# Maths Ex 13.4 Class 10 NCERT Solution

Maths Ex 13.4 Class 10 of NCERT Textbook consists of total 5 questions related to frustum. The solution of these questions are provided below

Contents

## Class 10 Maths Chapter 13

Chapter 13 of Maths Class 10 is about Surface area and volume. The exercise 13.4 is about Frustum. It contains questions related to surface area and volume of a Frustum.
When we slice (or cut) through a cone with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part that is now left over on the other side of the plane is called a frustum of the cone.

Class 10 Maths Exercise 13.1
Class 10 Maths Exercise 13.2
Class 10 Maths Exercise 13.3
Class 10 Maths Exercise 13.4

## Maths Ex 13.4 Class 10 NCERT Solutions

Ex 13.4 Class 10 Maths Question 1
Q1. 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.

Solution
According to question ,
Height of glass (h) = 14 cm
Radius of upper part of glass = R = 4/2 = 2 cm
Radius of Lower part of glass = r = 2/2 = 1 cm
Capacity of Glass = Volume of Glass in the shape of frustum
= 1/3 pi h ( R2 + r2 + R r )
= 1/3 (22/7) * ( 22 + 12 + 1*2 )
= 102.66 cm3

Ex 13.4 Class 10 Maths Question 2
Q2. 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 curved surface area of the frustum.

Solution
Let us take the lower radius of frustum = r and Upper radius = R
According to question ,
Circumference of Upper end of frustum = 18 cm
2 pi R = 18 cm
R = 9/pi cm

Circumference of lower end of frustum = 6 cm
2 pi r = 6 cm
r = 3/pi cm

The slant height of frustum (l) = 4 cm

Curved Surface area of Frustum = pi ( r +R ) * l
= pi ( 9/pi + 6/pi ) * 4
= 12 * 4 = 48 cm2

Ex 13.4 Class 10 Maths Question 3
Q3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution
According to question,
Radius of Open end (R) =10cm
Radius of upper Base (r) = 4 cm
Slant height = 15 cm
Area of material used= Curved surface area of frustum + area of upper base
= pi (R +r) * l + pi r2
= pi ( 10 + 4 ) * 15 + pi (4)2
= 709.64 cm2

Ex 13.4 Class 10 Maths Question 4
Q4. 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 20 per litre. Also find the cost of metal sheet used to make the container, if it costs 8 per 100 cm2 . (Take π = 3.14)

Solution
According to question ,
Radius of upper end of frustum (R) =20 cm
Radius of lower end of frustum (r) = 8 cm
Height of Frustum (h) = 16 cm
Volume of container in the shape of frustum = ( 1/3 pi * h ) ( R2 +r2 +R r )
= 1/3 *3.14 *16 ( 2 +82 +20* 8)
= 10449.92 cm3 = 10.44 litres

Also,
Cost of 1 litre of milk = Rs 20
Cost of 10.44 lite of milk = 20 * 10.44 = Rs 209

Surface area of frustum = pi (r +R ) * l + pi r2
= 3.14 * ( 20+8 ) * 20 + 3.14 * 8 * 8
= 1959.36 cm2

Area of metal sheet required = 1959.36 cm2
Cost of 100cm2 metal sheet = Rs 8
Cost of 1cm2 metal sheet = Rs 8/100 = Rs 0.08
Cost of 1959.36cm2 metal sheet = 1959.36 * 0.08 = Rs 156.75

Ex 13.4 Class 10 Maths Question 5
Q5. 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 be drawn into a wire of diameter 1 /16 cm, find the length of the wire.

Solution
According to question,
angle = 60 degree
Height of cone before cut = 20 cm
The cone is cut into two parts
Height of Frustum after cut = 10 cm
Now using trigonometric formulas to get the value of upper end and lower end radius of frustum obtained after cutting of the cone.
tan 30 = R/20
R = 20/sqrt(3)

Also ,
tan 30 = r/10
r = 10/sqrt(3)

Radius of wire = 1/32 cm
Let , Length of wire be ‘L’
According to question,
Volume of frustum = volume of wire
=1/3 pi * h (r2 +R2 +R r ) = pi r2 L
= 1/3 pi * 10 ( [10/sqrt(3)]2 + [20/sqrt(3)]2 + [10/sqrt(3)] * [20/sqrt(3)] ) = pi * 1/32 *1/32 *L
On Solving , we get
L = 796444.4 cm
L = 7964.44 m