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CDS Mathematics

Mensuration (2D & 3D)

📐 Mensuration · MC10 CDS Elementary Mathematics 🎯 High Priority

Mensuration is the second largest chapter in CDS, contributing 12–16 questions. Unlike Geometry (which is theorem-based), Mensuration is entirely formula-driven. Know the formula, substitute correctly, and the answer follows. The key skill is identifying the correct shape and choosing the right formula — area vs perimeter vs volume vs surface area.

📌 CDS exam focus: (1) Area and perimeter of all 2D shapes — triangle, quadrilateral, circle; (2) Combining shapes — "figure that can be split up into these figures"; (3) Surface area and volume of cuboid, cylinder, cone, sphere; (4) Conversion of solids — melting and recasting (volume constant); (5) Path around or inside a figure — finding area of borders.

Topics at a Glance

① 2D: Triangle

Area, perimeter, Heron's formula

② 2D: Quadrilaterals

Square, rect, parallelogram, rhombus, trapezium

③ 2D: Circle

Area, circumference, sector, ring

④ 3D: Cuboid & Cube

LSA, TSA, volume, diagonal

⑤ 3D: Cylinder, Cone, Sphere

CSA, TSA, volume — all formulas

⑥ Conversion of Solids

Volume constant when recast

1. 2D Mensuration

1.1

Complete Reference — All 2D Shapes

Memorise: Area formulas are the most tested. Perimeter = sum of all sides.

Shape	Area	Perimeter / Circumference	Special Formula
Triangle	½ × base × height	a + b + c	Heron's: √s(s−a)(s−b)(s−c), s=(a+b+c)/2
Equilateral △	(√3/4)a²	3a	Height = (√3/2)a
Right △	½ × base × height	a+b+c	c²=a²+b² (Pythagoras)
Square	a²	4a	Diagonal = a√2
Rectangle	l × b	2(l+b)	Diagonal = √(l²+b²)
Parallelogram	base × height	2(a+b)	Area = ab sinθ (included angle)
Rhombus	½ × d₁ × d₂	4a	Side = √((d₁/2)²+(d₂/2)²)
Trapezium	½(a+b)×h	a+b+c+d	h = perpendicular distance between ∥ sides
Circle	πr²	2πr	Diameter = 2r; Area with π=22/7 or 3.14
Sector	(θ/360)×πr²	2r + arc length	Arc = (θ/360)×2πr
Ring (Annulus)	π(R²−r²)	2π(R+r)	R=outer, r=inner radius

Worked Example — Path Around Rectangle

A rectangular garden 40m×30m has a 2m wide path around it. Find area of path.
Outer dimensions: (40+4)×(30+4)=44×34=1496 m². Inner=40×30=1200 m².
Path area=1496−1200=296 m².

2. 3D Mensuration

2.1

Surface Area & Volume — All 3D Solids

LSA/CSA = lateral/curved surface only. TSA includes all faces.

⚡ Complete 3D Mensuration Reference

CUBOID (l × b × h): LSA = 2h(l+b) TSA = 2(lb+bh+hl) Volume = l×b×h Diagonal = √(l²+b²+h²) CUBE (side a): LSA = 4a² TSA = 6a² Volume = a³ Diagonal = a√3 CYLINDER (radius r, height h): CSA = 2πrh TSA = 2πr(h+r) Volume = πr²h CONE (radius r, height h, slant l = √(r²+h²)): CSA = πrl TSA = πr(l+r) Volume = ⅓πr²h SPHERE (radius r): SA = 4πr² Volume = (4/3)πr³ HEMISPHERE (radius r): CSA = 2πr² TSA = 3πr² Volume = (2/3)πr³ CONVERSION RULE: Volume is conserved when a solid is melted and recast. V₁ = V₂ (total volume of new shapes = original shape volume)

For cone: always find slant height l = √(r²+h²) first. For sphere: 4πr² is TOTAL surface (no separate LSA).

Worked Example — Conversion of Solids

A solid sphere of radius 6 cm is melted and recast into small spheres of radius 1 cm. How many small spheres?
V(large)=(4/3)π×216. V(small)=(4/3)π×1. Number=216/1=216 small spheres.

A cylinder (r=6, h=14) is melted into a cone (r=7). Find height of cone.
Volume cylinder = π×36×14 = 504π. Volume cone = ⅓×π×49×h = 504π → h = 504×3/49 = ~30.86 cm.

📋 TOPIC-WISE PYQ

Mensuration — CDS Questions

Q1. The area of an equilateral triangle with side 8 cm is:

(a) 16√3 cm² (b) 24 cm² (c) 32 cm² (d) 8√3 cm²

Answer: (a) 16√3 cm²
Area=(√3/4)×8²=(√3/4)×64=16√3 cm².

Q2. The volume of a cube is 512 cm³. Find its surface area.

(a) 384 cm² (b) 256 cm² (c) 192 cm² (d) 512 cm²

Answer: (a) 384 cm²
a³=512 → a=8 cm. TSA=6a²=6×64=384 cm².

Q3. The area of a trapezium with parallel sides 12 m and 8 m and height 6 m is:

(a) 60 m² (b) 72 m² (c) 80 m² (d) 96 m²

Answer: (a) 60 m²
Area=½(12+8)×6=½×20×6=60 m².

