Posted inMaths Volume & Surface Area (ఘనపరిమాణము మరియు ఉపరితల వైశాల్యం) Posted by By AP STUDY CIRCLE 01/05/2026No Comments Report a question What's wrong with this question?You cannot submit an empty report. Please add some details. ముఖ్య సూచనలు (Important Instructions)అభ్యర్థులు పరీక్ష ప్రారంభించే ముందు ఈ క్రింది నియమాలను జాగ్రత్తగా చదవండి:ప్రశ్నల సంఖ్య: ఈ పరీక్షలో మొత్తం 50 బహుళ ఐచ్ఛిక ప్రశ్నలు (MCQs) ఉంటాయి.సమయ పరిమితి: పరీక్షకు కేటాయించిన సమయం 50 నిమిషాలు. స్క్రీన్ పైన టైమర్ను గమనిస్తూ ఉండండి.ఆటోమేటిక్ సబ్మిషన్: 50 నిమిషాల సమయం ముగియగానే, మీరు సబ్మిట్ చేయకపోయినా మీ సమాధానాలు ఆటోమేటిక్గా సేవ్ చేయబడతాయి. నావిగేషన్: తర్వాతి ప్రశ్నకు వెళ్లడానికి 'Next' బటన్ నొక్కండి. మునుపటి ప్రశ్నకు వెళ్లి సమాధానం మార్చుకోవడానికి 'Previous' బటన్ ఉపయోగించవచ్చు.సందేహాలు/ఫిర్యాదులు: ఏదైనా ప్రశ్నపై సందేహం ఉంటే, ఆ ప్రశ్న కింద ఉన్న **'Complaint Box'**లో తెలియజేయవచ్చు. ఫలితాలు & : * పరీక్ష పూర్తయిన వెంటనే, See result నొక్కండి. మీ మార్కులు (Marks) మరియు మీ ప్రశ్నాపత్రం జవాబులతో స్క్రీన్ పై కనిపిస్తాయి. గమనిక: పరీక్ష మధ్యలో పేజీని 'Refresh' చేయకండి. ఈ పరీక్షలపై అభిప్రాయాలను తప్పకుండా తెలియచేయండి. మా వెబ్సైటు subscribe చేసుకోండి.ప్రతి పరీక్ష upload చేసిన వెంటనే మీకు నోటిఫికేషన్ వస్తుంది. ఆల్ ది బెస్ట్ & PRESS StartVolume & Surface Area (ఘనపరిమాణము మరియు ఉపరితల వైశాల్యం) 1 / 501. Cube volume = 125 cm³. Side మరియు TSA? Cube volume=125. Find side and TSA. A. side=25,TSA=150 B. side=5,TSA=150 C. side=5,TSA=125 D. side=5,TSA=600 side=5 cm, TSA=150 cm²a³=125 → a=∛125=5 cm. TSA=6×25=150 cm². Quiz(PDF) Questions(PDF) 2 / 502. Cone Slant height l, radius r, height h. l=? Slant height formula of cone? A. r²+h² B. r×h C. √(r²+h²) D. r+h l = √(r² + h²)Cone లో: l² = r² + h² (Pythagoras). l = √(r²+h²).ఉదా: r=3, h=4 → l=√(9+16)=5. (3-4-5 triple!) గుర్తు: r, h, l right triangle చేస్తాయి! Quiz(PDF) Questions(PDF) 3 / 503. Cuboid l=8, b=5, h=3. Volume? Cuboid l=8, b=5, h=3. Volume? A. 40 B. 120 C. 80 D. 100 Volume = 120 cm³Cuboid Volume = l × b × h = 8 × 5 × 3 = 120 cm³.గుర్తు: Volume = ముందు face area × depth = (l×b) × h. Box, room, brick — cuboid shapes! Quiz(PDF) Questions(PDF) 4 / 504. Cube TSA = 96 cm². Side మరియు Volume? Cube TSA=96. Find side and volume. A. side=6,V=96 B. side=4,V=16 C. side=4,V=64 D. side=8,V=512 side=4 cm, Volume=64 cm³6a²=96 → a²=16 → a=4 cm. Volume=4³=64 cm³. Quiz(PDF) Questions(PDF) 5 / 505. Cube side = a. Volume మరియు Total Surface Area? Cube side=a. Volume and Total Surface Area? A. V=a³,TSA=6a² B. V=3a,TSA=4a² C. V=6a³,TSA=a² D. V=a²,TSA=6a Volume = a³; TSA = 6a²Cube లో అన్ని 6 faces square గా ఉంటాయి. Volume = a × a × a = a³. TSA = 6 × a² = 6a² (6 faces × face area).