Chapter 9 — Surface Area & Volume — Notes & Full Solutions

20 Most Important — 1 Mark Questions

Q1. Formula for curved surface area (CSA) of a cone?
A1. \(\text{CSA}_{\text{cone}}=\pi r l\), where \(r\) is base radius and \(l\) slant height.
Q2. Formula for total surface area (TSA) of a cone?
A2. \(\text{TSA}_{\text{cone}}=\pi r (l+r)\).
Q3. Formula for volume of a cone?
A3. \(\displaystyle V_{\text{cone}}=\frac{1}{3}\pi r^{2}h\).
Q4. Relation between slant height \(l\), radius \(r\) and height \(h\) of cone?
A4. \(l^{2}=h^{2}+r^{2}\) (Pythagoras in right triangle formed by apex, centre of base and a point on rim).
Q5. Formula for surface area of a sphere?
A5. \(\text{SA}_{\text{sphere}}=4\pi r^{2}\).
Q6. Formula for volume of a sphere?
A6. \(\displaystyle V_{\text{sphere}}=\frac{4}{3}\pi r^{3}\).
Q7. Formula for total surface area of a hemisphere (solid)?
A7. \(\text{TSA}_{\text{hemisphere}}=3\pi r^{2}\) (curved \(2\pi r^{2}\) + base \(\pi r^{2}\)).
Q8. CSA of a cylinder?
A8. \(\text{CSA}_{\text{cyl}}=2\pi r h\).
Q9. Total surface area of cylinder?
A9. \(\text{TSA}_{\text{cyl}}=2\pi r(r+h)\).
Q10. Volume of a hemisphere?
A10. \(\displaystyle V_{\text{hemisphere}}=\frac{2}{3}\pi r^{3}\).
Q11. For a cone if slant height \(l=10\) and radius \(r=6\), what is \(h\)?
A11. \(h=\sqrt{l^{2}-r^{2}}=\sqrt{100-36}=\sqrt{64}=8.\)
Q12. If sphere radius \(r=7\), surface area?
A12. \(4\pi r^{2}=4\pi(7^{2})=196\pi\). With \(\pi=\tfrac{22}{7}\): \(196\cdot\frac{22}{7}=616.\)
Q13. If cube edge \(a\), TSA?
A13. \(\text{TSA}=6a^{2}\).
Q14. If cuboid \(l,b,h\), volume?
A14. \(V=lbh\).
Q15. Relationship: how many cones fill a cylinder of same base & height?
A15. \(3\) cones fill \(1\) cylinder (equal base-radius and height): \(V_{\text{cyl}}=3V_{\text{cone}}\).
Q16. If TSA of cone equals \(4\pi r^{2}\), what does that imply about \(l\)?
A16. Because TSA \(=\pi r(l+r)=4\pi r^{2}\Rightarrow l+r=4r\Rightarrow l=3r\).
Q17. If \(\pi\) used as \(3.14\), compute \(2\pi r h\) for \(r=2,h=3\).
A17. \(2\cdot3.14\cdot2\cdot3=75.36\).
Q18. What is circumference of base of cone of radius \(r\)?
A18. \(2\pi r\).
Q19. If sphere volume is \(\dfrac{4}{3}\pi r^{3}\) and equals \(904.32\) with \(\pi=3.14\), what's \(r\)?
A19. Solve: \(r^{3}=\dfrac{904.32\cdot3}{4\cdot3.14}=216\Rightarrow r=6.\)
Q20. If cone height \(h\) doubles, volume changes by factor?
A20. Volume proportional to \(h\) → doubles (factor 2), provided \(r\) unchanged.

20 Most Important — 2 Marks Questions

Q1. Cuboid with \(l=20\,\text{cm}, b=12\,\text{cm}, h=10\,\text{cm}\). Find area of vertical faces and TSA.
A1. Area of vertical faces \(=2(l+b)h=2(20+12)\cdot10=2\cdot32\cdot10=640\ \text{cm}^2.\) TSA \(=2(lb+bh+lh)=2(20\cdot12 +12\cdot10+20\cdot10)=2(240+120+200)=1120\ \text{cm}^2.\)
Q2. If TSA cuboid \(=500\), \(b=6,h=5\). Find \(l\).
