3D Geometry Important Question
Duration: 3 min
This video lesson is available to enrolled students.
Inside: a video lesson and guided study material.
Module outline
- Quantitative Aptitude: Number System, Divisibility Rules, HCF LCM, Arithmetic, Percentage, Ratio and Proportion (Ratios), Average, Powers and Exponents (Surds and Indices), Profit and Loss, Simple Interest and Compound Interest, Mixture and Alligation, Age Problems, Partnership, Algebra, Permutation and Combination, Probability, Logarithm, Mensuration and Geometry, Time and Work, Speed Time and Distance, Data Interpretation, Sequence and Series (Series), Gamified / Game Based Aptitude, Speed Mathematics and Basic Mathematics
- Problem Solving & Reasoning: Coding Decoding, Data Sufficiency, Seating Arrangements, Puzzle Solving (Puzzles), Data Arrangement, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Inequality, Venn Diagram, Ranking, Order and Sequence, TCS NQT Advanced Reasoning Ability, Crypt Arithmetic, Analytical Reasoning (Counting Figures Reasoning), Series (Number and Letter Series) (Numerical Relations and Reasoning), Logical Sequence Of Words
- Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Completion (Fill in the blanks), Sentence Correction (Error Correction), Subject Verb Agreement (Verb Noun Agreement), Tenses, Articles, Preposition, Modals, Vocabulary
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
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.
2:00 – 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.