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NCERT · Class 10 · Maths

Chapter 12: Surface Areas and Volumes

Practice each question, then open the answer or worked solution.

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Exercise: 13.4 (

5 questions
Q1Written answer

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solution 1

Radius (r1) of the upper base = 4/2 = 2 cm

Radius (r2) of lower the base = 2/2 = 1 cm

Height = 14 cm

Now, the capacity of glass = Volume of the frustum of the cone

So, Capacity of glass = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×(14)(22+12+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm3

Q2Written answer

The slant height of a frustum of a cone is 4 cm, and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solution 2

Given,

Slant height (l) = 4 cm

Circumference of upper circular end of the frustum = 18 cm

∴ 2πr1 = 18

Or, r1 = 9/π

Similarly, the circumference of the lower end of the frustum = 6 cm

∴ 2πr2 = 6

Or, r2 = 3/π

Now, the surface area of the frustum = π(r1+r2) × l

= π(9/π+3/π) × 4

= 12×4 = 48 cm2

Q3Written answer

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, the radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.

Ncert solutions class 10 chapter 13-23

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Worked solution

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solution 3

Given,

For the lower circular end, radius (r1) = 10 cm

For the upper circular end, radius (r2) = 4 cm

Slant height (l) of frustum = 15 cm

Now,

The area of material to be used for making the fez = CSA of frustum + Area of the upper circular end

CSA of frustum = π(r1+r2)×l

= 210π

And, the Area of the upper circular end = πr22

= 16π

The area of material to be used for making the fez = 210π + 16π = (226 x 22)/7 = 710 2/7

∴ The area of material used = 710 2/7 cm2

Q4Written answer

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container at the rate of Rs. 20 per litre. Also, find the cost of the metal sheet used to make the container if it costs Rs. 8 per 100 cm2.

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 Solution 4

Given,

r1 = 20 cm,

r2 = 8 cm and

h = 16 cm

∴ Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

= 1/3 ×3.14 ×16((20)2+(8)2+(20)(8))

= 1/3 ×3.14 ×16(400 + 64 + 160) = 10449.92 cm3 = 10.45 lit

It is given that the rate of milk = Rs. 20/litre

So, the cost of milk = 20×volume of the frustum

= 20×10.45

= Rs. 209

Now, the slant height will be

Ncert solutions class 10 chapter 13-26

l = 20 cm

So, CSA of the container = π(r1+r2)×l

Ncert solutions class 10 chapter 13-27

= 1758.4 cm2

Hence, the total metal that would be required to make the container will be = 1758.4 + (Area of the bottom circle)

= 1758.4+πr2 = 1758.4+π(8)2

= 1758.4+201 = 1959.4 cm2

∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157

Q5Written answer

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire.

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Worked solution

The diagram will be as follows

NCERT Solutions Class 10 chapter 13-27

Consider AEG

Ncert solutions class 10 chapter 13-29

Radius (r1) of upper end of frustum = (10√3)/3 cm

Radius (r2) of lower end of container = (20√3)/3 cm

Height (r3) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

Ncert solutions class 10 chapter 13-30

Solving this, we get

Volume of the frustum = 22000/9 cm3

The radius (r) of wire = (1/16)×(½) = 1/32 cm

Now,

Let the length of the wire be “l”.

The volume of wire = Area of cross-section x Length

= (πr2)xl

= π(1/32)2x l

Now, Volume of frustum = Volume of wire

22000/9 = (22/7)x(1/32)2x l

Solving this, we get,

l = 7964.44 m