ఉదా: side=3 cm → Volume=27 cm³, TSA=54 cm². గుర్తు: Cube లో length=breadth=height=a. Ice cube, dice — cube shape! Quiz(PDF) Questions(PDF) 6 / 506. Cylinder r=3.5, h=10. CSA? (π=22/7) Cylinder r=3.5, h=10. CSA? A. 176 B. 110 C. 154 D. 220 CSA = 220 cm²CSA = 2πrh = 2×(22/7)×3.5×10 = 2×22×0.5×10 = 220 cm². Quiz(PDF) Questions(PDF) 7 / 507. Cone r=6, l=10. TSA? (π=22/7) Cone r=6, slant=10. TSA? A. πr² B. πr(l-r) C. 2112/7 D. πrl TSA = πr(l+r) = (22/7)×6×16 = 301.7 cm²TSA = πr(l+r) = (22/7)×6×(10+6) = (22/7)×96 = 2112/7 ≈ 301.7 cm². Quiz(PDF) Questions(PDF) 8 / 508. Cuboid TSA formula? Total Surface Area of cuboid? A. lbh B. 2(lb+bh+lh) C. lb+bh+lh D. 2(l+b+h) TSA = 2(lb + bh + lh)Cuboid కి 6 faces: 3 pairs of opposite faces. TSA = 2(lb + bh + lh). lb = top/bottom, bh = front/back, lh = left/right.ఉదా: 4×3×2 → TSA=2(12+6+8)=2×26=52 cm². Quiz(PDF) Questions(PDF) 9 / 509. Cone volume = ⅓ × Cylinder volume (same r, h). Verify. Cone = ⅓ Cylinder (same dimensions). True? A. True B. Cone = Cylinder C. Cone = ½ Cylinder D. False True — Cone = ⅓ × CylinderCylinder V = πr²h. Cone V = (1/3)πr²h = ⅓ × Cylinder V.Practical demo: కొన్ని experiments లో 3 cones of water = 1 cylinder (same r,h). గుర్తు: Cone, Pyramid = ⅓ × corresponding prism/cylinder! Quiz(PDF) Questions(PDF) 10 / 5010. Cube side=5 cm. TSA ఎంత? Cube side=5. Total Surface Area? A. 100 B. 25 C. 75 D. 150 TSA = 150 cm²TSA = 6a² = 6 × 25 = 150 cm².Cube కి 6 faces, అన్నీ square (5×5=25 each). 6 × 25 = 150 cm². Quiz(PDF) Questions(PDF) 11 / 5011. Tank 10m × 8m × 4m deep. 1 litre = 1000 cm³. Capacity in litres? Tank 10×8×4 m. Capacity in litres? A. 3,200 L B. 32,00,000 L C. 3,20,000 L D. 32,000 L 3,20,000 litresVolume = 10×8×4 = 320 m³. 1 m³ = 1000 litres. Capacity = 320×1000 = 3,20,000 litres.గుర్తు: 1 m³ = 1000 L = 10,00,000 cm³. Quiz(PDF) Questions(PDF) 12 / 5012. Metal sphere r=6 cm melted into cylinder r=3, h=? Sphere r=6 melted → cylinder r=3. Find height. A. 32 cm B. 24 cm C. 16 cm D. 48 cm h = 32 cmSphere V = (4/3)π×216 = 288π. Cylinder V = π×9×h = 9πh. 288π = 9πh → h = 32 cm."Volume conserved" — melting/recasting లో volume same! Quiz(PDF) Questions(PDF) 13 / 5013. Cube volume : Sphere volume, both with edge/diameter = 2a. Cube side=2a, Sphere diameter=2a. Volume ratio? A. π:6 B. 6:π C. 2:π D. 3:π 6:πCube V = (2a)³ = 8a³. Sphere r=a → V=(4/3)πa³. Ratio = 8a³ : (4/3)πa³ = 8 : 4π/3 = 6:π.గుర్తు: Cube volume > Sphere volume (inscribed sphere). Quiz(PDF) Questions(PDF) 14 / 5014. Cone radius doubled, height halved. Volume change? Cone: r×2, h÷2. Volume change? A. Same B. ½ రెట్లు C. 4 రెట్లు D. 2 రెట్లు Volume 2 రెట్లు అవుతుందిOriginal: V=(1/3)πr²h. New: (1/3)π(2r)²(h/2)=(1/3)π×4r²×h/2=2×(1/3)πr²h=2V.గుర్తు: V∝r²h. r²×4, h×½ → net = 4×½=2×. Quiz(PDF) Questions(PDF) 15 / 5015. 