A2. \(2(lb+bh+lh)=500\Rightarrow lb+bh+lh=250.\) Substitute \(b=6,h=5\): \(6l+30+5l=250\Rightarrow11l=220\Rightarrow l=20.\)
Q3. Cube side \(4.5\). Vertical faces area and TSA?
A3. \(l^2=4.5^2=20.25.\) Vertical faces \(=4l^2=81\ \text{cm}^2.\) TSA \(=6l^2=121.5\ \text{cm}^2.\)
Q4. TSA cube \(=5400\). Vertical faces area?
A4. \(6l^2=5400\Rightarrow l^2=900\Rightarrow l=30.\) Vertical faces \(=4l^2=3600\ \text{sq.cm}.\)
Q5. Cuboid volume \(=34.50\ \text{m}^3\), \(b=1.5,h=1.15\). Find length.
A5. \(l=\dfrac{V}{bh}=\dfrac{34.5}{1.5\times1.15}=\dfrac{34.5}{1.725}=20\ \text{m}.\)
Q6. Volume of cube with edge \(7.5\) cm?
A6. \(7.5^3=421.875\ \text{cm}^3.\)
Q7. Cylinder \(r=20,h=13\), \(\pi=3.14\). Find CSA & TSA.
A7. CSA \(=2\pi r h=2\cdot3.14\cdot20\cdot13=1632.8\ \text{cm}^2.\) TSA \(=2\pi r(r+h)=2\cdot3.14\cdot20(20+13)=4144.8\ \text{cm}^2.\)
Q8. Cylinder CSA \(=1980\ \text{cm}^2\), \(r=15,\pi=\tfrac{22}{7}\). Find \(h\).
A8. \(1980=2\pi r h=\dfrac{660}{7}h\Rightarrow h=\dfrac{1980\cdot7}{660}=21\ \text{cm}.\)
Q9. Derive formula \(\text{CSA}_{\text{cone}}=\pi r l\) from sector-net idea (short).
A9. Net of curved surface is a sector of circle radius \(l\) whose arc \(=\) circumference of base \(=2\pi r\). Sector area \(=\dfrac{\text{arc}}{2\pi l}\cdot\pi l^{2}=\pi r l\).
Q10. Cone and cylinder with same base & height: how many cones fill the cylinder?
A10. \(V_{\text{cyl}}=\pi r^2 h\). \(V_{\text{cone}}=\dfrac{1}{3}\pi r^{2}h\). So \(3\) cones fill \(1\) cylinder.
Q11. Compute slant, CSA and TSA for cone \(r=12,h=16,\pi=3.14\) (from solved example).
A11. \(l=\sqrt{12^2+16^2}=20.\) CSA \(=\pi r l=3.14\cdot12\cdot20=753.6\ \text{cm}^2.\) TSA \(=\pi r(l+r)=3.14\cdot12(20+12)=1205.76\ \text{cm}^2.\)
Q12. If TSA cone \(=704,\ r=7,\ \pi=\tfrac{22}{7}\). Find slant \(l\).
A12. \(704=\pi r(l+r)=22/7\cdot7(l+7)=22(l+7)\Rightarrow l+7=32\Rightarrow l=25.\)
Q13. Cone base area \(=1386\ \text{cm}^2,\ h=28,\ \pi=\tfrac{22}{7}\). Find surface area.
A13. Area base \(\pi r^2=1386\Rightarrow r^{2}=\dfrac{1386\cdot7}{22}=441\Rightarrow r=21.\) \(l=\sqrt{21^2+28^2}=35.\) CSA \(=\pi r l=\frac{22}{7}\cdot21\cdot35=2310\ \text{cm}^2.\)
Q14. If cone CSA \(=251.2,\ r=8,\ \pi=3.14\). Find \(l,h\).
