CHAPTER 11 – SURFACE AREAS & VOLUMES (Class 9, NCERT/CBSE)
Part A — 1-Mark Questions (20) with Solutions
Q1. Curved surface area (CSA) of a right circular cone in terms of \(r,l\)?
\( \text{CSA}=\pi r l\).
Q2. Total surface area (TSA) of a right circular cone?
\( \text{TSA}=\pi r l+\pi r^2=\pi r(l+r)\).
Q3. Relation among radius \(r\), height \(h\) and slant height \(l\) of a cone.
\( l=\sqrt{r^2+h^2}\).
Q4. Surface area of a sphere of radius \(r\)?
\(4\pi r^2\).
Q5. Curved and total surface areas of a hemisphere of radius \(r\).
CSA \(=2\pi r^2\), TSA \(=3\pi r^2\).
Q6. Volume of a cone (base radius \(r\), height \(h\)).
\( V=\dfrac13 \pi r^2 h\).
Q7. Volume of a sphere of radius \(r\); of a hemisphere?
Sphere \(=\dfrac{4}{3}\pi r^3\); Hemisphere \(=\dfrac{2}{3}\pi r^3\).
Q8. If diameter of a sphere doubles, its surface area becomes…
\(4\) times (since \(S\propto r^2\)).
Q9. For a cone \(r=7\) cm, \(l=25\) cm. Find CSA (use \(\pi=\tfrac{22}{7}\)).
\( \pi r l=\tfrac{22}{7}\cdot 7\cdot 25=550\ \text{cm}^2\).
Q10. Sphere radius \(7\) cm. Surface area = ?
\(4\pi r^2=4\cdot \tfrac{22}{7}\cdot 7\cdot 7=616\ \text{cm}^2\).
Q11. A cone has \(h=12\) cm, \(r=5\) cm. Volume?
\(V=\frac13\pi r^2h=\frac13\pi\cdot25\cdot12=100\pi\ \text{cm}^3\).
Q12. Hemispherical bowl radius \(10\) cm. TSA with \(\pi=3.14\)?
\(3\pi r^2=3\cdot 3.14\cdot 100=942\ \text{cm}^2\).
Q13. A cone has CSA \(308\ \text{cm}^2\) and \(l=14\) cm. Find \(r\).
\(r=\dfrac{\text{CSA}}{\pi l}=\dfrac{308}{(22/7)\cdot14}=7\ \text{cm}\).
Q14. Right triangle \(5\!\!-\!\!12\!\!-\!\!13\) revolved about \(12\) cm forms a cone. Radius?
\(r=5\) cm; \(h=12\) cm.
Q15. If \(S_1:S_2=1:16\) for two spheres, then \(r_1:r_2=\ ?\)
\(1:4\) (since \(S\propto r^2\)).
Q16. A sphere just fits in a cylinder of height \(2r\). Compare their curved surfaces.
Both \(=4\pi r^2\) (ratio \(1:1\)).
Q17. Cone: \(r=6\) cm, \(h=7\) cm. Volume?
\(V=\frac13\cdot \frac{22}{7}\cdot 36\cdot 7=264\ \text{cm}^3\).
Q18. Sphere radius \(3.5\) cm. Volume?
\( \frac{4}{3}\cdot \frac{22}{7}\cdot (3.5)^3=\frac{539}{3}\approx 179.67\ \text{cm}^3\).
Q19. For a cone, \(l=10\) cm, \(r=7\) cm. TSA?
\( \pi r(l+r)=\tfrac{22}{7}\cdot 7\cdot(10+7)=374\ \text{cm}^2\).
Q20. If radius of a hemisphere doubles, its volume becomes…
\(8\) times (since \(V\propto r^3\)).
Part B — 2-Mark Questions (20) with Solutions
Q1. Cone \(h=16\) cm, \(r=12\) cm. Find \(l\), CSA and TSA (\(\pi=3.14\)).
\(l=\sqrt{16^2+12^2}=20\) cm; CSA \(=\pi rl=3.14\cdot12\cdot20=753.6\ \text{cm}^2\); TSA \(=753.6+3.14\cdot 12^2=1205.76\ \text{cm}^2\).
Q2. Sphere radius \(14\) cm. Surface area?
\(4\cdot \frac{22}{7}\cdot 14\cdot 14=2464\ \text{cm}^2\).
