3D Geometry Important Question
Duration: 5 min
This video lesson is available to enrolled students.
Inside: a video lesson and guided study material.
Module outline
- TCS One Shot Free Videos & Notes: TCS Preparation
- Numerical Ability (Foundation Section): Ratio and Proportion (Ratios), Divisibility Rules, Data Interpretation, Mixture and Alligation, Permutation and Combination, Logarithm, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Percentage, Average, Powers and Exponents (Surds and Indices), Algebra, Probability, Profit and Loss, Simple Interest and Compound Interest, Time and Work, Pipe and Cistern, Mensuration and Geometry, Mental Ability, Age Problems, Statistics, Arithmetic, Partnership, Speed Mathematics and Basic Mathematics
- Reasoning Ability (Foundation Section): Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Seating Arrangements, Blood Relations, Directions (Direction Test), Analytical Reasoning (Counting Figures Reasoning), Puzzle Solving (Puzzles), Calendar, Clock, Cubes & Dices, Selection Decision Table (Decision Making), Crypt Arithmetic, Venn Diagram, Data Arrangement, Inequality, Analogy, Ranking, Order and Sequence, Logical Sequence Of Words
- New Pattern-Verbal Ability (Foundation Section): Sentence Completion (Fill in the blanks), English Writing Ability
- Old Pattern-Verbal Ability (Foundation Section): Vocabulary, Noun, Subject Verb Agreement (Verb Noun Agreement), Adjectives, Tenses, Pronoun, Preposition, Articles, Direct and Indirect Speech, Conjunction, Modals, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Correction (Error Correction), Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Verbal Analogy (Word Based Analogy), Active Passive Voice (Voice), Determiner, Facts, Inference and Judgements, Adverb, Cloze Test (Cloze Passage), Sentence Construction, Verb
- Advanced Quantitative Ability: TCS NQT Advanced Quantitative Ability
- Advanced Reasoning Ability: TCS NQT Advanced Reasoning Ability
- Advanced Coding: Coding Questions by Yash Sir, C Programming by Yash Sir, Python for Placements by Yash Sir, DSA using Python, Object Oriented Programming (OOPS), Programming in C++, Java
- TCS Full Length Mock Test:
- TCS Interview & Resume Prep: MR & HR Interview Preparation, TCS Interview Preparation, Mock Interview, Interview Recordings, Resume Preparation, Resume Review
- Mega Test:
- Grand 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
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.
2:00 – 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.

