*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 12.
NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.4 and all the other exercises are provided in this article in downloadable PDFs for the students to access easily. The NCERT Class 10 Solutions have been written and explained by our expert teachers in Maths subject, keeping in mind students’ understanding skills.
We have prepared these solutions based on the topics covered in the chapter and as per the NCERT syllabus and guidelines of the CBSE Board. Therefore, students can prepare well with the help of these NCERT solutions for Maths Class 10 and score good marks.
NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.4
Access Other Exercise Solutions of Class 10 Maths Chapter 13 – Surface Areas and Volumes
Exercise 13.1 Solutions 9 Questions (7 long, 2 short)
Exercise 13.2 Solutions 8 Questions (7 long, 1 short)
Exercise 13.3 Solutions 9 Questions (9 long)
Exercise 13.5 Solutions 7 Questions (7 long)
Access Answers to NCERT Solutions Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.4
1. 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:
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.
2. 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.
Solution:
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, CSA of frustum = π(r1+r2) × l
= π(9/π+3/π) × 4
= 12×4 = 48 cm2
3. 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, 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.
Solution:
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.
4. 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.
Solution:
Given,
r1 = 20 cm
r2 = 8 cm
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 litre
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
l = 20 cm
So, CSA of the container = π(r1+r2)×l
= 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
∴ The total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157.
5. 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.
Solution:
The diagram will be as follows
Consider AEG
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)
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
Exercise 13.4 of Class 10 Maths has problems based on the topic Frustum of a Cone, which is a very important topic. Let us discuss here the formulas related to this topic:
- Volume of the frustum of the cone = ⅓ π h (r12+r22+r1r2)
- Curved surface area of the frustum of the cone = π(r1 + r2)l
where l = √h2+(r1 – r2)2
- Total surface area of the frustum of the cone = πl (r1 + r2 ) + πr12 + πr22,
where l = √h2+(r1 – r2)2
Based on the formulas given above, you will be solving questions in this exercise. Access and practise complete NCERT Solutions for Chapter 13 of Class 10 Maths here. Also, learn from other materials, such as notes, books, and question papers, along with some tips and methods, which will help students to do their best in the board exam.
The questions in Exercise 13.4 are prepared as per the topic given in the chapter, such as the frustum of a cone, and different situations in real life where we can use the mentioned formulas. NCERT Solutions gives proper answers for each and every question in the textbook in a detailed and step-by-step manner without any difficulty.
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