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Exam study plan

SSC CGL Preparation: Concepts and Practice

Free

Build your SSC CGL foundations with English and Hindi lessons, worked examples and explained practice across Quantitative Aptitude, General Intelligence and Reasoning, English Comprehension and General Awareness. Study arithmetic, algebra, geometry and trigonometry; practise analogy, classification, series, directions, ranking and clocks; strengthen grammar and constitutional basics. Use the linked topic tests to check understanding and review mistakes. Coverage is expanding subject by subject and does not yet represent the complete SSC CGL syllabus. See the module list and mock-test section for currently available material.

Lessons

Subject → Module → Lesson

English ComprehensionGeneral AwarenessGeneral Intelligence and ReasoningQuantitative Aptitude
Course outline 100
74 / 100
Lessons 51–100 · Page 2 of 2
Previous page

1 lesson
  1. 51
    Replacement and repeated dilution1 practice test

5 lessons
  1. 52
    Work, rate and efficiency
  2. 53
    Combined work and remaining work
  3. 54
    Efficiency ratios and worker equivalence
  4. 55
    Alternate-day and changing-team work
  5. 56
    Work and wages

6 lessons
  1. 57
    Speed, distance, time and unit conversions
  2. 58
    Average speed for unequal times and distances
  3. 59
    Relative speed and meeting or overtaking
  4. 60
    Trains crossing people, platforms and other trains
  5. 61
    Boats and streams
  6. 62
    Races and circular tracks

2 lessons
  1. 63
    Length, area and volume unit conversions
  2. 64
    Measurement accuracy and dimensional checks

5 lessons
  1. 65
    Perimeter and area of squares and rectangles
  2. 66
    Area and perimeter of triangles
  3. 67
    Parallelogram, rhombus and trapezium areas
  4. 68
    Circumference, circle and semicircle areas
  5. 69
    Composite figures, paths and shaded regions

6 lessons
  1. 70
    Surface area and volume of cubes and cuboids
  2. 71
    Surface area and volume of cylinders
  3. 72
    Surface area and volume of right circular cones
  4. 73
    Surface area and volume of spheres and hemispheres
  5. 74
    Right prisms and right pyramids with triangular or square bases
  6. 75
    Composite solids and volume-preserving conversions

7 lessons
  1. 76
    Variables, expressions and algebraic operations
  2. 77
    Standard algebraic identities
  3. 78
    Elementary factorisation
  4. 79
    Linear equations in one variable
  5. 80
    Pairs of linear equations
  6. 81
    Surds and simplification
  7. 82
    Graphs of linear equations

9 lessons
  1. 83
    Lines, angles and parallel-line relationships
  2. 84
    Triangle angle and side properties
  3. 85
    Medians, altitudes, angle bisectors and triangle centres
  4. 86
    Congruence and similarity of triangles
  5. 87
    Pythagoras theorem and elementary applications
  6. 88
    Properties of quadrilaterals
  7. 89
    Interior and exterior angles of polygons
  8. 90
    Circle chords and angle properties
  9. 91
    Tangents and common tangents to circles

6 lessons
  1. 92
    Trigonometric ratios in a right triangle
  2. 93
    Standard-angle values
  3. 94
    Basic trigonometric identities
  4. 95
    Complementary-angle relationships
  5. 96
    Degrees and radians
  6. 97
    Elementary heights and distances

3 lessons
  1. 98
    Perfect squares and elementary square patterns
  2. 99
    Square roots by factorisation and division
  3. 100
    Estimating square roots

50 lessons across 10 modules

Lesson 74 of 100 | Quantitative Aptitude / Solid Mensuration / ठोस क्षेत्रमिति

Right prisms and right pyramids with triangular or square bases

Learning outcome

Calculate prism and pyramid areas and volumes, distinguishing perpendicular height from face slant height.

Concepts and assumptions

A right prism has parallel, congruent bases and perpendicular lateral edges. With base area B, perimeter p and height h, identical layers give volume B × h. Rectangular walls give lateral area p × h; total area = p × h + 2 × B.

A pyramid’s tapering sections give volume B × h ÷ 3, one-third of the matching prism. Height h is perpendicular to the base plane, not an altitude within the base triangle.

Here pyramids have square or equilateral bases, with the apex directly above their centre. Equal face altitudes s give lateral area p × s ÷ 2; total area adds B. If centre-to-side distance is ρ, s² = h² + ρ². The apex-to-vertex edge is different.

For arbitrary triangular bases, face altitudes may differ: add their individual triangular areas instead. Dimensions are exact centimetres; leave radicals exact without rounding.

Worked examples

Example 1 — Triangular prism. The base is a right triangle with perpendicular sides 6 cm and 8 cm and hypotenuse 10 cm; prism height is 15 cm. B = 6 × 8 ÷ 2 = 24 cm²; p = 24 cm. Volume = 24 × 15 = 360 cm³. Total area = 24 × 15 + 2 × 24 = 408 cm².

Example 2 — Square pyramid. Base side is 10 cm and perpendicular height is 12 cm. Centre-to-side distance = 5 cm, so s = √(144 + 25) = 13 cm. Lateral area = 40 × 13 ÷ 2 = 260 cm². Total area = 260 + 100 = 360 cm²; volume = 100 × 12 ÷ 3 = 400 cm³.

Example 3 — Regular triangular pyramid. Its base has side 6√3 cm, area 27√3 cm² and centre-to-side distance 3 cm. Perpendicular height is 4 cm. Thus s = √(16 + 9) = 5 cm. Base perimeter = 18√3 cm. Lateral area = 18√3 × 5 ÷ 2 = 45√3 cm²; total area = 45√3 + 27√3 = 72√3 cm². Volume = 27√3 × 4 ÷ 3 = 36√3 cm³.

Common mistakes

A triangular prism has rectangular walls, unlike a pyramid. A triangular base alone does not guarantee equal face slant heights.

Practice questions

  1. A right prism has a right-triangular base with sides 5, 12 and 13 cm, and height 10 cm. Find total area and volume.
  2. A right prism has square base side 7 cm and height 12 cm. Find total area and volume.
  3. A right square pyramid has base side 12 cm and perpendicular height 8 cm. Find total area and volume.
  4. A regular triangular pyramid has equilateral base side 8 cm and face slant height 10 cm. Find lateral and total areas.

Worked solutions

  1. B = 5 × 12 ÷ 2 = 30 cm²; p = 30 cm. Total area = 30 × 10 + 60 = 360 cm²; volume = 30 × 10 = 300 cm³.
  2. B = 49 cm²; p = 28 cm. Total area = 28 × 12 + 98 = 434 cm²; volume = 49 × 12 = 588 cm³.
  3. s = √(8² + 6²) = 10 cm. Total area = 48 × 10 ÷ 2 + 144 = 384 cm²; volume = 144 × 8 ÷ 3 = 384 cm³.
  4. Base altitude = √(64 - 16) = 4√3 cm, so B = 8 × 4√3 ÷ 2 = 16√3 cm². Lateral area = 24 × 10 ÷ 2 = 120 cm²; total area = 120 + 16√3 cm².
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