Chapter 13. SURFACE AREAS AND VOLUMES 

Chapter 13. Surface area and volumes

Exercise 13.1 complete solution  Exercise 13.2 complete solution  Exercise 13.3 complete solution  Exercise 13.4 complete solution  Exercise 13.5 complete solution

1. Surface area of Cube , Cuboid , Cylinder , Cone , Sphere and Hemisphere : (i) The Surface Area of a Cube  . (ii) The lateral surface area of a cube (iii) The surface area of a cuboid (iv) The lateral surface area of a cuboid  . (v)Curved surface area of a cylinder (vi) Total surface area of a cylinder   (vii) Curved surface area of a cone   (viii) Total surface area of a cone (ix) Surface Area of a sphere   (x) Curved surface area of a Hemisphere (xi) Total surface area of a hemisphere 2. The volume  of Cube , Cuboid , Cylinder , Cone , Sphere and Hemisphere : (i)  The volume of cube (ii) The volume of a cuboid (iii) The volume of a cylinder (iv) The volume of cone   (v) The volume of a sphere   (vi) The volume of hemisphere   7. The formulae involving the frustum of a cone are : (i) Volume of a frustum of a cone  . (ii) Curved surface area of a frustum of a cone  where   . (iii) Total surface area of frustum of a cone  ,where  vertical height of the frustum, slant height of the frustum ,  and are radii of the two bases (ends) of the frustum.

Class 10 Maths Chapter 13. SURFACE AREAS AND VOLUMES Exercise 13.1

 Unless stated otherwise, taken

Solution:  let  be the length of the cube .

A/Q,  

  

  

For cuboid : Here , 

The surface area of cuboid

 

 

 

Solution:  Here, Radius 

The height of the cylinder  

Area of the inner surface of the vessel

 Area of cylinder + Area of hemisphere

 

 

 

 

Solution: For cone :  Radius  3.5 cm       cm ,

Height  cm

The slant height

   

   

     12.5 cm

  The curve surface area of cone

     

 For hemisphere :  Radius  

 The curve surface area of hemisphere

 

   

The total surface area of the toy  

Solution:  The cubical block of side is 7 cm .

 So, the greatest diameter of the hemisphere is 7 cm .

 Here , Radius   ,

 The surface area of the solid

 Area of cubical block + Area of hemisphere  – Area of circular top

 

 

 

  

Solution: Here, diameter   and Radius 

The surface area of the remaining solid

 Area of cubical block + Area of hemisphere  – Area of circular top

 

 

Solution:  Here, Diameter  Radiusr 

 And height of cylinder  

  The surface area of the capsule

  S.A. of cylinder + Area of 2 hemisphere

 

 

 

 

 

 

 

 

Solution: Here , Height  , Diameter , Radius   and the slant height 

The area of the canvas of the tent

  Area of the cylinder + C.S.A. of cone

 

 

 

 

 

 

The cost of the canvas of the tent  .

Solution:  Here , Diameter,

 Radius , Height 

 The slant height

 

 

 The total surface area of the remaining solid

  C.S.A of cylinder + C.S.A. of cone + Area of circular top

   

   

  

  

  

   [appro.]

Solution: Here , Radius Height 

The total surface area of the article

  C.S.A. of cylinder + Area of 2 hemisphere

 

 

 

 

 

 

 

Class 10 Maths Chapter 13. SURFACE AREAS AND VOLUMES Exercise 13.2 Solutions

Unless stated otherwise , take  

Solution: Given ,  

The volume of solid  The volume of cone + The volume of hemisphere

 

 

 

 

Solution: Here, ,  ,

and

The volume of air contained in the model = The volume of the Cylinder  the volume of the cone

Solution:

Figure of gulab jamun : 

Here,  ,

and

The volume of a gulab jamuns = The volume of the hemisphere + The volume of the cylinder + The volume of the hemisphere

The volume of 45 gulap jamuns

The volume of gulap jamun contains in sugar syrup

Solution: For cuboid :

 Here , length  cm , breadth  cm and height cm

The volume of a cuboidal wood

 

For cone : Here , and

 The volume of a  cone

The volume of  4 cone

Therefore, the volume of wood in the entire stand

 

Solution:  For cone :  Here, Height of the cone  cm  and Radius  cm

 The volume of the cone

  For Sphere :  Here,  Radius  cm 

  The volume of the sphere

    

 The volume of the water that flows out of the cone

   (the volume of the cone)               