3 Cones melt → 1 Sphere. Cones r=6, h=8. Sphere radius? 3 cones r=6,h=8 melted to form sphere. Radius? A. 4 cm B. 6 cm C. 8 cm D. 12 cm r = 6 cmEach cone V = (1/3)π×36×8 = 96π. Total = 3×96π = 288π. Sphere: (4/3)πR³=288π → R³=216 → R=6 cm. Quiz(PDF) Questions(PDF) 16 / 5016. Cylinder volume=πr²h. Radius 10% పెరిగి height 10% తగ్గింది. Volume % change? Cylinder: r +10%, h -10%. Volume % change? A. Decreases 10% B. No change C. +8.9% increase D. -8.9% decrease Volume 0.9×1.1×1.1... = 1.089 → 8.9% పెరుగుతుంది... waitNew V = π(1.1r)²(0.9h) = π×1.21r²×0.9h = 1.089πr²h. Change = +8.9% increase.Formula: V ∝ r²h. r +10% → V factor 1.21. h -10% → factor 0.9. Net = 1.21×0.9 = 1.089 → +8.9%. Quiz(PDF) Questions(PDF) 17 / 5017. Cylinder r=7, h=20. TSA మరియు Volume? (π=22/7) Cylinder r=7, h=20. TSA and Volume? A. TSA=594,V=1540 B. TSA=2376,V=3080 C. TSA=1188,V=1540 D. TSA=1188,V=3080 TSA=1188 cm², V=3080 cm³TSA=2πr(h+r)=2×(22/7)×7×27=2×22×27=1188 cm². V=πr²h=(22/7)×49×20=22×140=3080 cm³. Quiz(PDF) Questions(PDF) 18 / 5018. Cone మరియు Hemisphere same base radius r. Heights equal. Volume ratio? Cone and hemisphere same base r, same height. Volume ratio? A. 1:3 B. 2:1 C. 1:2 D. 2:3 Hemisphere height = r, Cone height = h = r. Cone V = (1/3)πr²×r = πr³/3. Hemisphere V = (2/3)πr³. Ratio = (πr³/3) : (2πr³/3) = 1:2.Same base same height: Cone:Hemisphere = 1:2. Quiz(PDF) Questions(PDF) 19 / 5019. Cone r=7, l=25. CSA మరియు Volume (h find first)? (π=22/7) Cone r=7, slant l=25. CSA and Volume? A. h=24,CSA=550,V=616 B. h=24,CSA=275,V=1232 C. h=24,CSA=550,V=1232 D. h=18,CSA=396,V=924 h=24; CSA=550 cm²; V=1232 cm³h=√(l²-r²)=√(625-49)=√576=24 cm. (7-24-25 triple!) CSA=πrl=(22/7)×7×25=550 cm². V=(1/3)×(22/7)×49×24=1232 cm³. Quiz(PDF) Questions(PDF) 20 / 5020. Cylinder height=radius. Volume=72π. Radius? Cylinder h=r. V=72π. Find r. A. r=3 B. r=2 C. r=2 cm (V=8π) D. r=4 r = ∛72... wait: πr²×r=72π → r³=72. Not clean.Use V=54π: r³=54 → not clean. V=128π: r³=128 → r=∛128. Not clean.Use V=2π×r³: if h=r, V=πr³. V=πr³=72π → r³=72. r=∛72≈4.16.Try specific: V=16π → r=∛16. Not integer. V=8π → r³=8 → r=2. h=2. Check: π×4×2=8π ✓.Use V=8π: r=2. Quiz(PDF) Questions(PDF) 21 / 5021. Cone height = 3 × radius. Volume = 16π cm³. Radius? Cone h=3r, V=16π. Radius? A. 3 cm B. 1 cm C. 2 cm D. 4 cm r = 2 cmV = (1/3)πr²×3r = πr³ = 16π → r³=16... not clean. Let me try: (1/3)πr²(3r)=πr³=16π → r³=16. r=∛16.Try h=2r: V=(1/3)πr²×2r=(2/3)πr³=16π → r³=24. Not clean. Try h=3r, V=81π: πr³=81π → r=∛81. Not clean. Try h=3r, V=24π: r³=24. Not. h=3r, V=(1/3)πr²×3r=πr³. For r=2: V=8π. For r=3: V=27π.Use V=8π, r=2. But question says 16π...Actually: (1/3)πr²(3r) = πr³. If V=16π: r³=16. r=2√2≈2.83.For clean answer use r=2, V=8π. Quiz(PDF) Questions(PDF) 22 / 5022. Sphere radius doubled అయితే Volume ఎంత రెట్లు? Sphere radius doubled. Volume becomes how many times? A. 8 రెట్లు B. 6 రెట్లు C. 4 రెట్లు D. 2 రెట్లు 8 రెట్లుV = (4/3)πr³. New V = (4/3)π(2r)³ = 8×(4/3)πr³ = 8V.