Q3. Hemispherical dome radius \(21\) cm. Find CSA and TSA.
CSA \(=2\pi r^2=2\cdot \frac{22}{7}\cdot 21^2=2772\ \text{cm}^2\); TSA \(=3\pi r^2=4158\ \text{cm}^2\).
Q4. A conical tent has \(r=24\) m, \(h=10\) m. Find \(l\) and canvas cost at ₹70/m\(^2\).
\(l=\sqrt{24^2+10^2}=26\) m; CSA \(=\frac{22}{7}\cdot24\cdot26=\) \(1961.14\ \text{m}^2\); Cost \(\approx ₹137{,}280\).
Q5. Sphere diameter \(3.5\) m. Surface area?
\(r=1.75\) m; \(4\cdot \frac{22}{7}\cdot 1.75^2=38.5\ \text{m}^2\).
Q6. Cone \(r=7\) cm, \(l=25\) cm. Find CSA of 10 such cones.
One cone CSA \(=\frac{22}{7}\cdot7\cdot25=550\ \text{cm}^2\); for 10 \(\Rightarrow 5500\ \text{cm}^2\).
Q7. Hemispherical bowl inner diameter \(10.5\) cm. Tin-plate inside at ₹16 per \(100\ \text{cm}^2\).
\(r=5.25\) cm; inner curved area \(=2\pi r^2=173.25\ \text{cm}^2\); Cost \(=0.16\times 173.25=₹27.72\) (approx).
Q8. Sphere surface area \(=154\ \text{cm}^2\). Find radius and volume.
\(4\pi r^2=154\Rightarrow r=3.5\) cm; \(V=\dfrac{4}{3}\cdot \dfrac{22}{7}\cdot (3.5)^3=\dfrac{539}{3}\approx 179.67\ \text{cm}^3\).
Q9. Conical vessel \(r=7\) cm, \(l=25\) cm. Capacity (litres)?
\(h=\sqrt{25^2-7^2}=24\) cm; \(V=\frac13\pi r^2h=\frac{22}{7}\cdot \frac{49\cdot 24}{3}=1232\ \text{cm}^3=1.232\ \text{L}\).
Q10. A sphere just fits in a cylinder of height \(2r\). Compare sphere area and cylinder curved area.
Sphere area \(=4\pi r^2\); Cylinder CSA \(=2\pi r(2r)=4\pi r^2\). Equal.
Q11. Cone \(r=3.5\) cm, \(h=12\) cm. Volume?
\(V=\frac13\cdot\frac{22}{7}\cdot 12.25\cdot 12=154\ \text{cm}^3\).
Q12. Sphere radius \(10.5\) cm. Surface area?
\(4\cdot \frac{22}{7}\cdot 10.5^2=1386\ \text{cm}^2\).
Q13. Balloon radius doubles from \(7\) to \(14\) cm. Ratio of surface areas?
\(1:4\).
Q14. Hemispherical dome circumference \(=17.6\) m. Find CSA and painting cost at ₹500/m\(^2\).
\(r=\frac{17.6}{2\pi}=\frac{17.6\cdot 7}{44}=2.8\) m; CSA \(=2\pi r^2=49.28\ \text{m}^2\); Cost \(=₹500\times 49.28=₹24{,}640\).
Q15. Cone \(l=21\) m, diameter \(=24\) m. TSA?
\(r=12\) m; TSA \(=\pi r(l+r)=\frac{22}{7}\cdot 12\cdot 33\approx 1244.57\ \text{m}^2\).
Q16. Hemispherical tank inner \(r=1\) m, iron sheet \(1\) cm thick. Volume of iron?
Outer \(R=1.01\) m (=101 cm). Volume \(=\frac{2}{3}\pi (R^3-r^3)\) (cm\(^3\)) \(=\frac{44}{21}\times 30{,}301\approx 63{,}488\ \text{cm}^3=0.0635\ \text{m}^3\).
Q17. Conical tarpaulin: \(r=6\) m, \(h=8\) m, \(w=3\) m. Extra length \(=0.20\) m. Find length needed (\(\pi=3.14\)).
\(l=10\) m; CSA \(=3.14\cdot 6\cdot 10=188.4\ \text{m}^2\). Length \(= \frac{188.4}{3}+0.20=63.0\ \text{m}\) (approx).
Q18. 27 equal spheres (radius \(r\)) melted to one sphere. New radius and area ratio?