A14. \(l=\dfrac{\text{CSA}}{\pi r}=\dfrac{251.2}{3.14\cdot8}=10.\) Then \(h=\sqrt{l^2-r^2}=\sqrt{100-64}=6.\)
Q15. Cone volume \(=6280,\ r=30,\ \pi=3.14\). Find height.
A15. \(V=\dfrac{1}{3}\pi r^{2}h\Rightarrow h=\dfrac{3V}{\pi r^{2}}=\dfrac{3\cdot6280}{3.14\cdot900}=\dfrac{18840}{2826}=6.\overline{6}\ \text{(≈6.67 cm)}.\)
Q16. Cone CSA \(=2200,\ l=50,\ \pi=\tfrac{22}{7}\). Find TSA.
A16. From CSA \(=\pi r l\Rightarrow r=\dfrac{2200\cdot7}{22\cdot50}=14.\) TSA \(=\pi r(l+r)=\tfrac{22}{7}\cdot14(50+14)=2816\ \text{cm}^2.\)
Q17. Tent: 25 persons ×4 m² per person → ground area =100 m², height 18 m. Volume?
A17. Volume of cone \(=\tfrac{1}{3}\times\text{base area}\times h=\tfrac{1}{3}\cdot100\cdot18=600\ \text{m}^3.\)
Q18. Fodder heap: \(h=2.1\ \text{m}, d=7.2\ \text{m}\). Volume? (Use \(\pi=\tfrac{22}{7}\) or \(3.14\) — here use \(22/7\)).
A18. \(r=3.6.\) \(V=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\cdot\frac{22}{7}\cdot(3.6)^2\cdot2.1\approx28.52\ \text{m}^3.\)
Q19. For same fodder cone, minimum polythene needed ≈ lateral area?
A19. \(l=\sqrt{r^2+h^2}=\sqrt{3.6^2+2.1^2}\approx4.167.\) Lateral \(=\pi r l\approx\frac{22}{7}\cdot3.6\cdot4.167\approx47.12\ \text{m}^2.\)
Q20. Sphere with radius 21 cm: volume (use \(\pi=\tfrac{22}{7}\)).
A20. \(V=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\cdot\frac{22}{7}\cdot21^3=38808\ \text{cm}^3.\)

20 Most Important — 3 Marks Questions

Q1. Prove CSA cone \(=\pi r l\) by considering the sector-net (short proof).
A1. Net of curved surface is a circular sector radius \(l\) whose arc length \(=2\pi r\). Sector area \(=\dfrac{\text{arc}}{2\pi l}\cdot\pi l^{2}=\pi r l\). Hence CSA \(=\pi r l\).
Q2. A cone with \(r=6,h=8\). Compute slant, CSA, TSA and volume (use \(\pi=\tfrac{22}{7}\)).
A2. \(l=\sqrt{6^2+8^2}=10.\) CSA \(=\pi r l=\tfrac{22}{7}\cdot6\cdot10=188.571\approx188.57\). TSA \(=\pi r(l+r)=\tfrac{22}{7}\cdot6(10+6)=\tfrac{22}{7}\cdot6\cdot16=302.857\approx302.86.\) Volume \(=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\cdot\tfrac{22}{7}\cdot36\cdot8=<\!calculate\!>\approx804.571\ \text{(units}^3).\)
Q3. Sphere with SA \(=1256\) and \(\pi=3.14\). Find radius.
A3. \(4\pi r^{2}=1256\Rightarrow r^{2}=\dfrac{1256}{4\cdot3.14}=\dfrac{1256}{12.56}=100\Rightarrow r=10.\)
Q4. If \(\text{TSA}_{\text{cone}}=704,\ r=7\) (example), find \(l\) (worked earlier).
A4. From \(704=\pi r(l+r)=22/7\cdot7(l+7)=22(l+7)\Rightarrow l=25.\)
Q5. Volume of hemisphere equals twice the cone volume with same r & h—show formula for hemisphere volume.
A5. Hemisphere volume \(=\dfrac{2}{3}\pi r^{3}\). (Because hemisphere = half sphere and sphere \(=\dfrac{4}{3}\pi r^{3}\).)
