Question 1

Question 9A right circular cylinder just encloses a sphere of radius r. Find(i) surface area of the sphere,(ii) curved surface area of the cylinder,(iii) ratio of the areas obtained in (i) and (ii).

Accepted Answer

From figure,Radius of cylinder = radius of sphere = rHeight of cylinder = Diameter of sphere = 2r(i) By formula,Surface area of sphere = 4πr2Hence, surface area of sphere = 4πr2.(ii) By formula,Curved surface area of cylinder = 2πrh= 2πr × 2r= 4πr2.Hence, curved surface area of cylinder = 4πr2.(iii) Ratio = Surface area of sphereC.S.A. of cylinder=4πr24πr2=11\dfrac{	ext{Surface area of sphere}}{	ext{C.S.A. of cylinder}} = \dfrac{4πr^2}{4πr^2} = \dfrac{1}{1}C.S.A. of cylinderSurface area of sphere​=4πr24πr2​=11​.Hence, the ratio between these two surface area is 1 : 1.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 2

Question 2(i)Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm.

Accepted Answer

Given,Radius of the conical vessel (r) = 7 cmSlant height of the conical vessel (l) = 25 cmLet height of cone = h cmWe know that,⇒ l2 = r2 + h2⇒ h2 = l2 - r2⇒ h = l2−r2\sqrt{l^2 - r^2}l2−r2​Substituting values we get :⇒h=(25)2−(7)2⇒h=625−49⇒h=576=24 cm.\Rightarrow h = \sqrt{(25)^2 - (7)^2} \[1em] \Rightarrow h = \sqrt{625 - 49} \[1em] \Rightarrow h = \sqrt{576} = 24 \text{ cm}.⇒h=(25)2−(7)2​⇒h=625−49​⇒h=576​=24 cm.By formula,Capacity of the conical vessel (V) = 13πr2h\dfrac{1}{3}πr^2h31​πr2hSubstituting values we get :V=13×227×72×24=13×227×49×24=22×7×8=1232 cm3.=1232×(11000) l[∵1000 cm3=1 litre] =1.232 l.V = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 24 \[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 24 \[1em] = 22 \times 7 \times 8 \[1em] = 1232 \text{ cm}^3.\[1em] = 1232 \times \Big(\dfrac{1}{1000}\Big) \text{ l} \quad [∵ 1000 \text{ cm}^3 = 1 \text{ litre] }\[1em] = \text{1.232} \text{ l}.V=31​×722​×72×24=31​×722​×49×24=22×7×8=1232 cm3.=1232×(10001​) l[∵1000 cm3=1 litre] =1.232 l.Hence, capacity of the conical vessel = 1.232 l.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 3

Question 2(ii)Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm.

Accepted Answer

Given,Height of the conical vessel (h) = 12 cmSlant height of the conical vessel (l) = 13 cmLet radius of the vessel = r cmWe know that,⇒ l2 = r2 + h2⇒ r2 = l2 - h2⇒ r = l2−h2\sqrt{l^2 - h^2}l2−h2​Substituting values we get :r=(13)2−(12)2=169−144=25=5 cmr = \sqrt{(13)^2 - (12)^2} \[1em] = \sqrt{169 - 144} \[1em] = \sqrt{25} = 5 \text{ cm}r=(13)2−(12)2​=169−144​=25​=5 cmBy formula,Capacity of the conical vessel (V) = 13πr2h\dfrac{1}{3}πr^2h31​πr2hSubstituting values we get :V=13×227×52×12=13×227×300=660021=22007=22007×11000 l[∵1000 cm3=1l]=2270 l=1135 l.V = \dfrac{1}{3} \times \dfrac{22}{7} \times 5^2 \times 12 \[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 300 \[1em] = \dfrac{6600}{21} \[1em] = \dfrac{2200}{7} \[1em] = \dfrac{2200}{7} \times \dfrac{1}{1000} \text{ l} \quad [∵ 1000 \text{ cm}^3 = 1 l] \[1em] = \dfrac{22}{70} \text{ l} \[1em] = \dfrac{11}{35} \text{ l}.V=31​×722​×52×12=31​×722​×300=216600​=72200​=72200​×10001​ l[∵1000 cm3=1l]=7022​ l=3511​ l.Hence, capacity of the conical vessel = 1135\dfrac{11}{35}3511​ l.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 4

