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Important Questions

Surface areas and Volumes

Telangana Board · Class 9 · Mathematics

Most important questions from Surface areas and Volumes for Telangana Board Class 9 Mathematics board exam 2026. MCQs, short answer, and long answer questions with marks.

45 questions20 flashcards5 concepts

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45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

Find the volume of a cylinder with base radius 4 cm and height 9 cm. (Take π = 22/7)

Show answer

1584/7 cm³

Step 1: Given radius (r) = 4 cm, height (h) = 9 cm, π = 22/7. Step 2: Apply volume formula V = πr²h. Step 3: Substitute values: V = (22/7) × 4² × 9. Step 4: Calculate: V = (22/7) × 16 × 9 = (22 × 144)/7. Step 5: Simplify: V = 3168/7 = 1584/7 cm³. This equals approximately 226.3 cm³.

2multiple choice
1 marks

A cone has base radius 6 cm and height 8 cm. What is its slant height?

Show answer

10 cm

Step 1: Given radius (r) = 6 cm and height (h) = 8 cm. Step 2: Use the relationship l² = h² + r² where l is slant height. Step 3: Substitute values: l² = 8² + 6² = 64 + 36 = 100. Step 4: Take square root: l = √100 = 10 cm. Step 5: Verify using Pythagorean theorem - this forms a 6-8-10 right triangle, which is a multiple of the 3-4-5 triangle.

3multiple choice
1 marks

Find the curved surface area of a cone with radius 7 cm and slant height 15 cm. (Take π = 22/7)

Show answer

330 cm²

Step 1: Given radius (r) = 7 cm, slant height (l) = 15 cm, π = 22/7. Step 2: Apply curved surface area formula CSA = πrl. Step 3: Substitute values: CSA = (22/7) × 7 × 15. Step 4: Simplify: The 7s cancel out, so CSA = 22 × 15 = 330 cm². Step 5: This represents the area of the sector that would form the cone when unfolded.

4multiple choice
1 marks

What is the volume of a cone with base radius 5 cm and height 12 cm? (Take π = 22/7)

Show answer

1100/7 cm³

Step 1: Given radius (r) = 5 cm, height (h) = 12 cm, π = 22/7. Step 2: Apply volume formula V = (1/3)πr²h. Step 3: Substitute values: V = (1/3) × (22/7) × 5² × 12. Step 4: Calculate: V = (1/3) × (22/7) × 25 × 12 = (22 × 300)/(7 × 3). Step 5: Simplify: V = 6600/21 = 1100/7 cm³, which equals approximately 157.14 cm³.

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Frequently Asked Questions

What are the important topics in Surface areas and Volumes for Telangana Board Class 9 Mathematics?
Surface areas and Volumes covers several key topics that are frequently asked in Telangana Board Class 9 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Surface areas and Volumes — Telangana Board Class 9 Mathematics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many important questions are there in Surface areas and Volumes?
There are 45 practice questions available for Surface areas and Volumes. These cover multiple question types including MCQs, short answer, and long answer questions.

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