Q6. Compute SA & volume for sphere \(r=4\) using \(\pi=3.14\).
A6. SA \(=4\pi r^{2}=4\cdot3.14\cdot16=200.96\ \text{cm}^2.\) Volume \(=\dfrac{4}{3}\cdot3.14\cdot64=268.083\approx268.083\ \text{cm}^3.\)
Q7. Show that volume of sphere \(=\dfrac{4}{3}\pi r^{3}\) equals \(2\times\) volume hemisphere.
A7. By definition \(V_{\text{hemisphere}}=\dfrac{1}{2}V_{\text{sphere}}=\dfrac{1}{2}\cdot\dfrac{4}{3}\pi r^{3}=\dfrac{2}{3}\pi r^{3}.\)
Q8. A cone with \(r:h=5:12\) has volume \(314\ \text{m}^3\) and \(\pi=3.14\). Find \(h\) & \(l\).
A8. Let \(r=5k,h=12k\). \(V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(25k^{2})(12k)=100\pi k^{3}=314\). Using \(\pi=3.14\): \(100\cdot3.14\cdot k^{3}=314\Rightarrow k^{3}=1\Rightarrow k=1.\) So \(h=12\), \(r=5\), \(l=\sqrt{5^{2}+12^{2}}=13.\)
Q9. Road roller: diameter \(0.9\ \text{m}\), length \(1.4\ \text{m}\). Area pressed in 500 rotations? (\(\pi=3.14\)).
A9. One rotation area \(= \text{circumference}\times\text{length}=\pi d\times L.\) Total \(=500\cdot\pi\cdot0.9\cdot1.4=500\cdot3.14\cdot1.26=1978.2\ \text{m}^2.\)
Q10. Glass fish tank outer dims \(60.4\times40.4\times40.2\) cm with 2 mm gauge. Max water volume?
A10. Thickness \(=2\ \text{mm}=0.2\ \text{cm}.\) Inner dims: \(L=60.4-0.4=60.0,\ B=40.4-0.4=40.0,\ H=40.2-0.4=39.8\) cm. Volume \(=60.0\times40.0\times39.8=95{,}520\ \text{cm}^3\) (≈95.52 litres).\)
Q11. If sphere SA \(=314\) and \(\pi=3.14\), find volume.
A11. \(4\pi r^{2}=314\Rightarrow r^{2}=\dfrac{314}{12.56}=25\Rightarrow r=5.\) Volume \(=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\cdot3.14\cdot125\approx523.33\ \text{cm}^3.\)
Q12. If hemisphere volume \(=18000\pi\) cubic cm, find diameter.
A12. Hemisphere \(=\dfrac{2}{3}\pi r^{3}=18000\pi\Rightarrow r^{3}=18000\cdot\frac{3}{2}=27000\Rightarrow r=30.\) Diameter \(=60\ \text{cm}.\)
Q13. Find SA & V of spheres with \(r=9\) using \(\pi=3.14\).
A13. SA \(=4\pi r^{2}=4\cdot3.14\cdot81=1017.36\ \text{cm}^2.\) V \(=\dfrac{4}{3}\cdot3.14\cdot9^{3}=3052.08\ \text{cm}^3.\)
Q14. Road roller example variant: length \(2.1\), diameter \(1.4\), 500 rotations. Area and cost @ Rs.7/m².
A14. Area \(=500\cdot\pi\cdot1.4\cdot2.1=1470\pi\approx1470\cdot3.14=4615.8\ \text{m}^2.\) Cost \(=4615.8\times7\approx Rs.\ 32{,}310.6.\)
Q15. Sphere radius 21 (π=22/7): compute SA and V (as solved earlier).
A15. SA \(=4\pi r^{2}=4\cdot\frac{22}{7}\cdot441=5544\ \text{cm}^2.\) V \(=\dfrac{4}{3}\cdot\frac{22}{7}\cdot21^{3}=38808\ \text{cm}^3.\)
Q16. If TSA of cone \(=616\) and \(l=3r\), find \(l\).
