Module 3 · SSC CGL Quantitative Aptitude
Mensuration
Area, volume, surface area.
Triangles · Circles · Cylinders · Spheres
Learning Objectives
- Understand the fundamental concepts of Mensuration
- Apply key formulas and techniques to solve problems
- Practice with exam-level questions to build speed and accuracy
Key Concepts
1. Comprehensive Mensuration Formula Table
Master this formula table for quick revision. Bookmark this page for last-minute reference before the SSC CGL exam.
2. Triangles
Triangles are among the most frequently tested shapes in SSC CGL. Understanding their properties and formulas is critical.
Tip 1 — Triangle Angle Sum
The sum of all three interior angles of any triangle is always 180°. This is the foundation for solving many geometry problems.
Right Triangle — One angle = 90°. The side opposite the right angle is the hypotenuse. Area = ½ × base × height (the two legs). Perimeter = a + b + c. Pythagoras theorem: c2 = a2 + b2 where c is the hypotenuse.
Example 1 — Right Triangle Area
Find the area of a right triangle with legs 6 cm and 8 cm.
Area = ½ × 6 × 8 = 24 cm2. Hypotenuse = √(36 + 64) = 10 cm.
Isosceles Triangle — Two sides equal. Area = ¼ b × √(4a2 − b2) where a is the equal side and b is the base. Perimeter = 2a + b.
Tip 2 — Isosceles Height
Height of an isosceles triangle = √(a2 − b2/4). The altitude from the apex bisects the base.
Example 2 — Isosceles Triangle
An isosceles triangle has equal sides 10 cm and base 12 cm. Find its area.
Height = √(100 − 36) = √64 = 8 cm. Area = ½ × 12 × 8 = 48 cm2.
Equilateral Triangle — All sides equal. Area = (√3/4)a2, Perimeter = 3a, Height = (√3/2)a. Inradius = a/(2√3), Circumradius = a/√3.
Tip 3 — Equilateral Shortcut
For an equilateral triangle of side a: Height = √3/2 a, Area = √3/4 a2. Memorize these ratios: if side = 2, height = √3, area = √3.
Example 3 — Equilateral Triangle
Find the area of an equilateral triangle with side 14 cm.
Area = (√3/4) × 142 = (√3/4) × 196 = 49√3 cm2.
Heron's Formula — For any triangle with sides a, b, c: Area = √[s(s − a)(s − b)(s − c)] where s = (a + b + c)/2 is the semi-perimeter.
Example 4 — Heron's Formula
Find the area of a triangle with sides 13 cm, 14 cm, 15 cm.
s = (13 + 14 + 15)/2 = 21. Area = √[21(21−13)(21−14)(21−15)] = √(21 × 8 × 7 × 6) = √7056 = 84 cm2.
3. Quadrilaterals
Quadrilaterals include squares, rectangles, rhombus, parallelogram, and trapezium. Each has distinct area and perimeter formulas.
Square — All sides equal, all angles 90°. Area = a2, Perimeter = 4a, Diagonal = a√2. Inradius = a/2, Circumradius = a/√2.
Tip 4 — Square Diagonal
Diagonal of a square = a√2. This is the shortest distance between opposite corners. If diagonal is given, side = diagonal/√2.
Example 5 — Square
The diagonal of a square is 16√2 cm. Find its area.
Side = diagonal/√2 = 16 cm. Area = 162 = 256 cm2.
Rectangle — Opposite sides equal. Area = l × b, Perimeter = 2(l + b), Diagonal = √(l2 + b2).
Example 6 — Rectangle
A rectangle has length 12 cm and diagonal 13 cm. Find its area.
Breadth = √(132 − 122) = √25 = 5 cm. Area = 12 × 5 = 60 cm2.
Rhombus — All sides equal, diagonals bisect at 90°. Area = ½ × d1 × d2, Perimeter = 4a. Side a = ½ √(d12 + d22).
Tip 5 — Rhombus Area via Side & Angle
Area of rhombus can also be found using side and an included angle: A = a2 sin θ. Use whichever data is given.
Example 7 — Rhombus
The diagonals of a rhombus are 16 cm and 12 cm. Find its area and side length.
Area = ½ × 16 × 12 = 96 cm2. Side = ½ × √(256 + 144) = ½ × 20 = 10 cm.
Parallelogram — Opposite sides parallel and equal. Area = b × h, Perimeter = 2(a + b). Diagonals bisect each other.
Tip 6 — Parallelogram Area
Area = base × height (not side length). The height is the perpendicular distance between the two bases.
Example 8 — Parallelogram
A parallelogram has base 18 cm and height 11 cm. Find its area.