Question 6The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find(i) height of the cone(ii) slant height of the cone(iii) curved surface area of the cone

Accepted Answer

Given,Volume of cone = 9856 cm3Diameter of the cone (d) = 28 cmRadius of the cone (r) = Diameter2=282\dfrac{\text{Diameter}}{2} = \dfrac{28}{2}2Diameter​=228​ = 14 cm.(i) Let height of cone be h cm.Volume of the cone = 9856 cm3Substituting values we get :⇒13πr2h=9856⇒h=9856×3πr2⇒h=9856×3227×142⇒h=9856×3×722×196⇒h=2069764312⇒h=48 cm.\Rightarrow \dfrac{1}{3}πr^2h = 9856 \[1em] \Rightarrow h = 9856 \times \dfrac{3}{πr^2} \[1em] \Rightarrow h = 9856 \times \dfrac{3}{\dfrac{22}{7} \times 14^2} \[1em] \Rightarrow h = 9856 \times \dfrac{3 \times 7}{22 \times 196} \[1em] \Rightarrow h = \dfrac{206976}{4312} \[1em] \Rightarrow h = 48 \text{ cm}.⇒31​πr2h=9856⇒h=9856×πr23​⇒h=9856×722​×1423​⇒h=9856×22×1963×7​⇒h=4312206976​⇒h=48 cm.Hence, the height of the cone = 48 cm.(ii) Let slant height of the cone be l cm.By formula,⇒l=r2+h2⇒l=(14)2+(48)2⇒l=196+2304⇒l=2500=50 cm.\Rightarrow l = \sqrt{r^2 + h^2} \[1em] \Rightarrow l = \sqrt{(14)^2 + (48)^2} \[1em] \Rightarrow l = \sqrt{196 + 2304} \[1em] \Rightarrow l = \sqrt{2500} = 50 \text{ cm}.⇒l=r2+h2​⇒l=(14)2+(48)2​⇒l=196+2304​⇒l=2500​=50 cm.Hence, the slant height of the cone = 50 cm.(iii) By formula,Curved surface area of cone = πrlSubstituting values we get :⇒ Curved surface area of cone = 227\dfrac{22}{7}722​ × 14 × 50= 22 x 2 x 50= 44 x 50= 2200 cm2.Hence, curved surface area of cone = 2200 cm2.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 5

Question 9A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Accepted Answer

Given,Diameter of the conical heap (d) = 10.5 mRadius of the conical heap (r) = Diameter2=10.52\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}2Diameter​=210.5​ = 5.25 mHeight of the conical heap (h) = 3 mBy formula,Volume of the conical heap (V) = 13πr2h\dfrac{1}{3}πr^2h31​πr2hSubstituting values we get :V=13×227×(5.25)2×3=13×227×27.5625×3=13×227×82.6875=1819.12521=86.625 m3V = \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3 \[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 27.5625 \times 3 \[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 82.6875 \[1em] = \dfrac{1819.125}{21} \[1em] = \text{86.625 m}^3V=31​×722​×(5.25)2×3=31​×722​×27.5625×3=31​×722​×82.6875=211819.125​=86.625 m3By formula,Slant height (l) = r2+h2\sqrt{r^2 + h^2}r2+h2​Substituting values we get :l=(5.25)2+(3)2=27.5625+9=36.5625=6.047 ml = \sqrt{(5.25)^2 + (3)^2} \[1em] = \sqrt{27.5625 + 9} \[1em] = \sqrt{36.5625} \[1em] = 6.047 \text{ m}l=(5.25)2+(3)2​=27.5625+9​=36.5625​=6.047 mThe area of the canvas required to cover the heap of wheat (A) = Curved surface area of conical heap = πrlSubstituting values we get :A=227×5.25×6.047=698.42857=99.775 m2.A = \dfrac{22}{7} \times 5.25 \times 6.047 \[1em] = \dfrac{698.4285}{7} \[1em] = 99.775 \text{ m}^2.A=722​×5.25×6.047=7698.4285​=99.775 m2.Hence, the volume of the conical heap is 86.625 m3 and the area of the canvas required is 99.775 m2.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 6

Question 1(i)Find the volume of a sphere whose radius is 7 cm.