A16. TSA \(=\pi r(l+r)=\pi r(3r+r)=4\pi r^{2}=616.\) Using \(\pi=\tfrac{22}{7}\) gives \(4\cdot\frac{22}{7}r^{2}=616\Rightarrow r^{2}=49\Rightarrow r=7.\) So \(l=3r=21.\)
Q17. A hemisphere and cone with same \(r\) where cone height \(=r\). How many cones fill hemisphere?
A17. Hemisphere \(=\dfrac{2}{3}\pi r^{3}\). Cone \(=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\pi r^{3}.\) Ratio hemisphere:cone \(=2:1\). So two cones fill one hemisphere.
Q18. If sphere SA ≈154 then r≈? (Use π=3.14)
A18. \(4\pi r^{2}=154 \Rightarrow r^{2}=\dfrac{154}{12.56}\approx12.265\Rightarrow r\approx3.5.\)
Q19. Using 30-60-90 relation, give sin,cos,tan of 30°,60° quickly.
A19. \(\sin30^\circ=\tfrac12,\ \cos30^\circ=\tfrac{\sqrt3}{2},\ \tan30^\circ=\tfrac{1}{\sqrt3}.\) For \(60^\circ\) swap sin & cos.
Q20. If cone radius =14 and slant \(l=50\), TSA?
A20. TSA \(=\pi r(l+r)=\tfrac{22}{7}\cdot14(50+14)=2816\ \text{cm}^2\) (same compute as earlier).

Textbook Exercises — Complete Solutions

Practice 9.1 — Q1: Cuboid \(20\times12\times10\). (Solved)
A1. Vertical faces area \(=2(l+b)h=640\ \text{cm}^2.\) TSA \(=1120\ \text{cm}^2.\) (See two-mark Q1.)
Practice 9.1 — Q2: TSA cuboid 500, \(b=6,h=5\). Find \(l\).
A2. \(l=20\) units. (Worked in two-mark Q2.)
Practice 9.1 — Q3: Cube side \(4.5\) cm (vertical & total SA).
A3. Vertical faces \(=81\ \text{cm}^2\). TSA \(=121.5\ \text{cm}^2.\)
Practice 9.1 — Q4: Cube TSA \(=5400\). Vertical faces area?
A4. Vertical faces \(=3600\ \text{cm}^2.\)
Practice 9.1 — Q5: Cuboid V \(=34.50\ \text{m}^3,\ b=1.5,h=1.15\). Find length.
A5. \(l=20\ \text{m}.\)
Practice 9.1 — Q6: Volume cube edge \(7.5\) cm?
A6. \(421.875\ \text{cm}^3.\)
Practice 9.1 — Q7: Cylinder \(r=20,h=13,\pi=3.14\). CSA & TSA?
A7. CSA \(=1632.8\ \text{cm}^2\). TSA \(=4144.8\ \text{cm}^2.\)
Practice 9.1 — Q8: Cylinder CSA 1980, \(r=15,\pi=22/7\). Find \(h\).
A8. \(h=21\ \text{cm}.\)
Practice 9.2 — Q1: Cone \(h=12,l=13\). Find \(r\).
A1. \(r=\sqrt{l^2-h^2}=\sqrt{169-144}=5\ \text{cm}.\)
Practice 9.2 — Q2: TSA cone \(=7128,\ r=28,\pi=22/7.\) Find volume.
A2. From TSA: \(\pi r (l+r)=7128\Rightarrow 88(l+28)=7128\Rightarrow l=53.\) Then \(h=\sqrt{l^2-r^2}=\sqrt{53^2-28^2}=45.\) Volume \(=\dfrac{1}{3}\pi r^{2}h=\dfrac{1}{3}\cdot\frac{22}{7}\cdot784\cdot45=36960\ \text{cm}^3.\)
Practice 9.2 — Q3: Cone CSA \(=251.2,\ r=8,\pi=3.14\). Find \(l,h\).
A3. \(l=\dfrac{\text{CSA}}{\pi r}=\dfrac{251.2}{3.14\cdot8}=10,\ h=\sqrt{10^2-8^2}=6\ \text{cm}.\)
Practice 9.2 — Q4: Closed cone tin, \(r=6\,\text{m},l=8\,\text{m}\), cost Rs.10/m². Find cost (\(\pi=3.14\)).
