3D Geometry Important Question

Duration: 3 min

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Module outline

  1. Communication Round: Vocabulary, Subject Verb Agreement (Verb Noun Agreement), Tenses, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Cloze Test (Cloze Passage), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Direct and Indirect Speech, Active Passive Voice (Voice), Verbal Analogy (Word Based Analogy), Articles, Preposition, Conjunction, Sentence Construction, English Writing Ability, Communication Assessment
  2. Quantitative Aptitude: Ratio and Proportion (Ratios), Divisibility Rules, Data Interpretation, Mixture and Alligation, Permutation and Combination, Logarithm, Number System, HCF LCM, Speed Time and Distance, Percentage, Average, Powers and Exponents (Surds and Indices), Time and Work, Mensuration and Geometry, Age Problems, Pipe and Cistern, Probability, Profit and Loss, Simple Interest and Compound Interest, Algebra
  3. Analytical and Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Directions (Direction Test), Ranking, Order and Sequence, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Selection Decision Table (Decision Making), Logical Sequence Of Words, Puzzle Solving (Puzzles), Syllogisms, Seating Arrangements, Calendar, Clock, Blood Relations, Cubes & Dices, Venn Diagram, Analytical Reasoning (Counting Figures Reasoning)
  4. Cognizant Gaming Aptitude: Gamified / Game Based Aptitude
  5. Java (Cluster 1): Java
  6. ANSI SQL MySQL (Cluster 1,2,3): SQL
  7. HTML, CSS, JavaScript (Cluster 1,3): HTML, CSS, JavaScript
  8. Python (Cluster 2): Python for Placements by Yash Sir
  9. Cloud Fundamentals (Cluster 2): Emerging Tech Trends
  10. C# (Cluster 3): C#
  11. OOPs: Object Oriented Programming (OOPS)
  12. Technical Assessment (Coding): Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Pseudo Codes
  13. Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  14. DBMS: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, Relational Calculus, Transaction Management, Concurrency Control, Database Recovery, ORDBMS, Database Security & Authorization, Query Processing & Optimization, Enhanced Data Models, Data Warehousing & Mining, Big Data Systems, NoSQL
  15. Operating Systems: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Windows OS, Linux OS, Security, Distributed Systems, Virtual Machines
  16. Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security, Mobile Technology, Cloud Computing and IoT, Cyber and Info Security
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This short lecture solves a 3D geometry multiple-choice question on the volume of a regular right square pyramid. The problem states that the square base has side length 12 cm and the slant height is 10 cm, asking for the volume with options 384, 480, 576, and 640 cm³. The instructor first identifies the given values a = 12 cm and l = 10 cm, then uses the Pythagorean relation for a regular square pyramid, l² = h² + (a/2)², to find the vertical height. Substituting values gives h² = 100 − 36 = 64, so h = 8 cm. The base area is then calculated as a² = 12² = 144 cm². The volume formula V = (1/3) × base area × height is applied, giving V = (1/3) × 144 × 8. The instructor simplifies by dividing 144 by 3 first to get 48, then multiplies 48 × 8 = 384 cm³. The final answer is 384 cm³, matching option 1.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The slide titled 'Volume of a Regular Square Pyramid' presents the problem: a regular right-pyramid with square base side 12 cm and slant height 10 cm, asking for volume. Red underlines highlight 'a side length' and 'Its slant height is 10 cm,' with red arrows pointing to the pyramid's edges. The instructor writes a = 12cm and l = 10cm, then introduces the relation l² = h² + (a/2)². A small 3D pyramid diagram at lower left shows the slant edge and base side marked with red arrows. The working proceeds to h² = 100 − 36, h² = 64, and a boxed result h = 8.

  2. 2:00 – 2:57 02:00-02:57

    The instructor calculates the base area as 12² = 144 cm² and sets up the volume formula V = (1/3) × B × h with values 144 and 8. A long division of 144 by 3 yields 48, then 48 × 8 = 384 cm³. The final answer 384 cm³ is marked as option 1, the correct choice among options 384, 480, 576, and 640 cm³.

The lesson demonstrates a two-step method for finding the volume of a regular square pyramid when given base side and slant height. Step 1: Use the Pythagorean theorem in the cross-section triangle to find vertical height h from slant height l and half-base a/2, via l² = h² + (a/2)². Step 2: Compute base area a² and apply V = (1/3)a²h. The key teaching cue is simplifying the fraction by dividing base area by 3 before multiplying by height, making arithmetic easier. The worked example yields h = 8 cm and V = 384 cm³.

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