Accepted Answer

Given,Radius of the sphere (r) = 7 cmBy formula,Volume of the sphere (V) = 43πr3\dfrac{4}{3}πr^334​πr3Substituting values we get :V=43×227×73=43×227×343=8821×343=883×49=43123=143713 cm3.V = \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 343 \[1em] = \dfrac{88}{21} \times 343 \[1em] = \dfrac{88}{3} \times 49 \[1em] = \dfrac{4312}{3} \[1em] = 1437 \dfrac{1}{3} \text{ cm}^3.V=34​×722​×73=34​×722​×343=2188​×343=388​×49=34312​=143731​ cm3.Hence, the volume of sphere = 1437131437 \dfrac{1}{3}143731​ cm3.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 7

Question 1(ii)Find the volume of a sphere whose radius is 0.63 m.

Accepted Answer

Given,Radius of the sphere (r) = 0.63 mBy formula,Volume of the sphere = 43πr3\dfrac{4}{3}πr^334​πr3Substituting values we get :=43×227×(0.63)3=43×227×0.250047=8821×0.250047=22.00421=1.05 m3= \dfrac{4}{3} \times \dfrac{22}{7} \times (0.63)^3 \[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 0.250047 \[1em] = \dfrac{88}{21} \times 0.250047 \[1em] = \dfrac{22.004}{21} \[1em] = \text{1.05 m}^3=34​×722​×(0.63)3=34​×722​×0.250047=2188​×0.250047=2122.004​=1.05 m3Hence, the volume of sphere = 1.05 m3.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 8

Question 3The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?

Accepted Answer

Given,The diameter of the metallic ball (d) = 4.2 cmRadius of the metallic ball (r) = Diameter2=4.22\dfrac{\text{Diameter}}{2} = \dfrac{4.2}{2}2Diameter​=24.2​ cm = 2.1 cmThe density of the metal = 8.9 g per cm3We know that,Density = MassVolume\dfrac{\text{Mass}}{\text{Volume}}VolumeMass​Mass = Density × Volume ...........(1)By formula,Volume of the metallic ball (V) = 43πr3\dfrac{4}{3}πr^334​πr3Substituting values we get :V=43×227×(2.1)3=43×227×9.261=8821×9.261=814.96821=38.808 cm3V = \dfrac{4}{3} \times \dfrac{22}{7} \times (2.1)^3 \[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 9.261 \[1em] = \dfrac{88}{21} \times 9.261 \[1em] = \dfrac{814.968}{21} \[1em] = \text{38.808 cm}^3V=34​×722​×(2.1)3=34​×722​×9.261=2188​×9.261=21814.968​=38.808 cm3Substituting value in equation (1), we get :Mass of the metallic ball = 8.9 × 38.808= 345.39 gHence, the mass of the ball is 345.39 g.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 9

Question 4The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Accepted Answer

Given,⇒ Diameter of the moon (d) = 14\dfrac{1}{4}41​ × diameter of the earth (D)⇒ d2=14×D2\dfrac{\text{d}}{2} = \dfrac{1}{4} \times \dfrac{D}{2}2d​=41​×2D​⇒ The radius of the moon = 14\dfrac{1}{4}41​ × radius of the earthLet radius of moon be r and radius of earth be R.⇒ r = 14×R\dfrac{1}{4}\times R41​×R⇒ r = R4\dfrac{R}{4}4R​ .....(1)Since, earth and moon are spherical.Volume of the earth (V) = 43πR3\dfrac{4}{3}πR^334​πR3Volume of the moon (v) = 43πr3\dfrac{4}{3}πr^334​πr3Substituting value of r from equation (1) in above equation, we get :⇒v=43×π×(R4)3(From equation (1))⇒v=43×π×R4×R4×R4⇒v=164×43πR3⇒v=164×V.\Rightarrow v = \dfrac{4}{3}\times π \times \Big(\dfrac{R}{4}\Big)^3 (\text{From equation (1)}) \[1em] \Rightarrow v = \dfrac{4}{3} \times π \times \dfrac{R}{4} \times \dfrac{R}{4} \times \dfrac{R}{4} \[1em] \Rightarrow v = \dfrac{1}{64} \times \dfrac{4}{3}π R^3 \[1em] \Rightarrow v = \dfrac{1}{64} \times V.⇒v=34​×π×(4R​)3(From equation (1))⇒v=34​×π×4R​×4R​×4R​⇒v=641​×34​πR3⇒v=641​×V.Hence, the volume of the moon is 164\dfrac{1}{64}641​ times the volume of the earth.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 10