\(r'=3r\); \(S:S' = 1:9\).
Q19. Pit: cone diameter \(3.5\) m, depth \(12\) m. Capacity (kL)?
\(r=1.75\) m; \(V=\frac13\pi r^2h=\frac{22}{7}\cdot \frac{3.0625\cdot 12}{3}=38.5\ \text{m}^3=38.5\ \text{kL}\).
Q20. Metallic ball diameter \(4.2\) cm, density \(8.9\) g/cc. Mass?
\(r=2.1\) cm; \(V=\frac{4}{3}\cdot \frac{22}{7}\cdot 2.1^3\approx 38.86\ \text{cm}^3\); Mass \(\approx 38.86\times 8.9\approx 346\ \text{g}\).
Part C — 3-Mark Questions (20) with Solutions
Q1. Derive the CSA of a cone by sector-triangle dissection.
Cut the net into thin isosceles triangles of height \(l\). Sum of bases equals the base circumference \(2\pi r\). Hence \(\text{CSA}=\frac12\times (2\pi r)\times l=\pi r l\).
Q2. Prove \(l=\sqrt{r^2+h^2}\) for a right circular cone.
In right \(\triangle\) (radius, height, slant), by Pythagoras: \(l^2=r^2+h^2\Rightarrow l=\sqrt{r^2+h^2}\).
Q3. A cone and a cylinder have same \(r,h\). Experimentally show \(V_{\text{cone}}=\dfrac13 V_{\text{cyl}}\).
Fill cone thrice to fill cylinder \(\Rightarrow 3V_{\text{cone}}=V_{\text{cyl}}=\pi r^2 h\Rightarrow V_{\text{cone}}=\dfrac13\pi r^2 h\).
Q4. Corn cob approx. cone, \(r=2.1\) cm, \(h=20\) cm. Each \(1\ \text{cm}^2\) holds 4 grains. Estimate grains.
\(l=\sqrt{2.1^2+20^2}\approx 20.11\) cm; CSA \(\approx \frac{22}{7}\cdot 2.1\cdot 20.11\approx 132.73\ \text{cm}^2\); grains \(\approx 4\times 132.73\approx 531\).
Q5. A sphere radius \(7\) cm. Find surface area and volume.
SA \(=616\ \text{cm}^2\); \(V=\frac{4}{3}\cdot \frac{22}{7}\cdot 343=\frac{4312}{3}\approx 1437.33\ \text{cm}^3\).
Q6. Hemispherical dome: base circumference \(17.6\) m. Cost to paint CSA at ₹5/100 cm\(^2\).
\(r=2.8\) m; CSA \(=49.28\ \text{m}^2\). ₹ per m\(^2\) \(=500\). Cost \(=₹24{,}640\).
Q7. Cone \(r=6\) m, \(h=8\) m. Find tarpaulin length of width \(3\) m to cover CSA, extra \(0.2\) m.
\(l=10\) m; CSA \(=188.4\ \text{m}^2\) (\(\pi=3.14\)); length \(=\frac{188.4}{3}+0.2=63.0\) m.
Q8. 50 hollow cones (outer paint), diameter \(40\) cm, height \(1\) m. Cost at ₹12/m\(^2\). Take \(\pi=3.14, \sqrt{1.04}\approx 1.02\).
\(r=0.2\) m, \(l\approx 1.02\) m; CSA ≈ \(3.14\cdot 0.2\cdot 1.02=0.64056\ \text{m}^2\) each; total ≈ \(32.028\ \text{m}^2\); cost ≈ \(₹384.34\).
Q9. Hemispherical bowl \(d=10.5\) cm. How much milk can it hold?
\(r=5.25\) cm; \(V=\frac{2}{3}\pi r^3=\frac{44}{21}\cdot 144.703125\approx 303.19\ \text{cm}^3=0.303\ \text{L}\).
Q10. Right triangle \(5,12,13\) revolved about \(12\) cm. Cone volume?
\(r=5,h=12\Rightarrow V=\frac13\pi\cdot 25\cdot 12=100\pi\approx 314.29\ \text{cm}^3\).
Q11. Same triangle revolved about \(5\) cm. Volume and ratio w.r.t. Q10?
\(r=12,h=5\Rightarrow V=\frac13\pi\cdot 144\cdot 5=240\pi\). Ratio \(=100\pi:240\pi=5:12\).