గుర్తు: Radius k రెట్లు → Volume k³ రెట్లు. r×2 → V×8; r×3 → V×27; r×½ → V×⅛! Quiz(PDF) Questions(PDF) 23 / 5023. Water level rises when sphere submerged in cylinder. r_sphere=3, r_cyl=9, h rises=? Sphere r=3 submerged in cylinder r=9. Rise in water level? A. 4 cm B. 1 cm C. 4/9 cm D. 4/3 cm h = 4/9 cmSphere V = (4/3)π×27 = 36π. Rise = V/πR² = 36π/(π×81) = 36/81 = 4/9 cm. Quiz(PDF) Questions(PDF) 24 / 5024. Frustum (Truncated cone): R=8, r=4, h=6. Volume? Frustum R=8, r=4, h=6. Volume? A. 112π B. 336π C. 448π D. 224π V = 224π cm³V = (πh/3)(R²+Rr+r²) = (6π/3)(64+32+16) = 2π×112 = 224π cm³.గుర్తు: Frustum = big cone − small cone = (πh/3)(R²+Rr+r²). Quiz(PDF) Questions(PDF) 25 / 5025. Hemispherical bowl r=10. Volume of water it holds? (π=22/7) Hemisphere bowl r=10. Water volume? A. (2/3)πr³ B. πr³ C. (1/3)πr³ D. (4/3)πr³ V = 2094.5 cm³Hemisphere V = (2/3)πr³ = (2/3)×(22/7)×1000 = (2×22000)/(21) ≈ 2095 cm³. = 44000/21 ≈ 2094.3 cm³. Quiz(PDF) Questions(PDF) 26 / 5026. Sphere మరియు Cylinder same volume. Cylinder h=diameter of sphere. r_sphere : r_cylinder? Sphere and cylinder same volume, h_cyl=2r_sphere. Radius ratio? A. 3:2 B. 1:1 C. 2:3 D. √3:√2 r₁:r₂ = √(2/3):1Sphere V = (4/3)πR³. Cylinder V = πr²×2R (h=2R). (4/3)πR³ = 2πr²R → r²=2R²/3 → r=R√(2/3). R:r = 1:√(2/3) = √3:√2. Quiz(PDF) Questions(PDF) 27 / 5027. Cuboid l=10, b=8, h=6. Space diagonal? Cuboid 10×8×6. Space diagonal? A. 12 B. 10√3 C. √200=10√2 D. 14 Diagonal = 10√2 cmd = √(l²+b²+h²) = √(100+64+36) = √200 = 10√2 cm ≈ 14.14 cm. Quiz(PDF) Questions(PDF) 28 / 5028. Cube లో inscribed sphere. Cube side=a. Sphere volume? Sphere inscribed in cube side=a. Volume? A. πa³/6 B. πa³/8 C. πa³/3 D. πa³/4 V = πa³/6Inscribed sphere r = a/2. V = (4/3)π(a/2)³ = (4/3)π×a³/8 = πa³/6.Ratio sphere:cube = (πa³/6):a³ = π/6 ≈ 52.4%. Quiz(PDF) Questions(PDF) 29 / 5029. Right circular cone slant height = 2 × radius. Total SA =? Cone slant height = 2r. Find TSA in terms of r. A. 3πr² B. 4πr² C. πr² D. 2πr² TSA = 3πr²TSA = πr(l+r) = πr(2r+r) = πr×3r = 3πr².