← Arithmetical & Numerical Ability · DSSSB AE Civil

Chapter 11 of 19

Mensuration & Trigonometry

In the DSSSB AE Civil syllabus under Arithmetical & Numerical Ability · 3 parts

Mensuration — 2D Figures · Mensuration — 3D Solids · Trigonometry, Heights & Distances

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (18 sections)

Part 1 of 3

Mensuration — 2D Figures

Last reviewed 22 Sept 2026 · 5 min read

Quadrilaterals

FormulaAreas and perimeters

Figure	Area	Perimeter
Rectangle	$l \times b$	$2 (l + b)$ , diagonal $l^{2} + b^{2}$
Square	$a^{2} = \frac{d ^{2}}{2}$	$4 a$ , diagonal $a 2$
Parallelogram	base $\times$ height	$2 (a + b)$
Rhombus	$\frac{1}{2} d_{1} d_{2}$	$4 a$ , with $a = \frac{1}{2} d_{1}^{2} + d_{2}^{2}$
Trapezium	$\frac{1}{2} (a + b) h$	sum of the four sides

Triangles

FormulaFour ways to an area

Base and height: $\frac{1}{2} bh$
Three sides (Heron): $s = \frac{a + b + c}{2}$ , area $= s (s - a) (s - b) (s - c)$
Equilateral of side $a$ : $\frac{3}{4} a^{2}$ , height $\frac{3}{2} a$
Right-angled: $\frac{1}{2} \times$ product of the two legs

Circle, sector, segment and ring

FormulaCircle formulas

For radius $r$ and an angle $θ$ in degrees:

Area = π r^{2}, Circumference = 2 π r

Arc = \frac{θ}{360} \times 2 π r, Sector area = \frac{θ}{360} \times π r^{2} = \frac{arc \times r}{2}

Minor segment = sector - triangle, Ring (annulus) = π (R^{2} - r^{2})

Use $π = \frac{22}{7}$ when the radius is a multiple of 7, and $3.14$ otherwise.

Paths and borders

A path of width $w$ outside a rectangle $l \times b$ : outer size $(l + 2 w) \times (b + 2 w)$ ; path area $=$ outer $-$ inner.
A path inside: inner size $(l - 2 w) \times (b - 2 w)$ .
Two crossing paths through the middle of a field: area $= l w + b w - w^{2}$ — the overlap is counted once.

Part 2 of 3

Mensuration — 3D Solids

Last reviewed 22 Sept 2026 · 5 min read

The standard solids

FormulaVolumes and surfaces

Solid	Volume	Curved / lateral surface	Total surface
Cuboid	$l bh$	$2 h (l + b)$	$2 (l b + bh + h l)$
Cube	$a^{3}$	$4 a^{2}$	$6 a^{2}$
Cylinder	$π r^{2} h$	$2 π r h$	$2 π r (r + h)$
Cone	$\frac{1}{3} π r^{2} h$	$π r l$	$π r (r + l)$
Sphere	$\frac{4}{3} π r^{3}$	—	$4 π r^{2}$
Hemisphere	$\frac{2}{3} π r^{3}$	$2 π r^{2}$	$3 π r^{2}$

Diagonal of a cuboid $= l^{2} + b^{2} + h^{2}$ ; of a cube $= a 3$ .

FormulaCone, frustum, prism and pyramid

Cone slant height: $l = r^{2} + h^{2}$
Frustum of a cone (radii $R$ and $r$ , height $h$ ): $V = \frac{π h}{3} (R^{2} + R r + r^{2})$ , slant $l = h^{2} + (R - r)^{2}$ , curved surface $= π l (R + r)$
Hollow cylinder (outer $R$ , inner $r$ ): $V = π h (R^{2} - r^{2})$
Prism: $V =$ base area $\times$ height; lateral surface $=$ base perimeter $\times$ height
Pyramid: $V = \frac{1}{3} \times$ base area $\times$ height; lateral surface $= \frac{1}{2} \times$ base perimeter $\times$ slant height

Capacity and units

1 m^{3} = 1000 litres, 1 litre = 1000 cm^{3}

Part 3 of 3

Trigonometry, Heights & Distances

Last reviewed 22 Sept 2026 · 5 min read

The six ratios

In a right-angled triangle, with respect to an acute angle $θ$ :

sin θ = \frac{opposite}{hypotenuse}, cos θ = \frac{adjacent}{hypotenuse}, tan θ = \frac{opposite}{adjacent}

with $csc$ , $sec$ and $cot$ as the reciprocals of $sin$ , $cos$ and $tan$ . Also $tan θ = \frac{sin θ}{cos θ}$ and $cot θ = \frac{cos θ}{sin θ}$ .

Standard angles

$θ$	$0°$	$30°$	$45°$	$60°$	$90°$
$sin θ$	0	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{3}{2}$	1
$cos θ$	1	$\frac{3}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	0
$tan θ$	0	$\frac{1}{3}$	1	$3$	undefined

Identities

FormulaThe three that matter

sin^{2} θ + cos^{2} θ = 1, 1 + tan^{2} θ = sec^{2} θ, 1 + cot^{2} θ = csc^{2} θ

Useful rearrangements: $sec^{2} θ - tan^{2} θ = 1$ and $csc^{2} θ - cot^{2} θ = 1$ , so

sec θ - tan θ = \frac{1}{sec θ + tan θ}

FormulaComplementary angles

sin (90° - θ) = cos θ, tan (90° - θ) = cot θ, sec (90° - θ) = csc θ

So $sin^{2} 20° + sin^{2} 70° = 1$ and $tan 1° tan 2° \dots tan 89° = 1$ , because the terms pair off.

Signs and larger angles

All ratios are positive in the first quadrant; only $sin$ and $csc$ in the second, only $tan$ and $cot$ in the third, only $cos$ and $sec$ in the fourth ("add sugar to coffee"). For values past $90°$ : $sin (180° - θ) = sin θ$ and $cos (180° - θ) = - cos θ$ .

Maximum and minimum

$- 1 \leq sin θ \leq 1$ and $- 1 \leq cos θ \leq 1$
$a sin θ + b cos θ$ lies between $- a^{2} + b^{2}$ and $a^{2} + b^{2}$
$sin θ cos θ = \frac{1}{2} sin 2 θ$ , so its maximum is $\frac{1}{2}$
$sec^{2} θ + csc^{2} θ \geq 4$ , with equality at $45°$

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