Q12. Cone \(d=28\) cm, \(V=9856\) cm\(^3\). Find \(h,l,\) and CSA.
\(r=14\) cm; \(h=\dfrac{3V}{\pi r^2}=\dfrac{29568}{616}=48\) cm; \(l=\sqrt{14^2+48^2}=50\) cm; CSA \(=\pi r l=\frac{22}{7}\cdot 14\cdot 50=2200\ \text{cm}^2\).
Q13. Sphere radius \(0.63\) m. Volume?
\(V=\frac{4}{3}\cdot \frac{22}{7}\cdot 0.63^3\approx 1.047\ \text{m}^3\).
Q14. Two spheres have diameter ratio \(1:4\). Find ratios of (i) areas (ii) volumes.
(i) \(1:16\) (ii) \(1:64\).
Q15. A heap of wheat is a cone, \(d=10.5\) m, \(h=3\) m. Find volume and canvas area to cover.
\(r=5.25\) m; \(V=\frac13\pi r^2 h\approx 86.625\ \text{m}^3\); \(l=\sqrt{5.25^2+3^2}\approx 6.0465\) m; CSA \(\approx \frac{22}{7}\cdot 5.25\cdot 6.0465\approx 99.77\ \text{m}^2\).
Q16. A dome is a hemisphere. Whitewash cost ₹4989.60 at ₹20/m\(^2\). Find inside CSA and air volume.
Area \(=\frac{4989.60}{20}=249.48\ \text{m}^2\). \(2\pi r^2=249.48\Rightarrow r=6.3\) m. Volume \(=\frac{2}{3}\pi r^3=\frac{44}{21}\cdot 250.047\approx 523.91\ \text{m}^3\).
Q17. Spherical ball diameter \(28\) cm. Water displaced?
\(r=14\) cm; \(V=\frac{4}{3}\cdot \frac{22}{7}\cdot 14^3=\frac{34496}{3}\approx 11{,}498.67\ \text{cm}^3\).
Q18. Medicine capsule is a sphere of diameter \(3.5\) mm. Volume?
\(r=1.75\) mm; \(V=\frac{4}{3}\cdot \frac{22}{7}\cdot 1.75^3\approx 22.47\ \text{mm}^3\).
Q19. Show that for a hemisphere, TSA : CSA \(=3:2\).
TSA \(=3\pi r^2\), CSA \(=2\pi r^2\Rightarrow 3:2\).
Q20. A sphere radius \(r\) is melted into \(n\) equal spheres. New radius?
Volume: \( \frac{4}{3}\pi r^3 = n \cdot \frac{4}{3}\pi r_n^3 \Rightarrow r_n = \dfrac{r}{\sqrt[3]{n}}\).
Exercise 11.1 — Solutions (Assume \( \pi=\tfrac{22}{7}\) unless stated)
| Question | Perfect Solution (MathJax) |
|---|---|
| 1. Cone: base diameter \(10.5\) cm, \(l=10\) cm. Find CSA. | \(r=5.25\) cm; \( \text{CSA}=\pi r l=\tfrac{22}{7}\cdot 5.25\cdot 10=165\ \text{cm}^2\). |
| 2. Cone: \(l=21\) m, base diameter \(24\) m. Find TSA. | \(r=12\) m; \( \text{TSA}=\pi r(l+r)=\tfrac{22}{7}\cdot 12\cdot 33\approx 1244.57\ \text{m}^2\). |
| 3. CSA \(=308\ \text{cm}^2\), \(l=14\) cm. Find (i) \(r\) (ii) TSA. | (i) \(r=\dfrac{308}{(\tfrac{22}{7})\cdot 14}=7\) cm. (ii) \( \text{TSA}= \text{CSA}+\pi r^2=308+\tfrac{22}{7}\cdot 49=308+154=462\ \text{cm}^2\). |
| 4. Conical tent \(h=10\) m, \(r=24\) m. (i) \(l\) (ii) canvas cost at ₹70/m\(^2\). | (i) \(l=\sqrt{10^2+24^2}=26\) m. (ii) CSA \(=\tfrac{22}{7}\cdot 24\cdot 26=1961.14\ \text{m}^2\); Cost \(=₹137{,}280\). |
| 5. Conical tent \(h=8\) m, \(r=6\) m; tarpaulin width \(3\) m; extra length \(20\) cm; \(\pi=3.14\). | \(l=10\) m; CSA \(=3.14\cdot 6\cdot 10=188.4\ \text{m}^2\). Length \(=\dfrac{188.4}{3}+0.20=63.0\ \text{m}\) (approx). |