A4. TSA \(=\pi r(l+r)=3.14\cdot6(8+6)=263.76\ \text{m}^2.\) Cost \(=263.76\times10=Rs.\ 2{,}637.60.\)
Practice 9.2 — Q5: Volume cone \(=6280,\ r=30,\pi=3.14\). Find \(h\).
A5. \(h=\dfrac{3V}{\pi r^{2}}=\dfrac{18840}{2826}=6.\overline{6}\ \text{cm (≈6.67 cm)}.\)
Practice 9.2 — Q6: TSA cone \(=188.4,\ l=10,\pi=3.14\). Find \(h\).
A6. \(188.4=3.14\cdot r(10+r)\Rightarrow r(10+r)=60\Rightarrow r^2+10r-60=0.\) Solving \(r=\dfrac{-10+\sqrt{100+240}}{2}=\dfrac{-10+\sqrt{340}}{2}\approx4.2195\ \text{cm}.\) Then \(h=\sqrt{10^{2}-r^{2}}\approx9.068\ \text{cm}.\)
Practice 9.2 — Q7: Volume cone \(=1212,\ h=24,\ \pi=22/7.\) Find SA approximately.
A7. Solve \(r^{2}=\dfrac{3V}{\pi h}=\dfrac{3\cdot1212}{(22/7)\cdot24}\Rightarrow r\approx6.944\ \text{cm}.\) \(l=\sqrt{r^{2}+h^{2}}\approx24.99.\) TSA \(\approx \pi r(l+r)\approx697\ \text{cm}^2\) (approx.).
Practice 9.2 — Q8: CSA cone \(=2200,l=50,\pi=22/7\). Find TSA (done earlier).
A8. \(r=14,\) TSA \(=2816\ \text{cm}^2.\)
Practice 9.2 — Q9: Tent problem (done): volume \(=600\ \text{m}^3\).
A9. (See two-mark Q17.)
Practice 9.2 — Q10: Fodder heap (done): volume ≈28.52 m³, polythene ≈47.12 m².
A10. (See two-mark Q18 & Q19.)
Solved Example — Surface area of sphere \(r=7\) (\(\pi=22/7\)).
A. \(4\pi r^{2}=4\cdot\frac{22}{7}\cdot49=616\ \text{cm}^2.\)
Solved Example — Volume sphere \(r=21\) (\(\pi=22/7\)).
A. \(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\cdot\frac{22}{7}\cdot21^{3}=38808\ \text{cm}^3.\)
Practice 9.3 — Q1: SA & V of spheres with radii (i)4 (ii)9 (iii)3.5, use π=3.14.
A1. (i) \(r=4:\) SA \(=4\cdot3.14\cdot16=200.96\ \text{cm}^2.\) V \(=\dfrac{4}{3}\cdot3.14\cdot64=268.083\ \text{cm}^3.\) (ii) \(r=9:\) SA \(=4\cdot3.14\cdot81=1017.36\ \text{cm}^2.\) V \(=\dfrac{4}{3}\cdot3.14\cdot729=3052.08\ \text{cm}^3.\) (iii) \(r=3.5:\) SA \(=4\cdot3.14\cdot12.25=153.86\ \text{cm}^2.\) V \(=\dfrac{4}{3}\cdot3.14\cdot42.875\approx179.59\ \text{cm}^3.\)
Practice 9.3 — Q2: Hemisphere radius 5. Curved & total SA (\(\pi=3.14\)).
A2. Curved SA \(=2\pi r^{2}=2\cdot3.14\cdot25=157.0\ \text{cm}^2.\) Total SA \(=3\pi r^{2}=235.5\ \text{cm}^2.\)
Practice 9.3 — Q3: Sphere SA \(=2826,\ \pi=3.14.\) Find volume.
