²
3 π r²
⅔ π r³
Frustum (R, r, h, L)
π (R + r) L
π[(R+r)L + R² + r²]
⅓ π h (R² + r² + Rr)
📌 Formula — Cone slant height:l = √(r² + h²); Frustum slant: L = √(h² + (R − r)²)
⚠️ PSC TRAP:Hemisphere TSA = 3 π r² (not 2 π r²). The 2 π r² is only the CURVED part — you must add the flat circular base π r² to get total surface.
🧠 Memory Trick:"Sphere is FULL Four — cube is SIX faces — Hemisphere needs THREE (curved 2 + base 1)"
Prism vs Pyramid — Quick Rule
Solid
Volume Rule
Prism
Base Area × Height (top & bottom faces equal)
Pyramid
⅓ × Base Area × Height (one apex)
🧮 PSC-Style Solved Problems
Apply the formulas to PSC favourite question patterns
Problem 1 — Volume Conservation (Recasting)
📐 WORKED EXAMPLE: How many small balls of radius 2 cm can be made by melting one big ball of radius 8 cm?
Step 1: Volume is conserved: n × V(small) = V(large). Step 2: n × (4/3) π × 2³ = (4/3) π × 8³ Step 3: n = 8³ / 2³ = 512 / 8 = 64 balls
Problem 2 — Surface Area Ratio (Cube cut into smaller cubes)
📐 WORKED EXAMPLE: A 5 cm cube is cut into 1 cm cubes. Find the ratio of SA of the large cube to total SA of all small cubes.
Step 1: Volume of big cube = 5³ = 125 cm³ → 125 small cubes. Step 2: SA of big cube = 6 × 5² = 150 cm². Step 3: Total SA of 125 small cubes = 125 × 6 × 1² = 750 cm². Step 4: Ratio = 150 : 750 = 1 : 5
Problem 3 — Joining Two Cubes
📐 WORKED EXAMPLE: Two identical cubes of TSA = 6 cm² each are joined end-to-end. Find the SA of the resulting cuboid.
Step 1: 6 a² = 6 → a = 1 cm. Step 2: When joined, 2 faces are hidden inside. Step 3: Visible faces = 12 − 2 = 10 → SA = 10 × 1² = 10 cm²
Problem 4 — Cylinder Volume
📐 WORKED EXAMPLE: Find volume of a cylinder with r = 7 cm, h = 10 cm. Use π = 22/7.
Step 1: V = π r² h. Step 2: V = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cm³
Problem 5 — Equilateral Triangle Area
📐 WORKED EXAMPLE: Find area of an equilateral triangle with side 4 cm.
⚠️ PSC TRAP — Recasting: When melting / recasting solids, only VOLUME is conserved, NOT surface area. The total surface area changes after recasting (usually increases when cutting, decreases when joining). Never use SA conservation in recasting problems.
📌 Note — π values: Use π = 22/7 when r is a multiple of 7 (clean answer). Use π = 3.14 in all other cases. PSC almost always sets r = 7 or 14 to make 22/7 work.