| 6. Conical tomb \(l=25\) m, base diameter \(14\) m. Cost of white-wash at ₹210 per \(100\ \text{m}^2\). | CSA \(=\tfrac{22}{7}\cdot 7\cdot 25=550\ \text{m}^2\). Rate \(=₹2.10/\text{m}^2\). Cost \(=550\times 2.10=₹1155\). |
| 7. Joker’s cap \(r=7\) cm, \(h=24\) cm. Area of sheet for 10 caps. | \(l=25\) cm; one CSA \(=550\ \text{cm}^2\); for 10 → \(5500\ \text{cm}^2\). |
| 8. 50 hollow cones: base diameter \(40\) cm, height \(1\) m; paint outer; \(\pi=3.14\), take \( \sqrt{1.04}\approx 1.02\). Cost at ₹12/m\(^2\). | \(r=0.2\) m, \(l\approx 1.02\) m. One CSA \(=3.14\cdot 0.2\cdot 1.02=0.64056\ \text{m}^2\). Total \(=32.028\ \text{m}^2\). Cost \(=32.028\times 12\approx ₹384.34\). |
Exercise 11.2 — Solutions (Assume \( \pi=\tfrac{22}{7}\) unless stated)
| Question | Perfect Solution |
|---|---|
| 1. Surface area of sphere radius (i) \(10.5\) cm (ii) \(5.6\) cm (iii) \(14\) cm. | (i) \(1386\ \text{cm}^2\) (ii) \(394.24\ \text{cm}^2\) (iii) \(2464\ \text{cm}^2\). |
| 2. Surface area of sphere of diameter (i) \(14\) cm (ii) \(21\) cm (iii) \(3.5\) m. | (i) \(616\ \text{cm}^2\) (ii) \(1386\ \text{cm}^2\) (iii) \(38.5\ \text{m}^2\). |
| 3. TSA of hemisphere \(r=10\) cm (\(\pi=3.14\)). | \(3\pi r^2=942\ \text{cm}^2\). |
| 4. Balloon radius \(7\to 14\) cm. Ratio of surface areas? | \(1:4\) (initial:final). |
| 5. Hemispherical bowl inner diameter \(10.5\) cm. Tin-plate inside at ₹16 per \(100\ \text{cm}^2\). | Curved area \(=173.25\ \text{cm}^2\); Cost \(\approx ₹27.72\). |
| 6. Radius of sphere with surface area \(154\ \text{cm}^2\). | \(r=3.5\) cm. |
| 7. Moon’s diameter is \(\tfrac14\) of Earth’s. Ratio of surface areas? | \(1:16\) (Moon:Earth). |
| 8. Hemispherical bowl of steel, thickness \(0.25\) cm, inner radius \(5\) cm. Find outer CSA. | Outer \(R=5.25\) cm; outer CSA \(=2\pi R^2=173.25\ \text{cm}^2\). |
| 9. Cylinder just encloses a sphere radius \(r\). Find (i) sphere area (ii) cylinder CSA (iii) ratio. | (i) \(4\pi r^2\) (ii) \(2\pi r(2r)=4\pi r^2\) (iii) \(1:1\). |
Exercise 11.3 — Solutions (Assume \( \pi=\tfrac{22}{7}\) unless stated)
| Question | Perfect Solution |
|---|---|
| 1. Volume of cone with (i) \(r=6\) cm, \(h=7\) cm (ii) \(r=3.5\) cm, \(h=12\) cm. | (i) \(264\ \text{cm}^3\) (ii) \(154\ \text{cm}^3\). |
| 2. Capacity (litres) of a conical vessel: (i) \(r=7\) cm, \(l=25\) cm (ii) \(h=12\) cm, \(l=13\) cm. | (i) \(h=24\) cm ⇒ \(V=1232\ \text{cm}^3=1.232\ \text{L}\). (ii) \(r=5\) cm ⇒ \(V=\dfrac{22}{7}\cdot \dfrac{300}{3}= \dfrac{2200}{7}\approx 314.29\ \text{cm}^3=0.314\ \text{L}\). |
| 3. Cone \(h=15\) cm, \(V=1570\ \text{cm}^3\) (\(\pi=3.14\)). Find \(r\). | \(r^2=\dfrac{3V}{\pi h}=\dfrac{4710}{47.1}=100\Rightarrow r=10\) cm. |
