Question

SURFACE AREAS AND VOLUMES Example 14: An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig. 15.23). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold. ( take π = 22/7) ​

Answer 1

Answer:Area of metallic sheet used = curved surface area of Frustum + curved surface area of Cylinder + area of circular Base.

Diameter of the bigger circular end = 45 cm

Radius, r1 = 452= 22.5 cm  

Diameter of the smaller circular end = 25 cm

Radius, r2 = 252=12.5 cm

Height of the frustum, h = Total height of the bucket (H) - Height of the circular base (h1)

= 40 - 6 = 34 cm

Slant Height of frustrum (l) =h2+(r1−r2)2−−−−−−−−−−−−√

⇒l=342+(22.5−12.5)2−−−−−−−−−−−−−−−−√

⇒l=1156+(10)2−−−−−−−−−−√

⇒l=1156+100−−−−−−−−−√

⇒l=1256−−−−√

⇒ Slant Height, l = 35.44 cm

Curved surface area of Frustum = π(r1+r2)l

= 227×(22.5+12.5)×35.44

⇒227×35×35.44

= 3898.4 cm2

Area of Circular Base with radius 252=12.5 is given by,

Area of circular base = πr22

=227×12.5×12.5

= 491.07 cm2

Curved surface area of Cylinder = 2πr2h1

= 2×227×12.5×6

= 471.428 cm2

Area of metallic sheet used = 3898.4 cm2 + 491.07 cm2 + 471.428 cm2

= 4860.898 cm2

Volume of water in bucket = Volume of Frustrum

= 13πh(r21+r22+r1r2)

= 13×227×34(22.52+12.52+22.5×12.5) = 33615.48 cm3

Now, 1 litre = 1000 cm

Thus, Volume of water in bucket in litres = 33.62 litres

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