Class 9 Math: Surface Areas and Volumes Important Exam Questions
Preparing for school exams or final CBSE boards requires practicing questions patterned after official grading schemes. We have compiled a list of crucial questions for Class 9 Surface Areas and Volumes to help you test your mastery.
Exam-Pattern Questions and Answers
Question 1 (MEDIUM)
This is a descriptive problem. Work out all steps on paper before checking the solution.
Solution Explanation & Steps: Radii of two cylinders are r1 = 2x and r2 = 3x respectively. Height of two cylinders are h1 = 5x and h2 = 3x respectively. Ratio of their total surface areas = 1 1 1 2 2 2 2 ( ) 2 (2 )(7 ) 7 7 : 92 ( ) 2 (3 )(6 ) 9 r r h x x r r h x x π π π π + = = =+ OR Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find its total surface area. Ans. Let radius of cone be r cm Slant height of cone = l = 14 cm curved surface area of cone = 308 cm2 rl = 308 22 7 × r × 14 = 308 r = 308 7 722 14 cm× =× Total surface area of cone = r(r + l) = 22 7 × 7 × (7 + 14) = 22 7 × 7 × 21 = 462 cm2 SECTION - C Questions 26 to 31 carry 3 marks each.
Question 2 (EASY)
- Option 1: 2496 cm³
- Option 2: 1248 cm³
- Option 3: 1664 cm³
- Option 4: 416 cm³
Solution Explanation & Steps: (d) 416 cm³ Volume of cone = 1/3 × πr² h = 1/3 × 156 × 8 (∵ Area of base = πr² = 156 cm²) = 416 cm³
Question 3 (HARD)
This is a descriptive problem. Work out all steps on paper before checking the solution.
Solution Explanation & Steps: Here, Height of cylinder = 12 - 4 = 8 cm Radius of cone/cylinder = 3/2 = 1.5 cm Height of cone = 2 cm Volume of cylinder = πr2h = π(1.5)2 × 8 = 18π cm3 Volume of cone = 1/3 πr2h = 1/3 π(1.5)2 × 2 = 1.5π cm3 Total volume = Volume of cylinder + (Volume of cone) × 2 = 18π + 1.5π × 2 = 18π + 3π = 21π = 21 × 22/7 = 66 cm3. OR There are two identical solid cubical boxes of side 7cm. From the top face of the first cube a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. Find (i) the ratio of the total surface area of the two new solids formed (ii) volume of each new solid formed. Ans: (i) SA for first new solid (S₁) = 6a2 + 2πr2 - πr2 = 6a2 + πr2 = 6×7×7 + 22 7 ×3.5 × 3.5 = 294 + 38.5 = 332.5 cm2 SA for second new solid (S₂) = 6a2 + 2πr2 - πr2 = 6a2 + πr2 = 6×7×7 + 22 7 ×3.5 × 3.5 = 294 + 38.5 = 332.5 cm2 So S₁ : S₂ = 1:1 (ii) Volume for first new solid (V₁) = a3 - 2 3 πr3 = 7 × 7 × 7 - 2 3 × 22 7 × 3.5× 3.5× 3.5 = 343 - 539 6 = 1519 6 cm3 Volume for second new solid (V₂) = a3 + 2 3 πr3 = 7 × 7 × 7 + 2 3 × 22 7 × 3.5× 3.5× 3.5 = 343 + 539 6 = 2597 6 cm3
Grading Step-Marking Guidelines
Remember that teachers award partial marks for writing down formulas, stating the given variables, and drawing diagrams. Never leave a question blank on an exam. Write down the relevant formulas and initial substitution steps to secure partial credits.