Area = 18 × 11 = 198 cm2.
Trapezium — One pair of parallel sides. Area = ½ × (sum of parallel sides) × height = ½ × (a + b) × h. Perimeter = a + b + c + d.
Example 9 — Trapezium
A trapezium has parallel sides 10 cm and 16 cm, and height 8 cm. Find its area.
Area = ½ × (10 + 16) × 8 = ½ × 26 × 8 = 104 cm2.
4. Circles
Circle problems appear regularly in SSC CGL. Master the formulas for area, circumference, arcs, sectors, segments, and rings.
Basic Circle — Area = πr2, Circumference = 2πr. Diameter = 2r.
Tip 7 — π Approximation
In SSC CGL, use π = 22/7 unless stated otherwise. For quick calculations, remember π ≈ 3.14 and π/4 ≈ 0.785.
Example 10 — Circle Area
Find the area of a circle with circumference 88 cm. (Use π = 22/7)
2πr = 88 ⇒ r = 88 × 7/(2 × 22) = 14 cm. Area = πr2 = (22/7) × 196 = 616 cm2.
Arc Length — For a central angle θ (in degrees): Arc Length = (θ/360) × 2πr. For θ in radians: Arc Length = rθ.
Sector Area — Area of sector = (θ/360) × πr2 (for degrees) = ½ r2θ (for radians).
Tip 8 — Sector vs Segment
Sector = slice of circle (like a pizza slice). Segment = region between a chord and the arc. Area of segment = Sector area − Triangle area.
Example 11 — Sector Area
Find the area of a sector with radius 21 cm and central angle 60°. (Use π = 22/7)
Sector area = (60/360) × (22/7) × 212 = (1/6) × (22/7) × 441 = (1/6) × 1386 = 231 cm2.
Segment Area — Area of segment = Sector area − area of triangle formed by the radii and chord. For angle θ in radians: Segment area = ½ r2(θ − sin θ).
Example 12 — Segment Area
A chord subtends an angle of 90° at the centre of a circle of radius 14 cm. Find the segment area. (Use π = 22/7)
Sector area = (90/360) × (22/7) × 196 = (1/4) × 616 = 154 cm2. Triangle area = ½ × 14 × 14 = 98 cm2. Segment area = 154 − 98 = 56 cm2.
Ring (Annulus) — Area between two concentric circles: A = π(R2 − r2) where R is the outer radius and r is the inner radius.
Tip 9 — Ring Area Difference of Squares
Ring area = π(R − r)(R + r). This factorisation often simplifies calculations when R + r or R − r is given directly.
5. Three-Dimensional Shapes
3D mensuration covers surface areas (CSA & TSA) and volumes of solids. SSC CGL typically asks 2–3 questions from this section.
Cube — All edges equal. TSA = 6a2, LSA = 4a2, Volume = a3, Diagonal = a√3, Face Diagonal = a√2.
Tip 10 — Cube Diagonal
The main diagonal of a cube = √3 × side. Memorise: if side = 1, diagonal ≈ 1.732. Face diagonal = √2 × side.
Example 13 — Cube
The total surface area of a cube is 864 cm2. Find its volume.
6a2 = 864 ⇒ a2 = 144 ⇒ a = 12 cm. Volume = 123 = 1728 cm3.
Cuboid — Length, breadth, height. TSA = 2(lb + bh + hl), LSA = 2h(l + b), Volume = l × b × h, Diagonal = √(l2 + b2 + h2).
Example 14 — Cuboid
A cuboid measures 10 m × 8 m × 6 m. Find its TSA and volume.
TSA = 2(10×8 + 8×6 + 6×10) = 2(80 + 48 + 60) = 2 × 188 = 376 m2. Volume = 10 × 8 × 6 = 480 m3.
Cylinder (Solid) — CSA = 2πrh, TSA = 2πr(r + h), Volume = πr2h.
Cylinder (Hollow) — CSA (outer) = 2πRh, CSA (inner) = 2πrh, Total CSA = 2π(R + r)h, TSA = 2π(R + r)(R − r + h), Volume = π(R2 − r2)h.
Tip 11 — Hollow Cylinder Volume
Volume of hollow cylinder = π(R2 − r2)h = π(R − r)(R + r)h. This is the volume of material, not the empty space inside.
Example 15 — Cylinder
A solid cylinder has radius 7 cm and height 10 cm. Find its CSA, TSA, and volume. (Use π = 22/7)
CSA = 2 × 22/7 × 7 × 10 = 440 cm2. TSA = 440 + 2 × (22/7) × 49 = 440 + 308 = 748 cm2. Volume = (22/7) × 49 × 10 = 1540 cm3.