Question 5How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Accepted Answer

Given,Diameter of the hemispherical ball (d) = 10.5 cmRadius of the hemispherical ball (r) = Diameter2=10.52\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}2Diameter​=210.5​ cm = 5.25 cmBy formula,Volume of the hemispherical ball (V) = 23πr3\dfrac{2}{3}πr^332​πr3Substituting values we get :V=23×227×(5.25)3=4421×144.703=6366.937521=303.1875=303.18751000(∵1000 cm3=1 L)=0.3031875≈0.303 lV = \dfrac{2}{3} \times \dfrac{22}{7} \times (5.25)^3 \[1em] = \dfrac{44}{21} \times 144.703 \[1em] = \dfrac{6366.9375}{21} \[1em] = 303.1875 \[1em] = \dfrac{303.1875}{1000} (∵ 1000 \text{ cm}^3 = 1 \text{ L}) \[1em] = 0.3031875 \[1em] \approx 0.303 \text{ l}V=32​×722​×(5.25)3=2144​×144.703=216366.9375​=303.1875=1000303.1875​(∵1000 cm3=1 L)=0.3031875≈0.303 lHence, the hemispherical bowl can hold 0.303 litres of milk.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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    .quick-quiz h4 { color: #e67e22; margin-top: 0; }

BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 11

Question 7Find the volume of a sphere whose surface area is 154 cm2.

Accepted Answer

Let the radius of the sphere be r.By formula,Surface area of the sphere = 4πr2Substituting values we get :⇒4πr2=154⇒r2=1544π⇒r2=1544×227⇒r2=154×74×22⇒r2=494⇒r=494⇒r=72 cm.\Rightarrow 4πr^2 = 154 \[1em] \Rightarrow r^2 = \dfrac{154}{4π} \[1em] \Rightarrow r^2 = \dfrac{154}{4 \times \dfrac{22}{7}} \[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{4 \times 22} \[1em] \Rightarrow r^2 = \dfrac{49}{4} \[1em] \Rightarrow r = \sqrt{\dfrac{49}{4}} \[1em] \Rightarrow r = \dfrac{7}{2} \text{ cm}.⇒4πr2=154⇒r2=4π154​⇒r2=4×722​154​⇒r2=4×22154×7​⇒r2=449​⇒r=449​​⇒r=27​ cm.By formula,Volume of a sphere (V) = 43πr3\dfrac{4}{3}πr^334​πr3V=43×227×(72)3=43×227×3438=113×49=5393=17923 cm3V = \dfrac{4}{3} \times \dfrac{22}{7} \times \Big(\dfrac{7}{2}\Big)^3 \[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{343}{8} \[1em] = \dfrac{11}{3} \times 49 \[1em] = \dfrac{539}{3} \[1em] = 179 \dfrac{2}{3} \text{ cm}^3V=34​×722​×(27​)3=34​×722​×8343​=311​×49=3539​=17932​ cm3Hence, volume of the sphere is 17923 cm3179 \dfrac{2}{3} \text{ cm}^317932​ cm3.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 12

Question 8A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the(i) inside surface area of the dome,(ii) volume of the air inside the dome.

Accepted Answer

(i) Given,Inside of dome was white-washed at the cost of ₹ 4989.60.Cost of white washing = ₹ 20 per square metre.Inside surface area of the dome= Total cost for white-washing the dome insideRate of white-washing\dfrac{\text{Total cost for white-washing the dome inside}}{\text{Rate of white-washing}}Rate of white-washingTotal cost for white-washing the dome inside​⇒ 4989.6020\dfrac{4989.60}{20}204989.60​ = 249.48 m2Hence, inner surface area of the dome = 249.48 m2.(ii) Let 'r' be the radius of a hemispherical dome.Inner surface area of the hemispherical dome = 2πr2Substituting values we get :⇒2πr2=249.48⇒r2=249.482π⇒r2=249.482×227⇒r2=249.48×744⇒r2=5.67×7⇒r2=39.69⇒r=39.69⇒r=6.3 m\Rightarrow 2πr^2 = 249.48 \[1em] \Rightarrow r^2 = \dfrac{249.48}{2π} \[1em] \Rightarrow r^2 = \dfrac{249.48}{2 \times \dfrac{22}{7}} \[1em] \Rightarrow r^2 = \dfrac{249.48 \times 7}{44} \[1em] \Rightarrow r^2 = 5.67 \times 7\[1em] \Rightarrow r^2 = 39.69 \[1em] \Rightarrow r = \sqrt{39.69} \[1em] \Rightarrow r = 6.3 \text{ m}⇒2πr2=249.48⇒r2=2π249.48​⇒r2=2×722​249.48​⇒r2=44249.48×7​⇒r2=5.67×7⇒r2=39.69⇒r=39.69​⇒r=6.3 mThe volume of the air inside the dome will be the same as the volume of the hemisphere.Now the volume of the air inside the dome (v) = 23πr3\dfrac{2}{3}πr^332​πr3v=23×227×6.3×6.3×6.3=23×227×6.3×39.69=23×227×250.047=4421×250.047=11002.06821=523.9 m3v = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 6.3 \times 6.3 \[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 39.69 \[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 250.047 \[1em] = \dfrac{44}{21} \times 250.047 \[1em] = \dfrac{11002.068}{21} \[1em] = 523.9 \text{ m}^3v=32​×722​×6.3×6.3×6.3=32​×722​×6.3×39.69=32​×722​×250.047=2144​×250.047=2111002.068​=523.9 m3Hence, the volume of the air inside the dome is 523.9 m3.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

