Our ncert solutions for Exercise 13.4 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 13.4 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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## Exercise 13.4 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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### You can Find Mathematics solution Class 10 Chapter 13. Surface Areas and Volumes

- All Chapter review quick revision notes for chapter 13. Surface Areas and Volumes Class 10
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### Exercise 13.4 class 10 Mathematics Chapter 13. Surface Areas and Volumes

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- Exercise 13.4 Class 10 Maths 13. Surface Areas And Volumes - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 13. Surface Areas And Volumes Class 10
- Solutions Class 10
- Chapter 13. Surface Areas And Volumes Exercise 13.4 Class 10

## 13. Surface Areas and Volumes

### | Exercise 13.4 |

## Exercise 13.4 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

**13. Surface Areas and Volumes Exercise 13.4**

**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: **

Diameter of upper end D = 4 cm

Radius of upper end R = 2 cm

Diameter of lower end d = 2 cm

Radius of lower end r = 1 cm

**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: **

Slant height (l ) of the frustum of the cone = 4 cm

Perimeter of upper end = 18 cm

Thus, the curved surface area of the frustum is 48 cm^{2}.

**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: **

Slant height (l) of the cap = 15 cm

Radius of open end (R) = 10 cm

Radius of upper end (r) = 4 cm

Area of material used for making it = CSA of Frustum + Area of upper part

**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 Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π = 3.14)**

**Solution: **

Height of the container (h) = 16 cm

Radius (R) of the top of the container = 20 cm

Radius of the lower end of the container (r) = 8 cm

Capacity in litre= 10.45 litre (Approx)

The cost of milk = 20 × 10.45 = ₹209.00

Area of used sheet = πl* *(R + r) + πr^{2}

= 3.14 × 20 (20 + 8) + 3.14 × 8 × 8

= 3.14 × 20 (28) + 3.14 × 64

= 3.14 (560 + 64)

= 3.14 (624)

= 3.14 (624)

= 1959.36 cm^{2}

The cost of sheet at the rate of ₹ 8 per 100 cm^{2}

Hence, the cost of the metal sheet is ₹156.75.

**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: **

AD = 20 cm

Then, AG = 10 cm (cut into two parts at the middle)

∠BAC = 60^{o }

AD bisects ∠BAC

Thus, ∠CAD = 30^{o}

In right angled triangle ΔAGF

Similarly, in right angled triangle DADC,

Lets the length of wire be H

Volume of the wire = Volume of the frustum obtained

H = 7964.44 m

Hence, the length of the wire is 7964.44 m.

##### Other Pages of this Chapter: 13. Surface Areas and Volumes

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