 Therefore, the number of lead shots

Solution :  For big cylinder : Here ,

 Diameter  , Radius   and Height

The volume of the big cylindrical pole

 

 

For small cylinder : Here ,

 Radius , Height

The volume of the big cylindrical pole

 

 

The volume of solid pole

 

The mass of the pole

 

 

 

Solution: For cone and hemisphere : Here,

 Radius  and Height  

The volume of solid

 The volume of cone + The volume of hemisphere

 

 

 

 

  

For Cylinder :

Here , Radius  and Height 

The volume of Cylinder

 

 The volume of water left in the cylinder

  Volume of Cylinder  – Volume of solid  

 

 

 

 

 

 

Solution:  For cylindrical neck :

 Here ,  

The volume of cylindrical neck

 

 

For  spherical part :

 Here ,  

The volume of spherical part

 

The volume of spherical glass vessel

 

 

She is not correct . The correct answer is  .

Class 10 Maths Chapter 13. SURFACE AREAS AND VOLUMES Exercise 13.3 Solutions

Take  , unless stated otherwise.

Solution:  For sphere : Here,  

The volume of the metallic sphere

For cylinder : let  be the height of the cylinder .

Here ,  

The volume of the cylinder

A/Q , The volume of the cylinder = The volume of the metallic sphere .

Therefore, the height of the cylinder is 2.744 cm .

Solution:   Here , cm , cm and cm  

and  let  be the new radius of the sphere .

A/Q ,  

 cm

Therefore, the radius of the resulting sphere is 12 cm .

Solution:  For well (Cylinder) : Here,  ,and

The volume of the deep well

For platform (Cuboid) : let  be the height of the platform .

Here, ,  

The volume of the platform

 

A/Q,  

 

Therefore, the height of the platform is 2.5 m .

Solution: 

Solution:  For well (Cylinder) :

Here, ,   and

The volume of the well

 

For embankment : let  be the height of the embankment .

Here, and

The volume of circular ring

A/Q,  

Therefore, the height of the embankment is 1.125 m .

Solution: For cylinder :   Diameter   , Radius and  Height   

 The volume of cylinder

 

For cone : Height cm ,  Diameter  cm and Radius   

The total volume  The volume of cone  The volume of hemisphere

 cm³

 The total number of such cones which can be filled with ice cream

Solution: For silver coin: Here,  ,

and

The volume of the silver coin (cylinder shape)

For cubiod : Here,  , and

The volume of the cuboid

The number of the silver coin

Solution:  For cylindrical bucket : Here,  cm and  cm

 The volume of cylindrical bucket

  

 For conical heap : Here,  cm

 The volume of conical heap

A/Q , The volume of cylindrical bucket  The volume of conical heap

  

 The slant height of the heap

Solution: Since the speed of the water  

  

 

So, the length of canal

 Here,  m , m  and m

 The volume of the canal (Cuboid structure)

 

let  m² be area is irrigate in 30 minutes .

A/Q ,

  

  

    [ Note : 1 hectare = 10000 m² ]

Solution: let it filled the tank in  hours .

 For pipe  : Here , Diameter  cm  and  Radius

The length of pipe

The volume of pipe

 For cylindrical tank : Here , Diameter    , Radius and  

The volume of cylindrical tank

 

A/Q, The volume of pipe  =  The volume of cylindrical tank 

Class 10 Maths Chapter 13. SURFACE AREAS AND VOLUMES Exercise 13.4 Solutions

Use  unless stated otherwise.

Solution:  Here, , and

The volume of a frustum of a cone

Therefore, the capacity of the glass is

Solution: Here,  ,

 and 

The curved surface area of the frustum

Solution:  Here,   ,  and  

 The area of material used for making the cap

Solution:   Here , cm , cm  and  cm 

  The volume of frustum  of a cone

The cost of the milk at the rate of Rs. 20 per litre

 

Slant height

  

Again, the total surface area of a frustum of a cone

 

  

The cost of metal sheet used to make the container

Solution: In given figure ,

Here,  the height  cm

 ,  ,  ,

and

In   We have,

In  We have,

 Let,  be length of the wire .

 The diameter of wire  , Radius

 A/Q,  The volume of a cylindrical wire   The volume of a frustum of a cone.

     [ 1m = 100 cm]

Therefore, the length of the wire is  m

EXERCISE 13.5 (Optional)*

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per   .

2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600  of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280  , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig. 13.25).

6. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
7. Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.