Question

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find : (i) the volume of water which can completely fill the bucket. (ii) the area of the metal sheet used to make the bucket. [Use  = 22/7]

Answer 1

Solution:-

Let 'R' be the radius of the upper circular end and 'r' be the radius of the lower circular end.
Volume of the water that can completely fill the bucket = 1/3πh{R² + r² + (R*r)}
= 1/3*22/7*24*{14² + 7² + (14*7)}
= 1/3*22/7*24*(196 + 49 + 98)
= 1/3*22/7*24*343
Volume of water = 8624 cu cm

Now,

l² = √(R - r)²+h²
l² = √(14- 7)²+24²
l² = √49+576
l² = √625
l = 25 cm
So, slant height of the bucket is 25 cm.
Curved surface area of the frustum = πl(R+r)
= 22/7*25*(14+7)
= 22/7*25*21
Curved surface area of the frustum = 1650 sq cm
Now, Area of the base = πr²
= 22/7*7*7
= 154 sq cm
Total area of the metal sheet used = Curved surface area + Area of the base
= 1650 + 154 
= 1804 sq cm

Answer 2

Answer:

1804 sq cm

Step-by-step explanation:

Let 'R' be the radius of the upper circular end and 'r' be the radius of the lower circular end.

Volume of the water that can completely fill the bucket = 1/3πh{R² + r² + (R*r)}

= 1/3*22/7*24*{14² + 7² + (14*7)}

= 1/3*22/7*24*(196 + 49 + 98)

= 1/3*22/7*24*343

Volume of water = 8624 cu cm

Now,

l² = √(R - r)²+h²

l² = √(14- 7)²+24²

l² = √49+576

l² = √625

l = 25 cm

So, slant height of the bucket is 25 cm.

Curved surface area of the frustum = πl(R+r)

= 22/7*25*(14+7)

= 22/7*25*21

Curved surface area of the frustum = 1650 sq cm

Now, Area of the base = πr²

= 22/7*7*7

= 154 sq cm

Total area of the metal sheet used = Curved surface area + Area of the base

= 1650 + 154 

= 1804 sq cm