NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.4 and all the exercises are provided here for students, in downloadable PDF format. The answers have been written and explained by our expert 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 NCERT syllabus and guidelines(CBSE Board). Therefore, students can prepare well with the help of solutions for Maths class 10 and score good marks.

### Download PDF of 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 Question (7 long, 2 short)

Exercise 13.2 Solutions 8 Question (7 long, 1 short)

Exercise 13.3 Solutions 9 Question (9 long)

Exercise 13.5 Solutions 7 Question (7 long)

### Access Answers to NCERT 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 (r_{1}) of the upper base = 4/2 = 2 cm

Radius (r_{2}) of lower the base = 2/2 = 1 cm

Height = 14 cm

Now, Capacity of glass = Volume of frustum of cone

So, Capacity of glass = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

= (⅓)×π×(14)(2^{2}+1^{2}+ (2)(1))

∴ The capacity of the glass = 102×(⅔) cm^{3}

**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

∴ 2πr_{1} = 18

Or, r_{1} = 9/π

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

∴ 2πr_{2} = 6

Or, r_{2} = 6/π

Now, CSA of frustum = π(r_{1}+r_{2}) × l

= π(9/π+6/π) × 4

= 12×4 = 48 cm^{2}

**3. 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, 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 (r_{1}) = 10 cm

For the upper circular end, radius (r_{2}) = 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 upper circular end

CSA of frustum = π(r_{1}+r_{2})×l

= 210π

And, Area of upper circular end = πr_{2}^{2}

= 16π

∴ The area of material used = 710 × (2/7) cm^{2}

**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 metal sheet used to make the container, if it costs Rs. 8 per 100 cm ^{2}. **

**Solution:**

Given,

r_{1} = 20 cm,

r_{2} = 8 cm and

h = 16 cm

∴ Volume of the frustum = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

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

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

= Rs. 209

Now, slant height will be

So, CSA of the container = π(r_{1}+r_{2})×l

= 1758.4 cm^{2}

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

= 1758.4+201 = 1959.4 cm^{2}

∴ 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 (r_{1}) of upper end of frustum = (10√3)/3 cm

Radius (r_{2}) of lower end of container = (20√3)/3 cm

Height (r_{3}) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

Solving this we get,

Volume of the frustum = 22000/9 cm^{3}

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

Now,

Let the length of wire be “l”.

Volume of wire = Area of cross-section x Length

= (πr^{2})xl

= π(1/32)^{2}x l

Now, Volume of frustum = Volume of wire

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

Solving this we get,

l = 7964.44 m

Exercise 13.4 of 10th standard Maths have 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 (r*_{1}^{2}+r_{2}^{2}+r_{1}r_{2}) - Curved surface area of the frustum of the cone =
*π(r*_{1 }+ r_{2})l

where *l = √h ^{2}+(r_{1 }– r_{2})^{2}*

- Total surface area of the frustum of the cone =
*πl (r*_{1}+ r_{2}) + πr_{1}^{2}+ πr_{2}^{2},

where *l = √h ^{2}+(r_{1 }– r_{2})^{2}*

Based on these formulas given above, you will be solving questions in this exercise. Learn complete solutions for chapter 13 of class 10 maths from here. Also, learn from other materials, such as notes, books, questions papers along with some tips and tricks, which will help students to do their best in the second term exam.

The questions in exercise 13.4 are prepared as per the topic, given in the chapter as Frustum of the cone and different situations in real life, where we can use the mentioned formulas. NCERT Solutions gives the best answers for each and every question in a detailed and a step by step manner without any difficulty.