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Surface areas and Volumes

Karnataka Board · Class 9 · Mathematics

Practice quiz for Surface areas and Volumes — Karnataka Board Class 9 Mathematics. MCQs and questions with answers to test your preparation.

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Quick Quiz: Surface areas and Volumes

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1

Find the curved surface area of a cone with base radius 7 cm and slant height 15 cm.

2

A sphere has radius 10.5 cm. What is its surface area?

3

Find the volume of a cone with base radius 6 cm and height 8 cm.

4

A hemisphere has radius 7 cm. Find its curved surface area.

45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

Find the total surface area of a cone with base radius 5 cm and slant height 13 cm.

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282.9 cm²

Step 1: Given r = 5 cm, l = 13 cm. Step 2: Total surface area = curved surface area + base area = πrl + πr². Step 3: Factor out π and r to get πr(l + r). Step 4: Substitute = (22/7) × 5 × (13 + 5) = (22/7) × 5 × 18 = 1980/7 = 282.9 cm². Total surface area includes both curved surface and circular base.

2multiple choice
1 marks

The volume of a sphere is 4851 cm³. Find its radius. (Use π = 22/7)

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10.5 cm

Step 1: Given volume = 4851 cm³. Step 2: Apply volume formula (4/3)πr³ = 4851. Step 3: Substitute π = 22/7: (4/3) × (22/7) × r³ = 4851. Step 4: Solve for r³: r³ = (4851 × 3 × 7)/(4 × 22) = 101871/88 = 1157.625. Step 5: Take cube root: r = 10.5 cm. Working backwards from volume to radius requires solving the cubic equation.

3multiple choice
1 marks

A cone has height 12 cm and volume 1232 cm³. Find the radius of its base.

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7 cm

Step 1: Given h = 12 cm, V = 1232 cm³. Step 2: Apply volume formula (1/3)πr²h = 1232. Step 3: Substitute known values: (1/3) × (22/7) × r² × 12 = 1232. Step 4: Solve for r²: r² = (1232 × 3 × 7)/(22 × 12) = 25872/264 = 49. Step 5: Therefore r = 7 cm. This is a reverse application of the volume formula.

4multiple choice
1 marks

Find the volume of a hemisphere with diameter 21 cm.

Show answer

2425.5 cm³

Step 1: Given diameter = 21 cm, so radius r = 10.5 cm. Step 2: Volume of hemisphere = (2/3)πr³. Step 3: Substitute values = (2/3) × (22/7) × (10.5)³. Step 4: Calculate = (2/3) × (22/7) × 1157.625 = (2 × 22 × 1157.625)/(3 × 7) = 2425.5 cm³. Remember hemisphere volume is half of sphere volume.

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What are the important topics in Surface areas and Volumes for Karnataka Board Class 9 Mathematics?
Surface areas and Volumes covers several key topics that are frequently asked in Karnataka 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 — Karnataka 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.

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