Cone — Slant height l = √(r2 + h2). CSA = πrl, TSA = πr(r + l), Volume = (⅓)πr2h.
Tip 12 — Cone Slant Height
Always compute the slant height first when solving cone problems: l = √(r2 + h2). Many TSA questions require l.
Example 16 — Cone
A cone has radius 6 cm and height 8 cm. Find its CSA and volume. (Use π = 3.14)
l = √(36 + 64) = 10 cm. CSA = 3.14 × 6 × 10 = 188.4 cm2. Volume = (1/3) × 3.14 × 36 × 8 = 301.44 cm3.
Sphere — SA = 4πr2, Volume = (4/3)πr3.
Hemisphere — CSA = 2πr2, TSA = 3πr2, Volume = (2/3)πr3.
Tip 13 — Hemisphere TSA
TSA of a solid hemisphere = 3πr2 (curved + base). Do not confuse with CSA = 2πr2 which is just the curved part.
Example 17 — Sphere & Hemisphere
A sphere has radius 21 cm. Find its SA and volume. Also find the TSA of a hemisphere of the same radius. (Use π = 22/7)
Sphere SA = 4 × (22/7) × 441 = 4 × 1386 = 5544 cm2. Volume = (4/3) × (22/7) × 9261 = (4/3) × 29106 = 38808 cm3. Hemisphere TSA = 3 × (22/7) × 441 = 3 × 1386 = 4158 cm2.
Frustum of a Cone — Slant height l = √[h2 + (R − r)2]. CSA = π(R + r)l, TSA = π(R2 + r2 + (R + r)l), Volume = (⅓)πh(R2 + r2 + Rr).
Example 18 — Frustum
A frustum has top radius 7 cm, bottom radius 14 cm, and height 24 cm. Find its volume. (Use π = 22/7)
Volume = (1/3) × (22/7) × 24 × (196 + 49 + 98) = (1/3) × (22/7) × 24 × 343 = (1/3) × (22/7) × 8232 = (1/3) × 25872 = 8624 cm3.
6. SSC CGL Shortcuts & Tricks
These shortcut formulas save precious time in the exam. Memorise them thoroughly.
Tip 14 — Percentage Change in Area
If each dimension of a 2D figure changes by x%, the area changes by (2x + x2/100)%. For example, if side of square increases by 10%, area increases by 2(10) + 100/100 = 20 + 1 = 21%.
Tip 15 — Percentage Change in Volume
If each dimension of a 3D figure changes by x%, the volume changes by (3x + 3x2/100 + x3/10000)%. For x = 10%, volume change = 30 + 3 + 0.1 = 33.1%.
Example 19 — Area Change Shortcut
The side of a square is increased by 20%. Find the percentage increase in its area.
Using shortcut: 2(20) + 202/100 = 40 + 4 = 44% increase in area.
Example 20 — Volume Change Shortcut
The edge of a cube is increased by 30%. Find the percentage increase in its volume.
Using shortcut: 3(30) + 3(900)/100 + 27000/10000 = 90 + 27 + 2.7 = 119.7% increase in volume.
Example 21 — Diagonal of Cube
The side of a cube is 10 cm. Find the length of its main diagonal.
Diagonal = √3 × side = √3 × 10 = 10√3 cm ≈ 17.32 cm.
Tip 16 — Max Area for Given Perimeter
For a given perimeter, a circle has the maximum area among all 2D shapes. For a given perimeter in rectangles, the square has the maximum area.
Tip 17 — Min Perimeter for Given Area
For a given area, a circle has the minimum perimeter. For rectangles with fixed area, the square has the minimum perimeter.
Example 22 — Max Area Rectangle
A wire of length 40 cm is bent to form a rectangle. Find the maximum possible area.
For max area with given perimeter, it should be a square. Side = 40/4 = 10 cm. Max area = 100 cm2.
Example 23 — Ratio Shortcut
If the ratio of radii of two spheres is 2 : 3, find the ratio of their volumes.
Volumes scale with cube of radius. Ratio of volumes = 23 : 33 = 8 : 27.
Advanced Problems
Advanced Problem 1
A copper wire of diameter 6 mm is wound around a cylinder of length 18 cm and radius 28 cm covering its curved surface area. Find the length of the wire used. (Use π = 22/7)
Height of cylinder = 18 cm, radius = 28 cm. Number of turns = height/diameter of wire = 18/(0.6) = 30 turns. Length of one turn = 2π(28) = 176 cm. Total length = 30 × 176 = 5280 cm = 52.8 m.
Advanced Problem 2
A solid iron sphere of radius 9 cm is melted and recast into a wire of radius 0.3 cm. Find the length of the wire.