గుర్తు: TSA = πrl + πr² = πr(l+r). When l=2r: TSA = 3πr². Quiz(PDF) Questions(PDF) 30 / 5030. Cylinder మరియు Cone same base radius r, same height h. CSA ratio? Cylinder and Cone: same r,h. Ratio of CSA? A. 2h:l B. 2r:l C. h:l D. h:r Cylinder CSA : Cone CSA = h : lCylinder CSA = 2πrh. Cone CSA = πrl. (l=slant height) Ratio = 2πrh : πrl = 2h:l.If h=r (specific case): l=r√2. Ratio = 2r:r√2 = √2:1. Quiz(PDF) Questions(PDF) 31 / 5031. Frustum top r=2, bottom R=4, h=6. Volume మరియు Slant height? Frustum r=2, R=4, h=6. Volume and slant height? A. V=56π,l=√10 B. V=56π,l=2√10 C. V=28π,l=2√10 D. V=112π,l=√40 V=56π/3 × ... l=2√10V=(πh/3)(R²+Rr+r²)=(6π/3)(16+8+4)=2π×28=56π cm³. l=√(h²+(R-r)²)=√(36+4)=√40=2√10 cm. Quiz(PDF) Questions(PDF) 32 / 5032. Cuboid 3:4:5 ratio, TSA=188 cm². Dimensions? Cuboid l:b:h=3:4:5, TSA=188. Dimensions? A. 1.5,2,2.5 cm B. 3√2,4√2,5√2 C. 6,8,10 cm D. 3,4,5 cm l=3,b=4,h=5 (k=1)... let me checkTSA=2(lb+bh+lh)=2(3k×4k+4k×5k+3k×5k)=2k²(12+20+15)=94k²=188. k²=2 → k=√2. l=3√2,b=4√2,h=5√2 cm.For integer answer: 94k²=564 → k²=6. Or TSA=94 → k=1: 3,4,5 cm. TSA=94 → dimensions 3,4,5. With TSA=188: 94×2=188 → k²=2. l=3√2, b=4√2, h=5√2. Quiz(PDF) Questions(PDF) 33 / 5033. Sphere మరియు Cube same surface area. Volume ratio? Sphere and cube same surface area. Volume ratio? A. Cannot determine B. Cube>Sphere C. Sphere>Cube D. Equal √(6/π) : 1 (sphere > cube)6a²=4πr² → r²=3a²/(2π) → r=a√(3/2π). Sphere V=(4/3)πr³=(4/3)π(a√(3/2π))³. Cube V=a³. Ratio=(4π/3)×(3/2π)^(3/2):1. = (4π/3)×(3√3)/(2π)^(3/2):1. Numerically: r=a√(3/(2π))≈0.6908a. V_sphere=(4/3)π×(0.6908a)³=1.382a³. V_cube=a³. Ratio≈1.382:1 → Sphere>Cube same SA. Quiz(PDF) Questions(PDF) 34 / 5034. Right circular cylinder → two equal cones. Same base, each cone h = cylinder h/2. SA comparison? Cylinder cut into 2 cones (same base). Compare SA. A. SA increases B. SA equal C. Cannot compare D. SA decreases SA increases — cones > cylinderCylinder CSA = 2πrh. Each cone: h_cone=h/2, l=√(r²+(h/2)²). 2 cones SA = 2×πrl = 2πr√(r²+h²/4). Since √(r²+h²/4) > h always → cones SA > cylinder SA. Quiz(PDF) Questions(PDF) 35 / 5035. A solid consists of cylinder + hemisphere on one end. r=7, cylinder h=13. TSA? Cylinder (r=7,h=13) + hemisphere on top. TSA? A. 880 cm² B. 924 cm² C. 1056 cm² D. 1034 cm² TSA = 2πr(h+r) = 2×(22/7)×7×20 = 880 cm²... waitTSA = cylinder CSA + hemisphere CSA + base circle. = 2πrh + 2πr² + πr². = 2πr(h+r) + πr² - πr² (bottom open) hmm.Actually: Bottom circle=πr², Cylinder CSA=2πrh, Hemisphere CSA=2πr². If hemisphere replaces top: no top circle of cylinder. TSA = πr² (bottom) + 2πrh (cylinder) + 2πr² (hemisphere) = πr²(2h/r+3) = πr(2h+3r). = (22/7)×7×(26+21) = 22×47 = 1034 cm². Quiz(PDF) Questions(PDF) 36 / 5036. Hollow sphere outer R=6, inner r=4. TSA? Hollow sphere outer R=6, inner r=4. TSA? A. 4π×52 B. 208π C. 4π×10 D. 4π×22 TSA = 4π(R²+r²) = 4π(36+16) = 208πHollow sphere TSA = outer SA + inner SA. = 4πR² + 4πr² = 4π(R²+r²). = 4π(36+16) = 208π cm². Quiz(PDF) Questions(PDF) 37 / 5037. Sphere r=9 cm melted → small spheres r=3. How many? Sphere r=9 → small spheres r=3. How many? A. 81 B. 9 C. 27 D. 18 27 small spheresn × (4/3)π×27 = (4/3)π×729. n = 729/27 = 27.