Surface Areas and Volumes — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Andhra Pradesh (SSC) Class 10 Mathematics solutions for "Surface Areas and Volumes" — important questions with detailed answers, download PDF for board exam preparation.
✓ 100% Free
✓ No Login Needed
✓ NCERT / CBSE Aligned
✓ Download as PDF
TL;DR: Free step-by-step Andhra Pradesh (SSC) Class 10 Mathematics solutions for "Surface Areas and Volumes" — important questions with detailed answers, dow…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.
Surface Areas and Volumes — Textbook Questions Answered (9)
Q1: A solid is formed by combining a cylinder and a hemisphere at one end. The cylinder has radius 3.5 cm and height 10 cm. The hemisphere has the same radius. Find the total surface area. (Use π = 22/7)
Step 1: Total surface area = Curved surface area of cylinder + Base area of cylinder + Curved surface area of hemisphere
Step 2: Curved surface area of cylinder = 2πrh = 2 × (22/7) × 3.5 × 10
Step 3: = 2 × (22/7) × 35 = 2 × 22 × 5 = 220 cm²
Step 4: Base area of cylinder = πr² = (22/7) × (3.5)²
Step 5: = (22/7) × 12.25 = 22 × 1.75 = 38.5 cm²
Step 6: Curved surface area of hemisphere = 2πr² = 2 × (22/7) × (3.5)²
Step 7: = 2 × (22/7) × 12.25 = 2 × 38.5 = 77 cm²
Step 8: Total surface area = 220 + 38.5 + 77 = 335.5 cm²
Final Answer: 335.5 cm²
Q2: A cone is placed on top of a cylinder. The cylinder has radius 5 cm and height 8 cm. The cone has the same radius and slant height 13 cm. Find the total volume and total curved surface area.
Step 1: Volume of cylinder = πr²h = (22/7) × 5² × 8
Step 2: = (22/7) × 25 × 8 = (22/7) × 200
Step 3: = 4400/7 ≈ 628.57 cm³
Step 4: For cone, slant height l = 13 cm, radius r = 5 cm
Step 5: Height of cone: h² = l² − r² = 13² − 5² = 169 − 25 = 144
Step 6: h = 12 cm
Step 7: Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 25 × 12
Step 8: = (1/3) × (22/7) × 300 = (22 × 100)/7 = 2200/7 ≈ 314.29 cm³
Step 9: Total volume = 4400/7 + 2200/7 = 6600/7 ≈ 942.86 cm³
Step 10: Curved surface area of cylinder = 2πrh = 2 × (22/7) × 5 × 8 = (22/7) × 80 = 1760/7 cm²
Step 11: Curved surface area of cone = πrl = (22/7) × 5 × 13 = 1430/7 cm²
Step 12: Total curved surface area = 1760/7 + 1430/7 = 3190/7 ≈ 455.71 cm²
Final Answer: Total volume ≈ 942.86 cm³, Total curved surface area ≈ 455.71 cm²
Q3: Find the volume of a frustum of a cone if the radii of the two circular ends are 5 cm and 3 cm, and the height is 6 cm. (Use π = 22/7)
Step 1: Volume of frustum = (1/3)πh(R² + r² + Rr)
Step 2: Here, R = 5 cm (larger radius), r = 3 cm (smaller radius), h = 6 cm
Step 3: Volume = (1/3) × (22/7) × 6 × (5² + 3² + 5×3)
Step 4: Volume = (1/3) × (22/7) × 6 × (25 + 9 + 15)
Step 5: Volume = (1/3) × (22/7) × 6 × 49
Step 6: Volume = (22/7) × 2 × 49
Step 7: Volume = (22/7) × 98
Step 8: Volume = 22 × 14 = 308 cm³
Final Answer: 308 cm³
Q4: A solid is made by combining a hemisphere and a cone. The hemisphere has radius 4 cm. The cone has the same base radius and height 3 cm. Find the total surface area of the solid. (Use π = 3.14)
Step 1: Total surface area = Curved surface area of hemisphere + Curved surface area of cone
Step 2: Curved surface area of hemisphere = 2πr² = 2 × 3.14 × 4²
Step 3: = 2 × 3.14 × 16 = 100.48 cm²
Step 4: For cone, r = 4 cm, h = 3 cm
Step 5: Slant height l = √(r² + h²) = √(16 + 9) = √25 = 5 cm
Step 6: Curved surface area of cone = πrl = 3.14 × 4 × 5 = 62.8 cm²
Step 7: Total surface area = 100.48 + 62.8 = 163.28 cm²
Final Answer: 163.28 cm²
Q5: Find the volume of a composite solid formed by joining a cylinder (radius 6 cm, height 10 cm) and a cone (same radius, height 8 cm) at the cylinder's top.
