NCERT Solutions class 10 maths chapter 13

Surface Areas and Volumes

NCERT Solutions class 10 maths chapter 13 Exercise 13.4

Exercise 13.1 Exercise 13.2 Exercise 13.3 Exercise 13.5

Question 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.

Answer 1

Radius r₁ of the upper end = 4/2 = 2 cm

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

Height = 14 cm

Capacity of glass = Volume of frustum of cone

Capacity of glass = (1/3)×π×h(r₁²+r₂²+r₁r₂)

= (1/3)× π × (14)(2²+1²+ (2)(1))

∴ The capacity of the glass = 102×(2/3) cm³

Question 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.

Answer 2

Given,

Slant height l = 4 cm

Circumference of upper end of the frustum = 18 cm

∴ 2πr₁ = 18

Or, r₁ = 9/π

circumference of lower end of the frustum = 6 cm

∴ 2πr₂ = 6

Or, r₂ = 6/π

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

= π(9/π+6/π) × 4

= 12×4 = 48 cm²

Question 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.

Answer 3

Given,

the radius of lower end,(r₁) = 10 cm

radius of upper end, (r₂) = 4 cm

Slant height l of frustum = 15 cm

Now,

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

Coverd surface area of frustum = π(r₁+r₂)×l

= 210π

And, Area of upper end = πr₂²

= 16π

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

Question 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 cm².

Answer 4

Given,

r₁ = 20 cm,

r₂ = 8 cm and

h = 16 cm

∴ Volume of the frustum = (1/3)×π×h(r₁²+r₂²+r₁r₂)

the rate of milk = Rs. 20/litre

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

= Rs. 209

slant height will be

So, CSA of the container = π(r₁+r₂)×l

l= 1758.4 cm²

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

= 1758.4+201 = 1959.4 cm²

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

Question 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.