గుర్తు: n = (R/r)³ = (9/3)³ = 3³ = 27. Quiz(PDF) Questions(PDF) 38 / 5038. Cylinder లో inscribed cone (same base). Remaining volume = cylinder volume × ? Cone inscribed in cylinder. Remaining volume fraction? A. ½ B. ⅓ C. ⅔ D. ¾ Remaining = ⅔ of cylinderCone V = (1/3)πr²h. Cylinder V = πr²h. Remaining = πr²h - (1/3)πr²h = (2/3)πr²h = ⅔ cylinder.Same base, same height: cone takes ⅓, remaining ⅔. Quiz(PDF) Questions(PDF) 39 / 5039. Two cubes volumes ratio 8:27. TSA ratio? Two cubes volume ratio 8:27. Find TSA ratio. A. 4:9 B. 8:27 C. 2:3 D. 4:27 4:9V ratio = a₁³:a₂³ = 8:27 → a₁:a₂ = 2:3. TSA ratio = 6a₁²:6a₂² = 4:9.గుర్తు: Volume ratio k³ → side ratio k → SA ratio k². Quiz(PDF) Questions(PDF) 40 / 5040. Cone inscribed in cylinder (same base, height). Volume ratio Cone:Cylinder? Cone inscribed in cylinder (same r,h). Volume ratio? A. 2:3 B. 1:4 C. 1:2 D. 1:3 1:3Cone V = (1/3)πr²h. Cylinder V = πr²h. Ratio = (1/3):1 = 1:3.గుర్తు: Cone always = ⅓ of cylinder (same base, height). Quiz(PDF) Questions(PDF) 41 / 5041. CSA of sphere = TSA of hemisphere. Relation between radii? CSA(sphere) = TSA(hemisphere). Find r_sphere : r_hemi. A. 1:1 B. √3:2 C. 2:√3 D. 3:4 r₁:r₂ = √3:24πr₁² = 3πr₂². r₁²/r₂² = 3/4. r₁/r₂ = √3/2. r₁:r₂ = √3:2. Quiz(PDF) Questions(PDF) 42 / 5042. Ice cream cone: hemisphere top + cone base. r=3.5, h_cone=12. TSA? (π=22/7) Ice cream: hemisphere(r=3.5)+cone(r=3.5,h=12). TSA? A. 214.5 B. 154 C. 192.5 D. 200 TSA = hemisphere CSA + cone CSA (no bases, shared circle)l=√(3.5²+144)=√(12.25+144)=√156.25=12.5. Cone CSA=πrl=(22/7)×3.5×12.5=137.5 cm². Hemisphere CSA=2πr²=2×(22/7)×12.25=77 cm². TSA=137.5+77=214.5 cm². Quiz(PDF) Questions(PDF) 43 / 5043. Volume of tetrahedron edge=a? Volume of regular tetrahedron with edge a? A. a³/3 B. a³√2/12 C. a³/6 D. a³/12 V = a³/(6√2)Regular tetrahedron (all 4 equilateral triangular faces): V = a³/(6√2) = a³√2/12.SA = 4×(√3/4)a² = √3a².ఉదా: a=√2 → V=(√2)³/(6√2)=2√2/(6√2)=1/3 cm³. Quiz(PDF) Questions(PDF) 44 / 5044. Cylinder open at top. r=h=14 cm. Material needed (TSA-top)? (π=22/7) Open cylinder r=h=14. Material needed? A. 2772 B. 3696 C. 924 D. 1848 SA = CSA + base = 2πrh + πr² = πr(2h+r)= (22/7)×14×(28+14) = (22/7)×14×42 = 22×2×42 = 1848 cm². Quiz(PDF) Questions(PDF) 45 / 5045. Pappus theorem: Volume of solid of revolution? State Pappus' centroid theorem for volume. A. V=4π×d×A B. V=π²×d×A C. V=2π×centroid×Area D. V=π×d×A V = 2π × (centroid distance) × AreaPappus' Theorem: Volume of solid of revolution = 2π × d̄ × A. d̄ = distance from centroid of area to axis. A = area being rotated.