Question

A milk container is made of metal sheet in the shape of frustum of cone whose volume is $10459\frac{3}{7}cm^{3}$ . The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm² .
(Use $(\pi=\frac{22}{7})$)

Answer 1

Answer:

The cost of metal sheet used in making the container is  ₹ 4224  

Step-by-step explanation:

SOLUTION :  

GIVEN :

Let h be the height of the container.

Radius of the lower end of the frustum of cone( R) =  20 cm

Radius of the upper end of the frustum of cone,r = 8 cm

Volume of the frustum of Cone = 10459 3/7 cm³ = 73216/7 cm³

Volume  frustum of Cone = π/3  (R² + r² + Rr) h

= ⅓ × π (20² + 8² + 20× 8)× h

= ⅓ π (400 + 64 + 160)× h

= ⅓ π × 624 × h

= 208π × h

73216/7 = 22/7 × 208 ×h

73216 = 22 × 208 ×h

h = 73216/ (22 × 208)

h = 16 cm

Height of the container is 16 cm.

Slant height of a frustum , l = √(R - r)² + h²

l =√(20 - 8)² + 16²

l = √12² + 256

l = √144 + 256

l = √400  

l = 20 cm

Surface area of the metal used to make the container = π(R + r)l + πR²

= π(20 + 8) × 20 + π(20)²  

= π(28× 20 + 400)

= π(560 + 400)

= 960 π cm²

Cost of making the container  is =  960 π cm² × 1.40  

= 960 × 22/7 × 1.40

= 960 × 22 × 0.2 = ₹ 4224

Hence, the cost of metal sheet used in making the container is

₹ 4224

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

ANSWER:------

{Volume of a frustum cone=volume of milk container}

{1/3πh(R²+r²+Rr)=10459.42857}

{1/3×22/7×h((20²)+8²+20×8)=10459.42857}

{1/3×22/7×h(400+64+160)=10459.42857}

{1/3×22/7×h×624=10459.42857}

{h=10459.42857×21/22×624=16cm

h=16cm}

[Slant height(l)=√h²(R-r)²]

=√16²(20-8)²

[√256+144]

[=√400]

[]=20cm[]

{Slant height(l)=20cm}

{Cost of a metal sheet used in a container =1.40 per}

{cm²=}3218.285714×1.40=

ANSWER=[Rs.4505.6 ]

HENCE PROVED:------

hope it helps:-------

T!—!ANKS!!!