Volume of sphere = (4/3)πr3 = (4/3)π(729) = 972π cm3. Wire volume = πr2h = π(0.09)h. Equating: 0.09πh = 972π ⇒ h = 972/0.09 = 10800 cm = 108 m.
Advanced Problem 3
A cylindrical tank of radius 7 m and height 10 m is filled with water. Water is pumped out at 5 litres/second. How long will it take to empty the tank? (1 m3 = 1000 L)
Volume = π(49)(10) = 490π m3. Using π = 22/7: V = 490 × 22/7 = 1540 m3 = 15,40,000 L. Time = 15,40,000 / 5 = 3,08,000 seconds = 85.56 hours.
Advanced Problem 4
A cone and a hemisphere have equal radius and equal volume. Find the ratio of the height of the cone to the radius.
Volume of cone = (1/3)πr2h. Volume of hemisphere = (2/3)πr3. Equating: (1/3)πr2h = (2/3)πr3 ⇒ h = 2r. Ratio h : r = 2 : 1.
Advanced Problem 5
A sphere of radius 6 cm is dropped into a cylindrical vessel of radius 8 cm containing water. Find the rise in water level.
Volume of sphere = (4/3)π(216) = 288π cm3. This volume displaces water in the cylinder. Volume of water displaced = π(64)h. So 64πh = 288π ⇒ h = 4.5 cm.
Advanced Problem 6
A wire is bent into a circle of radius 7 cm. It is then rebent into a square. Find the area of the square. (Use π = 22/7)
Circumference = 2 × (22/7) × 7 = 44 cm. This is the perimeter of the square. Side of square = 44/4 = 11 cm. Area = 121 cm2.
Advanced Problem 7
A rectangular sheet of paper 44 cm × 18 cm is rolled along its length to form a cylinder. Find the volume of the cylinder. (Use π = 22/7)
Length 44 cm becomes circumference: 2πr = 44 ⇒ r = 7 cm. Height = 18 cm. Volume = π(49)(18) = (22/7) × 882 = 2772 cm3.
Advanced Problem 8
The curved surface area of a cylinder is 440 cm2 and its volume is 1540 cm3. Find the ratio of its height to radius. (Use π = 22/7)
CSA = 2πrh = 440, Volume = πr2h = 1540. Divide volume by CSA: (πr2h)/(2πrh) = 1540/440 ⇒ r/2 = 3.5 ⇒ r = 7 cm. From CSA: 2(22/7)(7)h = 440 ⇒ 44h = 440 ⇒ h = 10 cm. h : r = 10 : 7.
Practice Questions
Attempt these MCQs to test your mensuration knowledge. Click an option to check your answer instantly.
Q8001. Find the area of a triangle with sides 26 cm, 28 cm, and 30 cm.
336 cm2
320 cm2
360 cm2
300 cm2
Solution: s = (26+28+30)/2 = 42. Area = √[42×16×14×12] = √112896 = 336 cm2.
Q8002. The diagonal of a square is 12√2 cm. What is its area?
100 cm2
144 cm2
121 cm2
169 cm2
Solution: Side = diagonal/√2 = 12 cm. Area = 144 cm2.
Q8003. A rectangle has length 15 cm and diagonal 17 cm. Find its perimeter.
40 cm
42 cm
44 cm
46 cm
Solution: b = √(289-225) = 8 cm. Perimeter = 2(15+8) = 46 cm.
Q8004. The area of a circle is 154 cm2. Find its circumference. (Use π = 22/7)
40 cm
42 cm
44 cm
46 cm
Solution: πr2 = 154 ⇒ r = 7 cm. Circumference = 2(22/7)(7) = 44 cm.
Q8005. An equilateral triangle has side 12 cm. Find its height.
5√3 cm
6 cm
6√3 cm
8√3 cm
Solution: Height = (√3/2) × 12 = 6√3 cm.
Q8006. If the radius of a sphere is doubled, by what factor does its volume increase?
2
4
8
16
Solution: Volume ∝ r3. If radius doubles, volume becomes 23 = 8 times.
Q8007. A rhombus has diagonals 10 cm and 24 cm. Find its perimeter.
48 cm
50 cm
52 cm
56 cm
Solution: Side = ½√(100+576) = ½×26 = 13 cm. Perimeter = 4×13 = 52 cm.
Q8008. The TSA of a cube is 294 cm2. Find its volume.
274 cm3
294 cm3
343 cm3
364 cm3
Solution: 6a2 = 294 ⇒ a = 7 cm. Volume = 73 = 343 cm3.