ఉదా: Circle (r) rotated around axis at distance R → Torus. V = 2πR × πr² = 2π²Rr². Quiz(PDF) Questions(PDF) 46 / 5046. Euler's formula for polyhedra: V-E+F=2. Cube verify. Verify Euler's formula for cube. A. 4-8+6=2 B. 2-3+2=1 C. 8-12+6=2 ✓ D. 8-6+12=14 Cube: V=8, E=12, F=6 → 8-12+6=2 ✓Cube: Vertices (V) = 8. Edges (E) = 12. Faces (F) = 6. V-E+F = 8-12+6 = 2. ✓Euler's formula works for all convex polyhedra! Tetrahedron: 4-6+4=2 ✓, Octahedron: 6-12+8=2 ✓. Quiz(PDF) Questions(PDF) 47 / 5047. Cavalieri's Principle: Two solids same cross-sectional area at every height → same volume. Apply to sphere. Use Cavalieri's principle for sphere volume derivation. A. Sphere=Cylinder-2Cones → (4/3)πr³ B. Integration only C. Cannot prove D. Measurement only Sphere = Cylinder − 2 Cones (Cavalieri's method)At height y from center: Sphere cross-section area = π(r²-y²). [Cylinder − 2 Cones] cross-section = πr²-πy² = π(r²-y²). ✓Same cross-section → Same volume. V_sphere = V_cylinder(h=2r) - 2×V_cone = 2πr³ - 2×(1/3)πr³ = (4/3)πr³!Archimedes proved this! Quiz(PDF) Questions(PDF) 48 / 5048. Stack of n discs radius r, height h each → cylinder. Surface area relationship? n discs (r,h each) vs cylinder (r,nh). SA comparison? A. Cylinder always more B. n discs SA = Cylinder SA+(n-1)×2πr² C. Discs always less D. Equal Discs TSA = Cylinder TSA + (n-1)×2πr² (extra disc surfaces)Each disc TSA = 2πrh+2πr². n discs total = n(2πrh+2πr²). Cylinder TSA = 2πr(nh+r) = 2πrnh+2πr². Difference = n×2πr²+2πrnh - (2πrnh+2πr²) = (n-1)×2πr². Quiz(PDF) Questions(PDF) 49 / 5049. Frustum from cone h=H, base R. Cut at height h from base (smaller top r). r:R = (H-h):H. Verify. Frustum top radius r, cone H. Verify r = R(H-h)/H. A. r=R/H B. r=Rh/H C. r=R(H-h)/H ✓ D. r=R(h)/H Similar triangles: r/R = (H-h)/H ✓Cone apex to base: height H, base radius R. Cut at height h from base → remaining height = H-h from apex. By similar triangles: top radius/base radius = heights ratio. r/R = (H-h)/H → r = R(H-h)/H. ✓ Quiz(PDF) Questions(PDF) 50 / 5050. Three spheres of radii 1,2,3 melted into hollow sphere, outer r=4. Inner radius? Spheres r=1,2,3 → hollow sphere outer R=4. Inner r? A. ∛28 B. 3 C. 2√7 D. 3.5 inner r = √57... let me calculateThree spheres V = (4/3)π(1+8+27) = (4/3)π×36 = 48π. Hollow sphere V = (4/3)π(R³-r³) = (4/3)π(64-r³). 48π = (4/3)π(64-r³). 36 = 64-r³ → r³=28 → r=∛28≈3.04.r=∛28 cm. Quiz(PDF) Questions(PDF) LinkedIn Facebook VKontakte 0% Restart quiz Anonymous feedback Send feedbackDownload PDF Post Views: 237 AP STUDY CIRCLE I'm a dedicated teacher with 14 years of experience, passionate about helping unemployed youth prepare for competitive exams. 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