NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes
NCERT Solutions Class 10 Maths Chapter 13 Surface Areas And Volumes: National Council of Educational Research and Training Class 10 Maths Chapter 13 Solutions – Surface Areas And Volumes. NCERT Solutions Class 10 Maths Chapter 13 PDF Download.
NCERT Solutions Class 10 Maths Chapter 13: Overview
| Board | NCERT |
| Class | 10 |
| Subject | Maths |
| Chapter | 13 |
| Chapter Name | Surface Areas And Volumes |
| Topic | Exercise Solutions |
NCERT Solutions Class 10 Maths
Chapter 13 – Surface Areas And Volumes
Exercise 13.1
(1)
(1) The volume of each cube = 64 cm3 = V
Length of each ides = 3√3 cm
= 3√64 cm
= 4 cm
If we draw two cube side wise them we get (6 – 1) = 5 surfaces are open
Now that resulting cuboid has (5 × 2) = 10
Squares surfaces having each sides 4 cm only
The surface area of the cuboid
= (10 × surface of each square) sq unit
= (10 × 42) cm2
= 10 × 16 cm2
= 160 cm2
(2) The diameter of the hemisphere = 14 cm
The radius of the hemisphere = r = 7 cm
The height of the hollow cylinder
= r = 7 cm
The height of the cylinder (h)
= height of the vessel – radius of the hemisphere
= 13 cm – 7 cm = 6 cm = h
The inner surface area of the vessel
= Surface area of hemisphere + curved surface of hemisphere
= 2 π r2 + 2 π r h) sq unit
= [2 π r2 × 22/7 × (7)2 + 2 × 22/7 × 7 × 6] cm2
= 2 × 22/7 (7 + 6) cm2
= 2 × 22 × 13 cm2
= 572 cm2
The inner surface area of the vessel is 572 cm2
(3) The radius of the cone = 3.5 cm
The radius of the hemisphere = 3.5 cm
The height of the hemisphere is the same of the radius
The Total height of the toy = 15.5 cm
The height of the cone = h
= Total height of the toy – the radius of the hemisphere
= 15.4 cm – 3.5 cm
h = 12 cm
The slant height of the cone = l = √h2 + r2 unit
= √122 + 3.52 cm
= √144 + 12.25 cm
= √156.25 cm
= 12.5 cm
The total surface area of the toy
= CSA of the cone + CSA of the hemisphere
[CSA = Curved surface area]
= [π r l + 2 π r2] sq unit
= π r [L + 2 r] cm2
= 22/7 × 3.5 [12.5 + 2 × 3.5] cm2
= 11 × (12.5 + 7) cm2 = 11 × 19. 5 cm2 = 214.5 cm2
The total surface area of the toy in 214.5 cm2
(4) The sides of the cubical black a = 7 cm
The hemisphere is surmounted on the cubical black
The greatest diameter which the hemisphere can have is equals to the edge of the cube.
So it have diameter (d) = the side of the black (a) diameter (d) = 7 cm, radius (a) = 7/2 cm
The total surface of the solid
= Curved surface area of the hemisphere + surface area of the cubical black – Flat surface of the hemisphere
= (2 π r2 + 6a2 – π r2) sq unit
= (π r2 + 6a2 ) sq unit
= [22/7 × (7/2)2 + 6 × 72] cm2
= (22/7 × 7/2 × 7/2 + 6 × 49) cm2
= (38.5 + 294) cm2
= 332.5 cm2
(5) The diameter of the hemisphere (d) = L unit
The radius of the hemisphere (r) = L/2 unit
The sides of the hemisphere
r = L/2 unit
The sides of the cube (a) = L unit
The surface area of the hemisphere + surface area of cube – Flat surface of the hemisphere
= (6a2 + 2 π r2 – πr2) sq unit
= (6a2 + π r2) sq unit
= (6l2 + π r2/4) sq unit
= L2/4 (π + 24) sq unit
(6) The capsule is in the shape of a cylinder with hemisphere in each ends
The diameter of each hemisphere (d)
= The diameter of the cylinder = 5 mm
The radius of the hemisphere (r) = 5/2 mm
Length of the cylinder = Length of the capsule – 2 × radius of hemisphere
= 14 mm – 2 × 5/2 mm
= 14 mm – 5 mm
= 9 mm
Surface area of the capsule