Q4. If the radius of a circle is increased by 10%, its area increases by:

(a) 10% (b) 20% (c) 21% (d) 25%

Answer: (c) 21%
New area=π(1.1r)²=1.21πr². Increase=21%.

Q5. A cone has slant height 13 cm and base radius 5 cm. Its volume is (π=22/7):

(a) 314.3 cm³ (b) 628.6 cm³ (c) 1256 cm³ (d) 200π cm³

Answer: (a) 314.3 cm³
h=√(13²−5²)=√144=12. V=⅓×(22/7)×25×12=(22×25×12)/(7×3)=6600/21≈314.3 cm³.

Q6. The diagonal of a rhombus are 24 cm and 10 cm. Find its perimeter.

(a) 52 cm (b) 68 cm (c) 60 cm (d) 48 cm

Answer: (a) 52 cm
Side=√(12²+5²)=√169=13 cm. Perimeter=4×13=52 cm.

🔥 TRICKY QUESTIONS

Mensuration — Classic CDS Traps

🧩 T1. A wire is bent into a circle of area 154 cm². If the same wire is bent into a square, find the area of the square. (π=22/7)

Solution: 121 cm².
πr²=154 → r²=49 → r=7. Circumference=2π×7=44 cm=wire length=side of square perimeter=4a → a=11. Area=121 cm².

🧩 T2. If the radius of a sphere is doubled, by what factor does the volume increase?

Solution: 8 times.
V=(4/3)πr³. New V=(4/3)π(2r)³=(4/3)π×8r³=8V. Volume increases by factor 8.
Trap: students say "doubles" (linear) or "quadruples" (area). Volume scales as cube of radius ratio.

🧩 T3. A room 8m×6m has a border of tiles 1m wide around the wall. How many square tiles of side 25 cm needed?

Solution: 96 tiles.
Border area = Total − Inner = 8×6 − 6×4 = 48−24 = 24 m².
Each tile = 0.25×0.25 = 0.0625 m². Number = 24/0.0625 = 384 tiles. (Border inside room: 1m wide on each side, so inner = (8−2)×(6−2).)

📐 Formula Sheet — MC10

2D: Triangle

½×b×h (general)
(√3/4)a² (equilateral)
Heron's: √s(s−a)(s−b)(s−c)
s=(a+b+c)/2

2D: Quadrilateral

Square: a², 4a, diag a√2
Rect: lb, 2(l+b)
Rhombus: ½d₁d₂, 4a
Trapezium: ½(a+b)h

2D: Circle

Area: πr²
Circumference: 2πr
Sector area: (θ/360)πr²
Ring: π(R²−r²)

3D: Cube & Cuboid

Cube: TSA=6a², V=a³, diag=a√3
Cuboid: TSA=2(lb+bh+hl)
V=lbh; diag=√(l²+b²+h²)

3D: Cylinder & Cone

Cyl: CSA=2πrh; TSA=2πr(h+r); V=πr²h
Cone: l=√(r²+h²); CSA=πrl; V=⅓πr²h

3D: Sphere

Sphere: SA=4πr², V=(4/3)πr³
Hemisphere: CSA=2πr², TSA=3πr²
V=⅔πr³
Conversion: V₁=V₂

⚡ Quick Revision Booster — MC10

2D Shapes

Triangle: ½bh
Equilateral: (√3/4)a²
Rhombus: ½d₁d₂
Trapezium: ½(a+b)h

Circle

Area=πr²; Circ=2πr
Sector=(θ/360)πr²
If r→kr: area→k²r²

Cube/Cuboid

Cube TSA=6a²; V=a³
Diag=a√3
Cuboid V=lbh

Cylinder/Cone

Cyl V=πr²h
Cone V=⅓πr²h
Slant l=√(r²+h²)

Sphere

SA=4πr²; V=(4/3)πr³
Double r → V×8
Conversion: V constant

🚨 Traps

Radius ×2 → area ×4, volume ×8
TSA includes base(s); CSA does not
Cone has 1 base; cylinder has 2

✏️ PRACTICE EXERCISE

Test Yourself — MC10

E1. Find the area of a triangle with sides 13, 14, and 15 cm using Heron's formula.

(a) 72 cm² (b) 84 cm² (c) 90 cm² (d) 96 cm²

💡 s=(13+14+15)/2=21. Area=√21×8×7×6.

E2. The curved surface area of a cylinder is 440 cm² and radius is 7 cm. Find height. (π=22/7)

(a) 8 cm (b) 10 cm (c) 12 cm (d) 15 cm

💡 CSA=2πrh=440. Solve for h.

E3. A sphere of radius 9 cm is melted into 27 smaller spheres of equal size. Find their radius.

(a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm

💡 Volume conserved: (4/3)π×9³=27×(4/3)πr³. Find r.

E4. A path 2 m wide surrounds a circular garden of radius 14 m. Find area of the path. (π=22/7)

(a) 176 m² (b) 200 m² (c) 184 m² (d) 192 m²

💡 Area of path = π(R²−r²) where R=16, r=14.

E5. The length of a room is 1.5 times its width. If perimeter is 25 m, find area.

(a) 30 m² (b) 37.5 m² (c) 50 m² (d) 25 m²

💡 Let width=x, length=1.5x. Perimeter=2(x+1.5x)=5x=25 → x=5. Area=5×7.5.

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