A3. \(4\pi r^{2}=2826\Rightarrow r^{2}=\dfrac{2826}{12.56}=225\Rightarrow r=15.\) Volume \(=\dfrac{4}{3}\cdot3.14\cdot15^{3}=14130\ \text{cm}^3.\)
Practice 9.3 — Q4: If volume of sphere \(=38808\) and \(\pi=22/7\), find SA.
A4. From solved example we get \(r=21.\) SA \(=4\pi r^{2}=4\cdot\frac{22}{7}\cdot441=5544\ \text{cm}^2.\)
Practice 9.3 — Q5: Volume hemisphere \(=18000\pi\) cubic cm. Find diameter.
A5. Hemisphere \(=\dfrac{2}{3}\pi r^{3}=18000\pi\Rightarrow r^{3}=27000\Rightarrow r=30\Rightarrow d=60\ \text{cm}.\)
Problem Set 9 — Q1: Road roller \(d=0.9\ m,L=1.4\ m\). Area in 500 rotations?
A1. Area \(=500\cdot\pi d L=500\cdot\pi\cdot0.9\cdot1.4=630\pi\approx1978.2\ \text{m}^2\) (π=3.14).
Problem Set 9 — Q2: Glass tank outer dims & gauge (worked earlier).
A2. Inner dims: \(60.0\times40.0\times39.8\) cm → Volume \(=95{,}520\ \text{cm}^3.\)
Problem Set 9 — Q3: Cone ratio \(5:12\) and \(V=314\ m^3\) (worked earlier).
A3. \(h=12\ \text{m},\ l=13\ \text{m}.\)
Problem Set 9 — Q4: Sphere volume \(=904.32\ \text{cm}^3,\ \pi=3.14\). Find radius.
A4. \(r=6\ \text{cm}\) (see 2-mark Q19 earlier).
Problem Set 9 — Q5: Cube TSA \(=864\). Find volume.
A5. \(6a^2=864\Rightarrow a^2=144\Rightarrow a=12.\) Volume \(=12^3=1728\ \text{cm}^3.\)
Problem Set 9 — Q6: Sphere SA \(=154\). Find volume (approx, π=3.14).
A6. \(r\approx3.5\Rightarrow V\approx\dfrac{4}{3}\cdot3.14\cdot3.5^3\approx179.6\ \text{cm}^3.\)
Problem Set 9 — Q7: TSA cone \(=616\), \(l=3r\). Find \(l\) (done earlier).
A7. \(r=7,\ l=21.\)
Problem Set 9 — Q8: Well inner diameter \(=4.20\ m,\ depth=10\ m\). Inner surface & plaster cost @ Rs.52/m².
A8. Inner surface area \(=2\pi r h=2\pi\cdot2.1\cdot10=42\pi.\) Using \(\pi=22/7\): \(42\cdot\frac{22}{7}=132\ \text{m}^2.\) Cost \(=132\times52=Rs.\ 6{,}864.\)
Problem Set 9 — Q9: Road roller variant (done earlier). Area & cost.
A9. Area \(=1470\pi\approx4615.8\ \text{m}^2.\) Cost (@Rs.7/m²) \(=4615.8\times7\approx Rs.\ 32{,}310.6.\)

If you prefer exact fractional answers using \(\pi=\tfrac{22}{7}\) or decimal \(\pi=3.14\), replace the value consistently — I have kept the same convention as the textbook examples where specified. For any problem above where you want extra step-by-step algebra or a downloadable PDF of this page, tell me & I will export the HTML to a print-ready file.

Quick Reminder — Formula Sheet

Q. Cone — summary
A. Base area \(=\pi r^{2}\). \\ Curved surface area \(=\pi r l\). \\ Total surface area \(=\pi r(l+r)\). \\ Volume \(=\dfrac{1}{3}\pi r^{2}h\). \\ \(l^2=h^2+r^2.\)
Q. Sphere / Hemisphere — summary
A. Sphere SA \(=4\pi r^{2}\). Volume \(=\dfrac{4}{3}\pi r^{3}\). \\ Hemisphere (curved) \(=2\pi r^{2}\), Total SA \(=3\pi r^{2}\), Volume \(=\dfrac{2}{3}\pi r^{3}\).
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