= 2 × curved surface area of hemisphere + curved surface area cylindrical part
= (2 × 2 π r2 + 2 π r h) sq unit
= 2 π r (2 r + h) sq unit
= 2 × π × 5/2 (2 × 5/2 + 9) mm2
= (22/7 × 5 × 14) mm2
= 220 mm2
Surface area of the capsule is 220 mm2
(7) The height of the cylindrical h = 2.1 m
The diameter of the Conceal top = the diameter of the cylinder (d)
= 4 m
Therefore, The radius of the cylinder and the conical top (r) = 4/2
= 2 m
The slant height of the conical Lop (L) = 2.8 M
The area of canvas
= Curved surface area of the conical top + curved surface area of the cylinder
= (π r L + 2πrh) sq unit
= π r (l + 2h) m2
= π × 2 (2.8 + 2 × 2.1) m2
= 22/7 × 2 × (2.8 + 4.2) m2
= 22/7 × 2 × 7 m2
= 44 m2
The area of the canvas is 44 m2
The cost of canvas of the tent for each sq meter = Rs 500
The cost of canvas for the tent
= 44m2 × Rs 500 per m2
= Rs (44 × 500)
= Rs 22000
The total cost of canvas Rs – 22000
(8) The solid has a cylindrical size and a conical cavity
The height of the cylinder
= The height of the conical cavity (h) = 2.4 cm
The diameter of the cylinder = 1.4 cm
The radius of the cylinder = the radius of the conical cavity (r) = 1.4/2 cm
= 0.7 cm
The slant height of the conical cavity (l)
= √h2 + r2
= √(2.4)2 + (0.7)2 cm
= √6.25 cm
= 2.5 cm
The surface area of the solid
= Total surface area of solid cylinder + Total surface area of the cylinder
= π r L + [2 π r h + r2]
= π r (L + 2h + r)
= 22/7 × 0.7 (2.5 + 4.8 + 0.7)
= 22 × 0.1 × (8)
= 2.2 × 8
= 17.6 cm²
So, the total surface area of the remaining solid is 17.6 cm²
(9) The solid is mode from a solid cylinder and two hemisphere had been scooped out from each end.
∴ The radius of the hemisphere (r)
= the radius of the cylinder (r) = 3.5 cm
∴ The height of the cylinder (h)
= 10 cm
The surface area of the solid
= curved surface area of the cylinder + 2 × curved surface area of hemisphere
= (2 π r h + 2 × 2 π r²) cm²
= 2 π r (h + 2r) cm²
= 2 × 22/7 × 3.5 (10 + 2 × 3.5) cm²
= 22 × (10 + 7) cm²
= 22 × 17 cm²
= 371 cm²
∴ The total surface area of the solid is 374 cm²
EXERCISE – 13.2
(1)
The height of the cone (h)
= the radius of the cone (r) = 1 cm
The volume of the solid
= the volume of the cone + the volume of the hemisphere
= (1/3 π r² h + 2/3 π r) cm³
= 1/3 π r² (h + 2r) cm³
= 1/3 × 22/7 × (1)² (1 + 2) cm³
= 1/3 × π × 3 cm³
= π cm³
∴ Total volume of solid = π cm³
In case you are missed :- NCERT Solution for Triangles
(2) The made shaped like cylinder with two cone each side
The diameter of the solid
= 3 cm
The radius of the cylinder
(r) = The radius of each cone
(r) = 3/2 cm
Total height of the solid = 12 cm
Height of each cone (h) = 2 cm
Height of the cylinder (h₂) = height
= Height of the solid – 2 × height of each cone
= 12 cm – 2 × 2 cm
= 12 cm – 4 cm
= 8 cm
The volume of the solid model
= the volume of the cylinder + 2 × volume of the cone
= (π r² h₂ + 2 × 1/3 π r² h1) cm³
= π r² (h₂ + 2/3 h1) cm³
= π × (3/2)2 – (8 + 2/3 × 2) cm³
= 22/7 × 9/8 × (24 + 4/3) cm³
= 22/7 × 9/4 × 28/3 cm³
= 66 cm³
∴ The volume of the model
= the volume of air contained in the model
= 66 cm³
(3) For one gulab jamun it made by a cylinder having two hemisphere in each end
The radius of each hemisphere (r)
Stand
= the of the cylinder (r)
= 2.8/2 cm
= 1.4 cm
= (π r² h + 2 × 2/3 π r ³) cm³
= π r² (h + 4/3 r) cm³
= 22/7 × 1.4 × 1.4 (2.2 + 4/3 × 1.4) cm³
= 22/7 × 1.4 × 1.4 (6.6 + 5.6/3) cm³
= 22/7 × 1.4 × 1.4 × 12.2/3 cm³
= 25.05 cm³
Volume of 45 gulab jamun