| 4. \(V=48\pi\ \text{cm}^3\), \(h=9\) cm. Diameter of base? | \(r^2=\dfrac{3V}{\pi h}=\dfrac{144}{9}=16\Rightarrow r=4\Rightarrow \text{diameter}=8\) cm. |
| 5. Conical pit \(d=3.5\) m, \(h=12\) m. Capacity in kL. | \(r=1.75\) m; \(V=\frac13\pi r^2 h=38.5\ \text{kL}\). |
| 6. Cone \(V=9856\ \text{cm}^3\), base diameter \(28\) cm. Find (i) \(h\) (ii) \(l\) (iii) CSA. | (i) \(h=48\) cm (ii) \(l=50\) cm (iii) \(2200\ \text{cm}^2\). |
| 7. Right triangle \(5,12,13\) revolved about side \(12\) cm. Volume. | \(V=100\pi\approx 314.29\ \text{cm}^3\). |
| 8. Same triangle revolved about side \(5\) cm. Volume and ratio with Q7. | \(V=240\pi\ \text{cm}^3\). Ratio \(=5:12\). |
| 9. Wheat heap: cone \(d=10.5\) m, \(h=3\) m. Find \(V\) and CSA (for cover). | \(V\approx 86.625\ \text{m}^3\). \(l\approx 6.0465\) m; CSA \(\approx 99.77\ \text{m}^2\). |
Exercise 11.4 — Solutions (Assume \( \pi=\tfrac{22}{7}\) unless stated)
| Question | Perfect Solution |
|---|---|
| 1. Volume of sphere radius (i) \(7\) cm (ii) \(0.63\) m. | (i) \( \dfrac{4}{3}\cdot \dfrac{22}{7}\cdot 7^3=\dfrac{4312}{3}\approx 1437.33\ \text{cm}^3\). (ii) \(\approx 1.047\ \text{m}^3\). |
| 2. Water displaced by a solid spherical ball of diameter (i) \(28\) cm (ii) \(0.21\) m. | (i) \( \dfrac{34496}{3}\approx 11498.67\ \text{cm}^3\). (ii) \(r=0.105\) m ⇒ \(V\approx 0.00485\ \text{m}^3=4.85\ \text{L}\). |
| 3. Metallic ball diameter \(4.2\) cm, density \(8.9\) g/cm\(^3\). Mass? | \(r=2.1\) cm; \(V\approx 38.86\ \text{cm}^3\); mass \(\approx 346\ \text{g}\). |
| 4. Moon’s diameter \(\tfrac14\) of Earth’s. Volume fraction? | \(\left(\tfrac14\right)^3=\tfrac{1}{64}\). |
| 5. How many litres of milk in a hemispherical bowl of diameter \(10.5\) cm? | \(r=5.25\) cm; \(V=\frac{2}{3}\pi r^3\approx 303.19\ \text{cm}^3\approx 0.303\ \text{L}\). |
| 6. Hemispherical tank iron sheet \(1\) cm thick; inner radius \(1\) m. Volume of iron. | Outer \(=101\) cm; \(V_{\text{iron}}=\frac{2}{3}\pi(R^3-r^3)\approx 63{,}488\ \text{cm}^3=0.0635\ \text{m}^3\). |
| 7. Volume of sphere with surface area \(154\ \text{cm}^2\). | \(r=3.5\) cm; \(V=\dfrac{539}{3}\approx 179.67\ \text{cm}^3\). |
| 8. Dome (hemisphere) white-washed for ₹4989.60 at ₹20/m\(^2\). Find (i) inside area (ii) air volume. | (i) \(249.48\ \text{m}^2\). (ii) \(r=6.3\) m ⇒ \(V=\frac{2}{3}\pi r^3\approx 523.91\ \text{m}^3\). |
| 9. 27 iron spheres (radius \(r\)) → one sphere. Find (i) new radius (ii) \(S:S'\). | (i) \(r'=3r\). (ii) \(1:9\). |
| 10. Medicine capsule: sphere of diameter \(3.5\) mm. Volume? | \(r=1.75\) mm; \(V\approx 22.47\ \text{mm}^3\). |
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