3D Geometry Important Question

Duration: 5 min

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Inside: a video lesson and guided study material.

Module outline

  1. English Language: Vocabulary, Noun, Subject Verb Agreement (Verb Noun Agreement), Adjectives, Tenses, Pronoun, Verb, Articles, Adverb, Preposition, Conjunction, Modals, Determiner, 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)
  2. Quantitative Aptitude: Ratio and Proportion (Ratios), Divisibility Rules, Data Interpretation, Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Algebra, Percentage, Powers and Exponents (Surds and Indices), Time and Work, Average, Profit and Loss, Pipe and Cistern, Speed Mathematics and Basic Mathematics, Age Problems, Simple Interest and Compound Interest, Partnership, Mensuration and Geometry, Probability, IBPS Specific Videos
  3. Reasoning Ability: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Puzzle Solving (Puzzles), Seating Arrangements, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Inequality, Blood Relations, Directions (Direction Test), Syllogisms, Ranking, Order and Sequence, Logical Sequence Of Words, Machine Input-Output
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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