Bright Tutorials | Hariom Nagar, Nashik Road | 9403781999 | brighttutorials.in

Question 13

Question 10A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Accepted Answer

Given,Diameter of the spherical capsule (d) = 3.5 mmRadius of the spherical capsule (r) = Diameter2=3.52\dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2}2Diameter​=23.5​ mm = 1.75 mmMedicine needed to fill the capsule = Volume of the spherical capsule∴ Medicine needed to fill the capsule (v) = 43πr3\dfrac{4}{3}πr^334​πr3v=43×227×(1.75)3=43×227×5.359375=8821×5.359375=471.62521=22.46 mm3v = \dfrac{4}{3} \times \dfrac{22}{7} \times (1.75)^3 \[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 5.359375 \[1em] = \dfrac{88}{21} \times 5.359375 \[1em] = \dfrac{471.625}{21} \[1em] = 22.46 \text{ mm}^3v=34​×722​×(1.75)3=34​×722​×5.359375=2188​×5.359375=21471.625​=22.46 mm3Hence, the medicine needed to fill the capsule = 22.46 mm3.

Surface Areas and Volumes - Interactive Study Notes | Bright Tutorials

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    .key-point { background: #eaf2f8; padding: 10px 15px; border-left: 4px solid var(--bt-primary); margin: 10px 0; border-radius: 4px; }
    .quick-quiz { background: #fff8e1; padding: 15px; border-radius: 8px; border: 2px solid #f39c12; margin: 15px 0; }
    .quick-quiz h4 { color: #e67e22; margin-top: 0; }

BRIGHT TUTORIALS

BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-2027
    
    9403781999Excellence in Education
  
  Mathematics | Surface Areas and VolumesWeb Content &bull; Interactive Notes
  
    Surface Areas and Volumes &mdash; Interactive Study Guide
    
      Formula Master Table
      
        SolidCSATSAVolume
        
          Cube (a)4a&sup2;6a&sup2;a&sup3;
          Cuboid (l,b,h)2h(l+b)2(lb+bh+hl)lbh
          Cylinder (r,h)2&pi;rh2&pi;r(r+h)&pi;r&sup2;h
          Cone (r,h,l)&pi;rl&pi;r(l+r)&frac13;&pi;r&sup2;h
          Sphere (r)4&pi;r&sup2;&frac43;&pi;r&sup3;
          Hemisphere (r)2&pi;r&sup2;3&pi;r&sup2;&frac23;&pi;r&sup3;

Common Traps
      Cone CSA uses slant height l, NOT height h! l = &radic;(r&sup2;+h&sup2;)
      Hemisphere TSA = CSA + base circle = 2&pi;r&sup2; + &pi;r&sup2; = 3&pi;r&sup2;
      Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check
      
        Volume of cube with side 5 cm? (125 cm&sup3;)
        CSA of cylinder with r=7, h=10? (2&times;22/7&times;7&times;10 = 440 cm&sup2;)
        Volume of cone with r=3, h=4? (&frac13;&times;&pi;&times;9&times;4 = 12&pi; cm&sup3;)

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Surface Areas and Volumes

Exercise 112

Exercise 113

Exercise 114