Q8009. A sector of radius 14 cm has central angle 45°. Find its area. (π = 22/7)
64 cm2
70 cm2
77 cm2
84 cm2
Solution: Area = (45/360) × (22/7) × 196 = (1/8) × 616 = 77 cm2.
Q8010. A cylinder has radius 7 cm and height 12 cm. Find its CSA. (π = 22/7)
484 cm2
528 cm2
554 cm2
616 cm2
Solution: CSA = 2πrh = 2(22/7)(7)(12) = 528 cm2.
Q8011. A trapezium has parallel sides 8 cm and 12 cm, and height 5 cm. Find its area.
40 cm2
50 cm2
60 cm2
80 cm2
Solution: Area = ½(8+12)×5 = ½×20×5 = 50 cm2.
Q8012. The side of a square is increased by 30%. Find the percentage increase in its area.
60%
63%
69%
72%
Solution: Change = 2(30) + 302/100 = 60 + 9 = 69%.
Q8013. A cone has radius 9 cm and height 12 cm. Find its slant height.
12 cm
15 cm
18 cm
21 cm
Solution: l = √(81+144) = √225 = 15 cm.
Q8014. A hemisphere has radius 7 cm. Find its TSA. (π = 22/7)
308 cm2
385 cm2
462 cm2
539 cm2
Solution: TSA = 3πr2 = 3(22/7)(49) = 462 cm2.
Q8015. The diagonals of a rhombus are in ratio 3:4 and its area is 96 cm2. Find the diagonals.
8 cm, 12 cm
10 cm, 14 cm
12 cm, 16 cm
14 cm, 18 cm
Solution: Let diagonals be 3k and 4k. Area = ½(3k)(4k) = 6k2 = 96 ⇒ k = 4. Diagonals = 12 cm, 16 cm.
Q8016. Find the number of bricks each measuring 25 cm × 12.5 cm × 7.5 cm needed for a wall 20 m × 2 m × 0.75 m.
10000
12800
14400
16000
Solution: Wall volume = 2000×200×75 = 3,00,00,000 cm3. Brick volume = 25×12.5×7.5 = 2343.75 cm3. Bricks = 3,00,00,000/2343.75 = 12800.
Q8017. The edge of a cube is 6 cm. What is the length of its main diagonal?
6 cm
6√2 cm
6√3 cm
12 cm
Solution: Diagonal = √3 × side = 6√3 cm.
Q8018. A hollow cylinder has outer radius 8 cm, inner radius 6 cm, height 14 cm. Find its volume of material. (π = 22/7)
616 cm3
1232 cm3
1848 cm3
2464 cm3
Solution: Volume = π(64-36)(14) = (22/7)(28)(14) = 1232 cm3.
Q8019. A circle of radius 7 cm is inscribed in a square. Find the area of the square.
121 cm2
196 cm2
154 cm2
144 cm2
Solution: Side of square = diameter = 14 cm. Area = 196 cm2.
Q8020. A cone and a cylinder have equal radius and equal height. Find the ratio of their volumes.
1:2
1:3
2:3
1:4
Solution: Cone volume = (1/3)πr2h, Cylinder volume = πr2h. Ratio = 1:3.
Q8021. A parallelogram has base 15 cm and area 180 cm2. Find its height.
10 cm
12 cm
14 cm
15 cm
Solution: Area = base × height ⇒ 180 = 15h ⇒ h = 12 cm.
Q8022. The volume of a sphere is 38808 cm3. Find its radius. (π = 22/7)
18 cm
21 cm
24 cm
27 cm
Solution: (4/3)(22/7)r3 = 38808 ⇒ r3 = 38808 × 3/4 × 7/22 = 9261 ⇒ r = 21 cm.
Q8023. A rectangular field has length 40 m and breadth 30 m. A path of width 2 m runs inside the field along its boundary. Find the area of the path.
240 m2
264 m2
280 m2
300 m2
Solution: Outer area = 40×30 = 1200. Inner area = 36×26 = 936. Path area = 1200-936 = 264 m2.
Q8024. The radius of a wheel is 35 cm. How many revolutions will it take to cover 11 km? (π = 22/7)
4000
4500
5000
5500
Solution: Circumference = 2(22/7)(35) = 220 cm = 2.2 m. Distance = 11000 m. Revolutions = 11000/2.2 = 5000.
Q8025. A right triangle has legs 9 cm and 12 cm. Find the area of the circumscribing circle.
100π cm2
144π cm2
56.25π cm2
81π cm2
Solution: Hypotenuse = √(81+144) = 15 cm. For a right triangle, circumdiameter = hypotenuse. So r = 7.5 cm. Circle area = π(7.5)2 = 56.25π cm2.