NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.4

NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.4

 

Q1. A drinking glass in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 24 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, Capacity of glass = Volume of frustum of 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

 

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 surface area of the frustum.

Solution:-

 

Curved Surface area of frustum = π(r1 + r2)l

= π(9/π + 6/π) x 4

= 12 x 4

= 48 cm2

 

Q3. A fez, the cap used by the turks, is shaped like the frustum of a cone 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 materials used for making it.

Solution:-

Ncert solutions class 10 chapter 13-23

 

 

 

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

CSA of frustum = π(r1 + r2)l

= 210π

Area of upper circular end = πr22

= 16π

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

Q4. A container , opened from the top and made up of a metal sheet , is in the form 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 liter. Also find the cost of metal sheet used to make the container if it costs Rs 8 per 100 cm2.

Solution:-

 

Volume of frustum = (1/3) x π x h(r12 + r22 + r1r2)

= 314 x 16 x 208/100000 litres

Cost of milk = 20 x Volume of frustum

= Rs 20 x 314 x 16 x 208/100000

= Rs 209.

CSA of container = π(r1 + r2)l

= 314/100(20 + 8)x20

= 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) x 1959.4 = Rs 157

 

Q5. A metallic right circular cone 20 cm and high and whose vertical angle is 600 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:-

NCERT Solutions Class 10 chapter 13-27

 

 

 

 

 

 

Volume of frustum = 22000/9 cm3

Volume of wire = Area of cross section x Length

= (πr2)l

= π(1/32)2 x l

Volume of frustum = Volume of wire

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

L = 7964.44 m