Step 1: Volume of cylinder = πr²h = (22/7) × 6² × 10
Step 2: = (22/7) × 36 × 10 = (22/7) × 360
Step 3: = 7920/7 ≈ 1131.43 cm³
Step 4: Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 6² × 8
Step 5: = (1/3) × (22/7) × 36 × 8
Step 6: = (1/3) × (22/7) × 288
Step 7: = (22 × 96)/7 = 2112/7 ≈ 301.71 cm³
Step 8: Total volume = 7920/7 + 2112/7 = 10032/7 ≈ 1433.14 cm³
Final Answer: 10032/7 cm³ or approximately 1433.14 cm³
Q6: A frustum of a cone has slant height 10 cm, and the radii of the two ends are 6 cm and 4 cm. Find the curved surface area of the frustum. (Use π = 22/7)
Step 1: Curved surface area of frustum = π(R + r)l
Step 2: Here, R = 6 cm (larger radius), r = 4 cm (smaller radius), l = 10 cm (slant height)
Step 3: Curved surface area = (22/7) × (6 + 4) × 10
Step 4: = (22/7) × 10 × 10
Step 5: = (22/7) × 100
Step 6: = 2200/7 ≈ 314.29 cm²
Final Answer: 2200/7 cm² or approximately 314.29 cm²
Q7: Find the total surface area of a hemisphere with radius 7 cm (including the flat base). (Use π = 22/7)
Step 1: Total surface area = Curved surface area + Base area
Step 2: Curved surface area of hemisphere = 2πr²
Step 3: = 2 × (22/7) × 7² = 2 × (22/7) × 49
Step 4: = 2 × 22 × 7 = 308 cm²
Step 5: Base area = πr² = (22/7) × 7² = 22 × 7 = 154 cm²
Step 6: Total surface area = 308 + 154 = 462 cm²
Final Answer: 462 cm²
Q8: A cone has height 24 cm and base radius 7 cm. Find its volume and lateral surface area. (Use π = 22/7)
Step 1: Volume of cone = (1/3)πr²h
Step 2: = (1/3) × (22/7) × 7² × 24
Step 3: = (1/3) × (22/7) × 49 × 24
Step 4: = (1/3) × 22 × 7 × 24
Step 5: = (1/3) × 3696 = 1232 cm³
Step 6: Slant height l = √(h² + r²) = √(24² + 7²) = √(576 + 49) = √625 = 25 cm
Step 7: Lateral surface area = πrl = (22/7) × 7 × 25
Step 8: = 22 × 25 = 550 cm²
Final Answer: Volume = 1232 cm³, Lateral surface area = 550 cm²
Q9: Find the volume of a frustum with height 7 cm, upper radius 2 cm, and lower radius 4 cm. (Use π = 22/7)
Step 1: Volume of frustum = (1/3)πh(R² + r² + Rr)
Step 2: Here, h = 7 cm, R = 4 cm (lower radius), r = 2 cm (upper radius)
Step 3: Volume = (1/3) × (22/7) × 7 × (4² + 2² + 4×2)
Step 4: Volume = (1/3) × (22/7) × 7 × (16 + 4 + 8)
Step 5: Volume = (1/3) × (22/7) × 7 × 28
Step 6: Volume = (1/3) × 22 × 28
Step 7: Volume = (616)/3 ≈ 205.33 cm³
Final Answer: 616/3 cm³ or approximately 205.33 cm³
Related
- NCERT Solutions: Surface Areas and Volumes — the CBSE version of this same topic.
- Real Numbers — more Andhra Pradesh (SSC) Class 10 Mathematics.
- Polynomials — more Andhra Pradesh (SSC) Class 10 Mathematics.
- Pair of Linear Equations in Two Variables — more Andhra Pradesh (SSC) Class 10 Mathematics.
- Quadratic Equations — more Andhra Pradesh (SSC) Class 10 Mathematics.
- All Andhra Pradesh (SSC) Class 10 Mathematics chapters →