= 45 × 25.05 cm³
It given that each gulab jamun contains 30% sugar syrup
So syrup quantity in 45 gulab jamun
= 30% × 45 × 25.05 cm³
= 30/100 × 45 × 25.05 cm³
= 3381.75/10= 338.175 cm³
∴ The quantity of 45 gulab jamun is 338.175 cm³
(4) The stand is in the shape of a cuboid with four cone shaped depression Stand
For the cuboid
Length = L
= 15 cm
Breadth = b
= 10 cm
Height = h
= 3.5 cm
For the cone shapes depression
radius = r
= 0.5 cm
depth = d
= 1.4 cm
∴ The volume of the wooden stand
= Volume of the cuboid – 4 × volume of each cone
= (L h b) cm³ – 4 × 1/3 π r² d cm³
= L h b cm³ – 4/3 π r² d cm³
= (15 × 10 × 3.5 – 4/3 π × (0.5)² × 1.4) cm³
= (525 – 4/3 × 22/7 × 0.25 × 1.4) cm³
= (525 – 22 × 0.2/3) cm³
= (525 – 4.4/3) cm³
= (525 – 1.47) cm³
= 523.53 cm³
∴ The volume of the wooden pen stand h
= 523.53 cm³
(5) The cone holds water its brim the radius of the cone (R) = 5 cm the height of the cone (h) = 8 cm
The water its hold is equivalent of its volume
The volume of the cone (v) = 1/3 π r² h cm³
= 1/3 × 22/7 × 5 × 5 × 8 cm³
= 22 × 25 × 8/3 × 7 cm³
How, the radius of each lead shots is
= 0.50 m
The volume of lead shot = 4/3 π r³
= 4/3 × 22/7 × (0.5)³ cm³
The number of lead shot
= The volume of water from out/volume of one lead shot
= 22/7 × 50/3/4/3 × 22/7 × 0.5 × 0.5
= 10 × 10 × 10 × 10/4 × 5 × 5
= 100
Number of lead shots is 100.
(6) In the pole one cylinder is surmounted on another cylinder
The diameter of base cylinder = 24 cm
The radius of base cylinder (r) = 24 / 2 = 12 cm
The Height of the base cylinder = 220 cm
The radius of the mounted cylinder (r) = 8 cm
The height of mounted cylinder (h) = 60 cm
The volume of the pole
= The volume of the base cylinder + the volume of the sur mounted cylinder
= (π r² h + π r² h) cm³
= π [r² h + r² h] cm³
= π [{12² × 220} + {(8)² × 60}] cm³
= π [(144 × 220) + (64 × 60)] cm³
= π [31680 + 3840] cm³
= 3.14 × 35520 cm³
= 111532.8 cm³
Now, we have give that 1 cm³ of iron has 8 g of mass
∴ Total mass of the iron pole is
= (8 × 11532.8) g
= 892262.4 g
= 892.262 kg
(7) The solid shape is made with a cone surmounted on a hemisphere
The water displaced by it is the same of its own volume
∴ The height of the cone (h) = 120 cm
The radius of the cone
= The radius of the hemisphere
= 60 cm
The radius of the cylinder = 60 cm
= (r)
The height of the cylinder (h) = 180 cm
The volume of the shape
= Volume of the cone + volume of the hemisphere
The volume of water left in the cylinder
= volume of the cylinder – the volume of the solid shape
= π r² h cm³ – (1/3 π r² h + 2/3 π r³) cm³
= {π r² h – 1/3 π r² (h + 2r)} cm³
= π r² (h – 1/3 h – 2/3 r) cm³
= 22/7 × (60)² (180 – 1/3 × 120 – 2/3 × 60) cm³
= 22/7 × 60 × 60 (180 – 40 – 40) cm³
= 22/7 × 60 × 60 × 100 cm³
= 792000/7 cm³
= 792000/7 × 1 L
= 1131.43 L (approx.)
(8) The shape is a spherical glass vessel and a cylinder surmounted on u
The diameter of the cylinder = 2 cm
The radius of the cylinder (r) = 1 cm
The height of the cylinder (h) = 8 cm
The diameter of the shape = 8.5 cm
The radius of the sphere (r) = 8.5/2 cm
The volume of the glass vessel
= volume of the cylinder + volume of the sphere
= (π r² h + 4/3 π r³) cm³
= {π × 1² × 8 + 4/3 π × (8.5/2)³} cm³
= π (8 + 4/3 × 8.5/2 × 8.5/2 × 8.5/2) cm³
= 3.14 (8 + 102.35) cm³
= 3.14 × 110.35 cm³
= 346.51 cm³
Volume obtained by child is incorrect, correct volume is 346.51 cm³
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