3D Geometry Important Question

Duration: 3 min

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Inside: a video lesson and guided study material.

Module outline

  1. General Intelligence & Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Analogy, Analytical Reasoning (Counting Figures Reasoning), Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Syllogisms, Puzzle Solving (Puzzles), Venn Diagram, Clerical Aptitude, Verbal Classification
  2. General Awareness: Polity, Physics, Biology, Chemistry, History, Geography, Indian Economy, Environment, GK- Current Affairs, Static General Knowledge
  3. Quantitative Aptitude: Ratio and Proportion (Ratios), Data Interpretation, Mixture and Alligation, Number System, Speed Time and Distance, Percentage, Average, Arithmetic, Simple Interest and Compound Interest, Profit and Loss, Partnership, Time and Work, Algebra, Mensuration and Geometry, Trigonometry, Linear Algebra
  4. English Comprehension: Vocabulary, Noun, Subject Verb Agreement (Verb Noun Agreement), Adjectives, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Pronoun, Preposition, Articles, Adverb, Interjection, Conjunction, Modals, Determiner, Active Passive Voice (Voice), Direct and Indirect Speech, Verbal Analogy (Word Based Analogy), Cloze Test (Cloze Passage), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks)
  5. MOCK TEST:
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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