3D Geometry Important Question

Duration: 5 min

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Inside: a video lesson and guided study material.

Module outline

  1. Section A | English: Vocabulary, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Preposition, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Articles, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Verbal Analogy (Word Based Analogy)
  2. Section A | Quant Aptitude: Ratio and Proportion (Ratios), Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Arithmetic, Algebra, Powers and Exponents (Surds and Indices), Average, Age Problems, Percentage, Profit and Loss, Partnership, Time and Work, Simple Interest and Compound Interest, Mensuration and Geometry, Probability, Pipe and Cistern
  3. Section A | Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Calendar, Clock, Blood Relations, Directions (Direction Test), Seating Arrangements, Ranking, Order and Sequence, Puzzle Solving (Puzzles)
  4. Section B | C Programming: C Programming by Yash Sir, Coding Questions by Yash Sir
  5. Section B | Data Structures: Algorithm Analysis, Greedy Algorithms, Time Complexity Analysis, Sorting Algorithms, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs
  6. Section B | OOP (C++): Object Oriented Programming (OOPS), Programming in C++
  7. Section B | OS: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management, Windows OS
  8. Section B | Networking: 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
  9. Section B | AI & Big Data Basics: AI & ML by Arun Sir, Big Data Systems
  10. Section C | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation, Analog-Digital Conversion
  11. Section A, C | Comp. Fundamentals, COA: Basics of COA, Floating Point Rep, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations, Multiprocessors, Programming the Basic Computer
  12. MOCK TEST:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This bilingual 3D geometry lesson solves a right-prism problem: the triangular base has sides 20 cm, 21 cm and 29 cm; volume is 7560 cm³; find lateral surface area. The instructor first computes the base perimeter P = 20 + 21 + 29 = 70 cm. A red right-triangle is sketched on the prism’s end face, and the Pythagorean check 20² + 21² = 29² (400 + 441 = 841) confirms the base is a right triangle, so its area is ½ × b × h = ½ × 20 × 21 = 210 cm². Using volume V = base area × prism height, the instructor finds h = 7560 / 210 = 36 cm. Finally, lateral surface area LSA = P × h = 70 × 36 = 2520 cm², matching option 4. The method emphasizes recognizing a right-triangle base from side lengths, computing area via ½bh, extracting prism height from volume, and applying LSA = perimeter × height.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The bilingual problem is displayed in English and Hindi: a right-prism with triangular base sides 20 cm, 21 cm, 29 cm and volume 7560 cm³; options are 2448, 2556, 2484, 2520 cm². A light-blue wireframe triangular prism appears at lower left. The instructor begins red handwritten work: perimeter P = 20 + 21 + 29 = 70 cm, then boxes the triangle area formula ½ × b × h and sketches a red right-triangle on the prism’s end face to mark base and height.

  2. 2:00 – 4:39 02:00-04:39

    The instructor verifies the base is a right triangle by writing 20² + 21² = 29² and 400 + 441 = 841, then computes base area as ½ × 20 × 21 = 210 cm². Using volume, the prism height is found: h = 7560 / 210 = 36 cm. The lateral surface area is then calculated with LSA = P × h = 70 × 36 = 2520 cm², and the answer is circled as option 4.

The lesson teaches a systematic approach to right-prism lateral surface area when the base is a triangle. Key steps: (1) compute perimeter from side lengths; (2) test for a right triangle using the Pythagorean theorem to justify area = ½bh; (3) compute base area; (4) derive prism height from V = base area × h; (5) apply LSA = perimeter × height. The worked example yields 2520 cm², reinforcing that lateral surface area depends only on base perimeter and prism height, not on the shape’s internal angles once area is known.

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